Type:
Solution Manual
Resource:
Physics
Edition:
9th Edition
Author(s):
John D. Cutnell Kenneth W. Johnson
CHAPTER 1 INTRODUCTION AND
MATHEMATICAL CONCEPTS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1.
(d) The resultant vector R is drawn from the tail of the first vector to the head of the last vector.
2.
(c) Note from the drawing that the magnitude R of the resultant vector R is equal to the shortest distance between the tail of A and the head of B. Thus, R is less than the magnitude (length) of A plus the magnitude of B.
3.
(a) The triangle in the drawing is a right triangle. The lengths A and B of the two sides are known, so the Pythagorean theorem can be used to determine the length R of the hypotenuse. 4.0 km (b) The angle is found by using the inverse tangent function, = tan −1 = 53 . 3.0 km (b) In this drawing the vector –C is reversed relative to C, while vectors A and B are not reversed.
4. 5.
6.
(c) In this drawing the vectors –B and –C are reversed relative to B and C, while vector A is not reversed.
7.
(e) These vectors form a closed four-sided polygon, with the head of the fourth vector exactly meeting the tail of the first vector. Thus, the resultant vector is zero.
8.
(c) When the two vector components Ax and Ay are added by the tail-to-head method, the sum equals the vector A. Therefore, these vector components are the correct ones.
9.
(b) The three vectors form a right triangle, so the magnitude of A is given by the 2
Pythagorean theorem as A =
( 2 A ) + ( 2 A ) = 4 A + 4 A = 2 A + A = 2 A. 2
A doubles:
2
Ax + Ay . If Ax and Ay double in size, then the magnitude of
x
2
y
2 x
2 y
2 x
2 y
−1
Ay . If Ax and Ax
10. (a) The angle is determined by the inverse tangent function, = tan
Ay both become twice as large, the ratio does not change, and remains the same. 11. (b) The displacement vector A points in the –y direction. Therefore, it has no scalar component along the x axis (Ax = 0 m) and its scalar component along the y axis is negative.
2
INTRODUCTION AND MATHEMATICAL CONCEPTS
12. (e) The scalar components are given by Ax = −(450 m) sin 35.0 = −258 m and Ay = −(450 m) cos 35.0 = −369 m.
13. (d) The distance (magnitude) traveled by each runner is the same, but the directions are different. Therefore, the two displacement vectors are not equal. 14. (c) Ax and Bx point in opposite directions, and Ay and By point in the same direction. 15. (d) 16. Ay = 3.4 m, By = 3.4 m 17. Rx = 0 m, Ry = 6.8 m 18. R = 7.9 m, = degrees
Chapter 1 Problems
3
CHAPTER 1 INTRODUCTION AND
MATHEMATICAL CONCEPTS PROBLEMS ____________________________________________________________________________________________
1.
REASONING We use the fact that 1 m = 3.28 ft to form the following conversion factor: (1 m)/(3.28 ft) = 1. SOLUTION To convert ft2 into m2, we apply the conversion factor twice:
(
)
1 m 1 m 2 Area = 1330 ft 2 3.28 ft = 124 m 3.28 ft _____________________________________________________________________________ 2.
REASONING a. To convert the speed from miles per hour (mi/h) to kilometers per hour (km/h), we need to convert miles to kilometers. This conversion is achieved by using the relation 1.609 km = 1 mi (see the page facing the inside of the front cover of the text). b. To convert the speed from miles per hour (mi/h) to meters per second (m/s), we must convert miles to meters and hours to seconds. This is accomplished by using the conversions 1 mi = 1609 m and 1 h = 3600 s. SOLUTION a. Multiplying the speed of 34.0 mi/h by a factor of unity, (1.609 km)/(1 mi) = 1, we find the speed of the bicyclists is
mi mi 1.609 km km Speed = 34.0 (1) = 34.0 = 54.7 h h 1 mi h b. Multiplying the speed of 34.0 mi/h by two factors of unity, (1609 m)/(1 mi) = 1 and (1 h)/(3600 s) = 1, the speed of the bicyclists is
mi mi 1609 m 1 h m Speed = 34.0 (1)(1) = 34.0 = 15.2 h s h 1 mi 3600s ______________________________________________________________________________ 3.
SSM REASONING We use the facts that 1 mi = 5280 ft, 1 m = 3.281 ft, and 1 yd = 3 ft. With these facts we construct three conversion factors: (5280 ft)/(1 mi) = 1, (1 m)/(3.281 ft) = 1, and (3 ft)/(1 yd) = 1.
4
INTRODUCTION AND MATHEMATICAL CONCEPTS
SOLUTION By multiplying by the given distance d of the fall by the appropriate conversion factors we find that
3 ft 1 m 5280 ft 1 m d = 6 mi + 551 yd = 10 159 m 1 yd 3.281 ft 1 mi 3.281 ft
(
4.
)
(
)
REASONING The word “per” indicates a ratio, so “0.35 mm per day” means 0.35 mm/d, which is to be expressed as a rate in ft/century. These units differ from the given units in both length and time dimensions, so both must be converted. For length, 1 m = 103 mm, and 1 ft = 0.3048 m. For time, 1 year = 365.24 days, and 1 century = 100 years. Multiplying the resulting growth rate by one century gives an estimate of the total length of hair a long-lived adult could grow over his lifetime. SOLUTION Multiply the given growth rate by the length and time conversion factors, making sure units cancel properly:
mm 1 m 1 ft 365.24 d 100 y Growth rate = 0.35 = 42 ft/century 3 1 y century d 0.3048 m 10 mm ______________________________________________________________________________ 5.
REASONING In order to calculate d, the units of a and b must be, respectively, cubed and squared along with their numerical values, then combined algebraically with each other and the units of c. Ignoring the values and working first with the units alone, we have m) ( a3 m32 m2 d= 2→ = = cb ( m/s )( s )2 m / s s 2 1 s 3
(
)
Therefore, the units of d are m2/s. SOLUTION With the units known, the numerical value may be calculated:
d=
6.
( 9.7)3 m2 /s = 0.75 m2 /s ( 69)( 4.2)2
REASONING The dimensions of the variables v, x, and t are known, and the numerical factor 3 is dimensionless. Therefore, we can solve the equation for z and then substitute the known dimensions. The dimensions L and T can be treated as algebraic quantities to determine the dimensions of the variable z.
Chapter 1 Problems
5
3v . We know the following dimensions: xt 2 v = L / T , x = L , and t = T . Since the factor 3 is dimensionless, z has the dimensions of L / T = 1 v = x t 2 L T 2 T 3 SOLUTION Since v = zxt 2 , it follows that z = 1 3
____________________________________________________________________________________________
7.
SSM REASONING This problem involves using unit conversions to determine the number of magnums in one jeroboam. The necessary relationships are
1.0 magnum= 1.5 liters 1.0 jeroboam= 0.792 U. S. gallons 1.00 U. S. gallon = 3.785 10 –3 m3 = 3.785 liters These relationships may be used to construct the appropriate conversion factors. SOLUTION By multiplying one jeroboam by the appropriate conversion factors we can determine the number of magnums in a jeroboam as shown below:
0.792 gallons
liters 1.0 magnum (1.0 jeroboam ) 1.0 jeroboam 3.785 = 2.0 magnums 1.0 gallon 1.5 liters
____________________________________________________________________________________________
8.
REASONING By multiplying the quantity 1.78 10−3 kg/ ( s m ) by the appropriate conversions factors, we can convert the quantity to units of poise (P). These conversion factors are obtainable from the following relationships between the various units:
1 kg = 1.00 103 g 1 m = 1.00 102 cm 1 P = 1 g/ ( s cm ) SOLUTION The conversion from the unit kg/ ( s m ) to the unit P proceeds as follows:
kg 1.00 103 g 1 m 1P −3 = 1.78 10−2 P 1.78 10 2 s m 1 kg 1.00 10 cm 1 g / s cm
(
)
_____________________________________________________________________________________________
6
INTRODUCTION AND MATHEMATICAL CONCEPTS
9.
REASONING Multiplying an equation by a factor of 1 does not alter the equation; this is the basis of our solution. We will use factors of 1 in the following forms:
1 gal = 1 , since 1 gal = 128 oz 128 oz
3.785 10−3 m3 = 1 , since 3.785 10−3 m3 = 1 gal 1 gal 1 mL = 1 , since 1 mL = 10−6 m3 −6 3 10 m
SOLUTION The starting point for our solution is the fact that Volume = 1 oz Multiplying this equation on the right by factors of 1 does not alter the equation, so it follows that 1 gal 3.785 10−3 m3 1 mL Volume = (1 oz )(1)(1)(1) = 1 oz = 29.6 mL 128 oz 10−6 m3 1 gal
(
)
Note that all the units on the right, except one, are eliminated algebraically, leaving only the desired units of milliliters (mL).
10. REASONING To convert from gallons to cubic meters, use the equivalence 1 U.S. gal = 3.785×10−3 m3. To find the thickness of the painted layer, we use the fact that the paint’s volume is the same, whether in the can or painted on the wall. The layer of paint on the wall can be thought of as a very thin “box” with a volume given by the product of the surface area (the “box top”) and the thickness of the layer. Therefore, its thickness is the ratio of the volume to the painted surface area: Thickness = Volume/Area. That is, the larger the area it’s spread over, the thinner the layer of paint. SOLUTION a. The conversion is
(
3.785 10−3 m3 0.67 U.S. gallons = 2.5 10−3 m3 U.S. gallons
)
b. The thickness is the volume found in (a) divided by the area,
Thickness =
Volume 2.5 10−3 m3 = = 1.9 10−4 m Area 13 m2
Chapter 1 Problems
11.
7
SSM REASONING The dimension of the spring constant k can be determined by first solving the equation T = 2 m / k for k in terms of the time T and the mass m. Then, the dimensions of T and m can be substituted into this expression to yield the dimension of k. SOLUTION Algebraically solving the expression above for k gives k = 4 2 m / T 2 . The term 4 2 is a numerical factor that does not have a dimension, so it can be ignored in this analysis. Since the dimension for mass is [M] and that for time is [T], the dimension of k is
Dimension of k =
M T 2
12. REASONING AND SOLUTION The following figure (not drawn to scale) shows the geometry of the situation, when the observer is a distance r from the base of the arch. The angle is related to r and h by tan = h/ r . Solving for r, we find h = 192 m
h 192 m r= = = 5.5 10 3 m = 5.5 km r tan tan 2.0 ______________________________________________________________________________
13. SSM REASONING The shortest distance between the two towns is along the line that joins them. This distance, h, is the hypotenuse of a right triangle whose other sides are ho = 35.0 km and ha = 72.0 km, as shown in the figure below. SOLUTION The angle is given by tan = ho / ha so that
35.0 km = 25.9 S of W 72.0 km
= tan −1
We can then use the Pythagorean theorem to find h.
h = ho2 + ha2 = (35.0 km) 2 + ( 72 .0 km) 2 = 80.1 km
8
INTRODUCTION AND MATHEMATICAL CONCEPTS
14. REASONING The drawing shows a schematic representation of the hill. We know that the hill rises 12.0 m vertically for every 100.0 m of distance in the horizontal direction, so that ho = 12.0 m and
ho
ha = 100.0 m . Moreover, according to Equation 1.3, the
tangent function is tan = ho / ha . Thus, we can use the
h
inverse tangent function to determine the angle θ.
a
SOLUTION With the aid of the inverse tangent function (see Equation 1.6) we find that h
12.0 m = 6.84 100.0 m
= tan −1 o = tan −1 ha
15.
REASONING Using the Pythagorean theorem (Equation 1.7), we find that the relation between the length D of the diagonal of the square (which is also the diameter of the circle) and the length L of one side of the square is D = L2 + L2 = 2 L . SOLUTION Using the above relation, we have
D 0.35 m = = 0.25 m 2 2 ______________________________________________________________________________ D = 2L
or
L=
16. REASONING In both parts of the drawing the line of sight, the horizontal dashed line, and the vertical form a right triangle. The angles a = 35.0° and b = 38.0° at which the person’s line of sight rises above the horizontal are known, as is the horizontal distance d = 85.0 m from the building. The unknown vertical sides of the right triangles correspond, respectively, to the heights Ha and Hb of the bottom and top of the antenna relative to the person’s eyes. The antenna’s height H is the difference between Hb and Ha: H = H b − H a . The horizontal side d of the triangle is adjacent to the angles a and b, while the vertical
sides Ha and Hb are opposite these angles. Thus, in either triangle, the angle is related to
h the horizontal and vertical sides by Equation 1.3 tan = o : ha H tan a = a d tan b =
Hb d
(1)
(2)
Chapter 1 Problems
9
H Hb
Ha
a
b d
d
(a)
(b)
SOLUTION Solving Equations (1) and (2) for the heights of the bottom and top of the antenna relative to the person’s eyes, we find that
H a = d tan a
and
H b = d tan b
The height of the antenna is the difference between these two values:
H = H b − H a = d tan b − d tan a = d ( tan b − tan a )
(
)
H = (85.0 m ) tan 38.0 − tan 35.0 = 6.9 m
17. REASONING The drawing shows the heights of the two balloonists and the horizontal distance x between them. Also shown in dashed lines is a right triangle, one angle of which is 13.3. Note that the side adjacent to the 13.3 angle is the horizontal distance x, while the side opposite the angle is the distance between the two heights, 61.0 m − 48.2 m. Since we know the angle and the length of one side of the right triangle, we can use trigonometry to find the length of the other side.
13.3
x
61.0 m
48.2 m
SOLUTION The definition of the tangent function, Equation 1.3, can be used to find the horizontal distance x, since the angle and the length of the opposite side are known:
tan13.3 =
length of opposite side length of adjacent side (= x)
Solving for x gives
x=
length of opposite side 61.0 m − 48.2 m = = 54.1 m tan13.3 tan13.3
____________________________________________________________________________________________
10
INTRODUCTION AND MATHEMATICAL CONCEPTS
18.
REASONING As given in Appendix E, the law of cosines is a
c
c2 = a 2 + b2 − 2ab cos
γ b
where c is the side opposite angle γ, and a and b are the other two sides. Solving for γ, we have a 2 + b2 − c 2 2ab
= cos −1
SOLUTION For c = 95 cm, a = 150 cm, and b = 190 cm −1 a
= cos
2
2 2 2 + b2 − c 2 −1 (150 cm ) + (190 ) − ( 95 cm ) = 30 = cos 2ab 2 (150 cm )(190 cm )
Thus, the angle opposite the side of length 95 cm is 30 . Similarly, for c = 150 cm, a = 95 cm, and b =190 cm, we find that the angle opposite the side of length 150 cm is 51 . Finally, for c = 190 cm, a = 150 cm, and b = 95 cm, we find that the angle opposite the side of length 190 cm is 99 . As a check on these calculations, we note that 30 + 51 + 99 = 180 , which must be the case for the sum of the three angles in a triangle. ____________________________________________________________________________________________
19. REASONING Note from the drawing that the shaded right triangle contains the angle , the side opposite the angle (length = 0.281 nm), and the side adjacent to the angle (length = L). If the length L can be determined, we can use trigonometry to find . The bottom face of the cube is a square whose diagonal has a length L. This length can be found from the Pythagorean theorem, since the lengths of the two sides of the square are known.
0.281 nm
0.281 nm
0.281 nm L
SOLUTION The angle can be obtained from the inverse tangent function, Equation 1.6, as = tan −1 ( 0.281 nm ) / L . Since L is the length of the hypotenuse of a right triangle whose sides have lengths of 0.281 nm, its value can be determined from the Pythagorean theorem:
Chapter 1 Problems
L=
11
( 0.281 nm )2 + ( 0.281 nm )2 = 0.397 nm
Thus, the angle is
0.281 nm −1 0.281 nm = tan = 35.3 L 0.397 nm ______________________________________________________________________________
= tan −1
h
20. REASONING a. The drawing shows the person standing on the earth and looking at the horizon. Notice the right triangle, the sides of which are R, the radius of the earth, and d, the distance from the person’s eyes to the horizon. The length of the hypotenuse is R + h, where h is the height of the person’s eyes above the water. Since we know the lengths of two sides of the triangle, the Pythagorean theorem can be employed to find the length of the third side.
d R
90
Horizon
R Earth
b. To convert the distance from meters to miles, we use the relation 1609 m = 1 mi (see the page facing the inside of the front cover of the text). SOLUTION a. The Pythagorean theorem (Equation 1.7) states that the square of the hypotenuse is equal to the sum of the squares of the sides, or ( R + h ) = d + R . Solving this equation for d 2
2
2
yields
d = ( R + h ) − R 2 = R 2 + 2 Rh + h 2 − R 2 2
(
)
= 2 Rh + h 2 = 2 6.38 106 m (1.6 m ) + (1.6 m ) = 4500 m 2
b. Multiplying the distance of 4500 m by a factor of unity, (1 mi)/(1609 m) = 1, the distance (in miles) from the person's eyes to the horizon is
1 mi d = ( 4500 m )(1) = 4500 m = 2.8 mi 1609 m ______________________________________________________________________________
(
)
12
INTRODUCTION AND MATHEMATICAL CONCEPTS
21. SSM REASONING The drawing at the right shows the location of each deer A, B, and C. From the problem statement it follows that
b = 62 m c = 95 m
= 180 − 51 − 77 = 52 Applying the law of cosines (given in Appendix E) to the geometry in the figure, we have a 2 − 2ab cos + (b 2 − c 2 ) = 0 2
which is an expression that is quadratic in a. It can be simplified to Aa + Ba + C = 0 , with
A=1 B= –2b cos = – 2(62 m) cos 52 = –76 m C = (b 2 − c 2 ) = (62 m) 2 − (95 m) 2 = –5181 m2
This quadratic equation can be solved for the desired quantity a. SOLUTION Suppressing units, we obtain from the quadratic formula
a=
− ( −76 ) ( −76 ) 2 − 4 (1)(–5181) = 1.2 10 2 m and – 43 m 2 (1)
Discarding the negative root, which has no physical significance, we conclude that the distance between deer A and C is 1.2 102 m .
22. REASONING The trapeze cord is L = 8.0 m long, so that the trapeze is initially h1 = L cos 41° meters below the support. At the instant he releases the trapeze, it is h2 = L cos meters below the support. The difference in the heights is d = h2 – h1 = 0.75 m. Given that the
41° h2
h1
L L
41°
trapeze is released at a lower elevation than the platform, we expect to find < 41°. SOLUTION Putting the above relationships together, we have
0.75 m
Chapter 1 Problems
d = h2 − h1 = L cos − L cos 41
13
d + L cos 41 = L cos
or
d + cos 41 L
cos =
d 0.75 m + cos 41 = cos −1 + cos 41 = 32 L 8.0 m
= cos−1
23. SSM REASONING a. Since the two force vectors A and B have directions due west and due north, they are perpendicular. Therefore, the resultant vector F = A + B has a magnitude given by the Pythagorean theorem: F2 = A2 + B2. Knowing the magnitudes of A and B, we can calculate the magnitude of F. The direction of the resultant can be obtained using trigonometry. b. For the vector F = A – B we note that the subtraction can be regarded as an addition in the following sense: F = A + (–B). The vector –B points due south, opposite the vector B, so the two vectors are once again perpendicular and the magnitude of F again is given by the Pythagorean theorem. The direction again can be obtained using trigonometry. SOLUTION a. The drawing shows the two vectors and the resultant vector. According to the Pythagorean theorem, we have F 2 = A2 + B 2
F=
A2 + B 2
F=
b445 N g+ b325 N g
F B
F
A 2
North
North
–B A
2
(b)
(a)
= 551 N Using trigonometry, we can see that the direction of the resultant is
tan =
B A
or
= tan −1
325 N I F G H445 N J K= 36.1 north of west
b. Referring to the drawing and following the same procedure as in part a, we find
b g or F = A + b− B g = b445 N g+ b−325 N g = 551 N B F325 N IJ= 36.1 south of west tan = or = tan G H445 N K A
F 2 = A2 + − B
2
2
2
2
2
−1
_____________________________________________________________________________________________
14
INTRODUCTION AND MATHEMATICAL CONCEPTS
24. REASONING Since the initial force and the resultant force point along the east/west line, the second force must also point along the east/west line. The direction of the second force is not specified; it could point either due east or due west, so there are two answers. We use “N” to denote the units of the forces, which are specified in newtons. SOLUTION If the second force points due east , both forces point in the same direction and the magnitude of the resultant force is the sum of the two magnitudes: F1 + F2 = FR. Therefore, F2 = FR – F1 = 400 N – 200 N = 200 N If the second force points due west , the two forces point in opposite directions, and the magnitude of the resultant force is the difference of the two magnitudes: F2 – F1 = FR. Therefore, F2 = FR + F1 = 400 N + 200 N = 600 N _____________________________________________________________________________________________
25. SSM REASONING For convenience, we can assign due east to be the positive direction and due west to be the negative direction. Since all the vectors point along the same eastwest line, the vectors can be added just like the usual algebraic addition of positive and negative scalars. We will carry out the addition for all of the possible choices for the two vectors and identify the resultants with the smallest and largest magnitudes. SOLUTION There are six possible choices for the two vectors, leading to the following resultant vectors:
F1 + F2 = 50.0 newtons + 10.0 newtons = +60.0 newtons = 60.0 newtons, due east F1 + F3 = 50.0 newtons − 40.0 newtons = +10.0 newtons = 10.0 newtons, due east F1 + F4 = 50.0 newtons − 30.0 newtons = +20.0 newtons = 20.0 newtons, due east F2 + F3 = 10.0 newtons − 40.0 newtons = −30.0 newtons = 30.0 newtons, due west F2 + F4 = 10.0 newtons − 30.0 newtons = −20.0 newtons = 20.0 newtons, due west F3 + F4 = −40.0 newtons − 30.0 newtons = −70.0 newtons = 70.0 newtons, due west The resultant vector with the smallest magnitude is F1 + F3 = 10.0 newtons, due east . The resultant vector with the largest magnitude is F3 + F4 = 70.0 newtons, due west .
Chapter 1 Problems
15
26. REASONING The Pythagorean theorem (Equation 1.7) can be used to find the magnitude of the resultant vector, and trigonometry can be employed to determine its direction. a. Arranging the vectors in tail-to-head fashion, we can see that the vector A gives the resultant a westerly direction and vector B gives the resultant a southerly direction. Therefore, the resultant A + B points south of west. b. Arranging the vectors in tail-to-head fashion, we can see that the vector A gives the resultant a westerly direction and vector –B gives the resultant a northerly direction. Therefore, the resultant A + (–B) points north of west. SOLUTION Using the Pythagorean theorem and trigonometry, we obtain the following results:
a.
Magnitude of A + B =
( 63 units )2 + ( 63 units )2 = 89 units
= tan −1
63 units = 45 south of west 63 units
b.
Magnitude of A − B =
( 63 units )2 + ( 63 units )2 = 89 units
= tan −1
63 units = 45 north of west 63 units ______________________________________________________________________________ 27. REASONING At the turning point, the distance to the campground is labeled d in the drawing. Note that d is the length of the hypotenuse of a right triangle. Since we know the lengths of the other two sides of the triangle, the Pythagorean theorem can be used to find d. The direction that cyclist #2 must head during the last part of the trip is given by the angle . It can be determined by using the inverse tangent function.
Turning point
N
d Campground
1950 m 1430 m Start
E W SOLUTION 1080 m a. The two sides of the triangle have lengths of 1080 m S and 520 m (1950 m − 1430 m = 520 m). The length d of the hypotenuse can be determined from the Pythagorean theorem, Equation (1.7), as
d=
(1080 m )2 + ( 520 m )2 = 1200 m
16
INTRODUCTION AND MATHEMATICAL CONCEPTS
b. Since the lengths of the sides opposite and adjacent to the angle are known, the inverse tangent function (Equation 1.6) can be used to find .
520 m = 26 south of east 1080 m ______________________________________________________________________________
= tan −1
28.
REASONING The triple jump consists of a double jump in one direction, followed by a perpendicular single jump, which we can represent with displacement vectors J and K (see the drawing). These two perpendicular vectors form a right triangle with their resultant D K = J + K, which is the displacement of the colored checker. In order to find the magnitude D of the displacement, we first need D to find the magnitudes J and K of the double jump and the single jump. As the three sides of a right triangle, J, K, and D (the hypotenuse) are related to one another by the Pythagorean theorem (Equation 1.7) The double jump moves the colored d s J checker a straight-line distance equal to the length of four s square’s diagonals d, and the single jump moves a length equal to two square’s diagonals. Therefore,
J = 4d
and
K = 2d
(1)
Let the length of a square’s side be s. Any two adjacent sides of a square form a right triangle with the square’s diagonal (see the drawing). The Pythagorean theorem gives the diagonal length d in terms of the side length s: d = s 2 + s 2 = 2s 2 = s 2
(2)
SOLUTION First, we apply the Pythagorean theorem to the right triangle formed by the three displacement vectors, using Equations (1) for J and K:
( 4d )2 + ( 2d )2 = 16d 2 + 4d 2 =
D = J 2 + K2 =
20d 2 = d 20
Substituting Equation (2) into Equation (3) gives
(
D = d 20 = s 2
) 20 = s 40 = ( 4.0 cm ) 40 = 25 cm
(3)
Chapter 1 Problems
29. REASONING Both P and Q and the vector sums K and M can be drawn with correct magnitudes and directions by counting grid squares. To add vectors, place them tail-to-head and draw the resultant vector from the tail of the first vector to the head of the last. The vector 2P is equivalent to P + P, and −Q is a vector that has the same magnitude as Q, except it is directed in the opposite direction.
17
P Q M −Q
8.00 cm
K
P The vector M runs 11 squares P horizontally and 3 squares vertically, and the vector K runs 4 squares horizontally and 9 squares vertically. These distances can be converted from grid squares to centimeters with the grid scale: 1 square = 4.00 cm. Once the distances are calculated in centimeters, the Pythagorean theorem (Equation 1.7) will give the magnitudes of the vectors. SOLUTION a. The vector M = P + Q runs 11 squares horizontally and 3 squares vertically, and these cm distances are equivalent to, respectively, 4.00 (11 squares ) = 44.0 cm and square
cm 4.00 ( 3 squares ) = 12.0 cm . Thus, the magnitude of M is square M=
( 44.0 cm)2 + (12.0 cm)2 = 45.6 cm
b. Similarly, the lengths of the horizontal and vertical distances of K = 2P − Q are 4 horizontal squares and 9 vertical squares, or 16.0 cm and 36.0 cm, respectively. The magnitude of K is then K=
(16.0 cm)2 + ( 36.0 cm)2 = 39.4 cm
30. REASONING a. and b. The following drawing shows the two vectors A and B, as well as the resultant vector A + B. The three vectors form a right triangle, of which two of the sides are known. We can employ the Pythagorean theorem, Equation 1.7, to find the length of the third side. The angle in the drawing can be determined by using the inverse cosine function, Equation 1.5, since the side adjacent to and the length of the hypotenuse are known.
18
INTRODUCTION AND MATHEMATICAL CONCEPTS
N B 15.0 units A+B
A
W
E
12.3 units S
c. and d. The following drawing shows the vectors A and −B, as well as the resultant vector A − B. The three vectors form a right triangle, which is identical to the previous drawing, except for orientation. Thus, the lengths of the hypotenuses and the angles are equal. N 12.3 units
W
A A−B
E
15.0 units
−B
S
SOLUTION a. Let R = A + B. The Pythagorean theorem (Equation 1.7) states that the square of the 2 2 2 hypotenuse is equal to the sum of the squares of the sides, so that R = A + B . Solving for B yields
(15.0 units )2 − (12.3 units )2 = 8.6 units
B = R 2 − A2 =
b. The angle can be found from the inverse cosine function, Equation 1.5:
12.3 units = 34.9 north of west 15.0 units
= cos−1
c. Except for orientation, the triangles in the two drawings are the same. Thus, the value for B is the same as that determined in part (a) above: B = 8.6 units d. The angle is the same as that found in part (a), except the resultant vector points south of west, rather than north of west: = 34.9 south of west ______________________________________________________________________________ 31.
SSM REASONING AND SOLUTION The single force needed to produce the same effect is equal to the resultant of the forces provided by the two ropes. The following figure shows the force vectors drawn to scale and arranged tail to head. The magnitude and direction of the resultant can be found by direct measurement using the scale factor shown in the figure.
Chapter 1 Problems
19
a. From the figure, the magnitude of the resultant is 5600 N . b. The single rope should be directed along the dashed line
in the text drawing.
____________________________________________________________________________________________
32. REASONING a. Since the two displacement vectors A and B have directions due south and due east, they are perpendicular. Therefore, the resultant vector R = A + B has a magnitude given by the Pythagorean theorem: R2 = A2 + B2. Knowing the magnitudes of R and A, we can calculate the magnitude of B. The direction of the resultant can be obtained using trigonometry. b. For the vector R = A – B we note that the subtraction can be regarded as an addition in the following sense: R = A + (–B). The vector –B points due west, opposite the vector B, so the two vectors are once again perpendicular and the magnitude of R again is given by the Pythagorean theorem. The direction again can be obtained using trigonometry. SOLUTION a. The drawing shows the two vectors and the resultant vector. According to the Pythagorean theorem, we have
R 2 = A2 + B 2 B=
or
B = R 2 − A2
A
( 3.75 km ) − ( 2.50 km ) = 2.8 km 2
2
North
North
R
R B
–B
(a)
(b)
A
Using trigonometry, we can see that the direction of the resultant is
tan =
B A
or
= tan −1
2.8 km I F G H2.50 km J K= 48 east of south
b. Referring to the drawing and following the same procedure as in part a, we find
b g or B = R − A = b3.75 kmg− b2.50 kmg = 2.8 km B F2.8 km IJ= 48 west of south tan = or = tan G H2.50 km K A
R 2 = A2 + − B
2
2
2
2
2
−1
_____________________________________________________________________________________________
20
INTRODUCTION AND MATHEMATICAL CONCEPTS
33. REASONING AND SOLUTION The following figure is a scale diagram of the forces drawn tail-to-head. The scale factor is shown in the figure. The head of F3 touches the tail of F1, because the resultant of the three forces is zero. a. From the figure, F3 must have a
magnitude of 78 N if the resultant force acting on the ball is zero.
60.0° F1 = 50.0 N
F3 60.0°
b. Measurement with a protractor indicates that the angle = 34 . Scale Factor:
F2 = 90.0 N
20.0 N
_____________________________________________________________________________________________
34.
REASONING The magnitude of the x-component of the force vector is the product of the magnitude of the force times the cosine of the angle between the vector and the x axis. Since the x-component points in the +x direction, it is positive. Likewise, the magnitude of the y component of the force vector is the product of the magnitude of the force times the sine of the angle between the vector and the x axis. Since the vector points 36.0º below the positive x axis, the y component of the vector points in the −y direction; thus, a minus sign must be assigned to the y-component to indicate this direction. SOLUTION The x and y scalar components are
35.
a.
Fx = (575 newtons) cos 36.0 = 465 newtons
b.
Fy = –(575 newtons) sin 36.0 = –338 newtons
SSM REASONING AND SOLUTION In order to determine which vector has the largest x and y components, we calculate the magnitude of the x and y components explicitly and compare them. In the calculations, the symbol u denotes the units of the vectors. Ax = (100.0 u) cos 90.0° = 0.00 u Bx = (200.0 u) cos 60.0° = 1.00 102 u Cx = (150.0 u) cos 0.00° = 150.0 u
Ay = (100.0 u) sin 90.0° = 1.00 102 u By = (200.0 u) sin 60.0° = 173 u Cy = (150.0 u) sin 0.00° = 0.00 u
a. C has the largest x component. b. B has the largest y component. ______________________________________________________________________________
Chapter 1 Problems
21
36. REASONING The triangle in the drawing is a right triangle. We know one of its angles is 30.0, and the length of the hypotenuse is 8.6 m. Therefore, the sine and cosine functions can be used to find the magnitudes of Ax and Ay. The directions of these vectors can be found by examining the diagram. SOLUTION a. The magnitude Ax of the displacement vector Ax is related
N
to the length of the hypotenuse and the 30.0 angle by the sine function (Equation 1.1). The drawing shows that the direction of Ax is due east.
W
#1
E S
30.0
Ax = A sin 30.0 = ( m ) sin 30.0 = m, due east
8.6 m
Ay
b. In a similar manner, the magnitude Ay of Ay can be found by using the cosine function (Equation 1.2). Its direction is due south.
A 90 #2
Ay = A cos30.0 = ( m ) cos30.0 = m, due south
Net
Ax
______________________________________________________________________________ 37. REASONING Using trigonometry, we can determine the angle from the relation tan = Ay /Ax: SOLUTION
Ay −1 12 m = tan = 45 12 m Ax A 12 m b. = tan −1 y = tan −1 = 35 17 m Ax A 17 m c. = tan −1 y = tan −1 = 55 12 m Ax ______________________________________________________________________________ a.
= tan −1
38. REASONING The drawing assumes that the horizontal direction along the ground is the x direction and shows the plane’s velocity vector v, along with its horizontal component vx and vertical component vy. These components, together with the velocity vector, form a right triangle, as indicated. Based on this right triangle, we
v
vy 90º
34º vx
22
INTRODUCTION AND MATHEMATICAL CONCEPTS
can use the cosine function to determine the horizontal velocity component. SOLUTION According to Equation 1.2, the cosine of an angle is the side of the right triangle adjacent to the angle divided by the hypotenuse. Thus, for the 34º angle in the drawing we have
cos 34 =
vx v
vx = v cos 34 = (180 m/s ) cos 34 = 150 m/s
or
_____________________________________________________________________________________________
39. SSM REASONING The x and y components of r are mutually perpendicular; therefore, the magnitude of r can be found using the Pythagorean theorem. The direction of r can be found using the definition of the tangent function. SOLUTION According to the Pythagorean theorem, we have r=
x 2 + y 2 = ( −125 m ) 2 + ( −184 m) 2 = 222 m
The angle is
184 m = 55.8 125 m
= tan −1
____________________________________________________________________________________________
40. REASONING AND SOLUTION a. From the Pythagorean theorem, we have F = (150 N) 2 + (130 N) 2 = 2.0 10 2 N
b. The angle is given by
= tan −1
130 N I F G H150 N J K= 41
_____________________________________________________________________________________________
41. REASONING Two vectors that are equal must have the same magnitude and direction. Equivalently, they must have identical x components and identical y components. We will begin by examining the given information with respect to these criteria, in order to see if there are obvious reasons why some of the vectors could not be equal. Then we will compare our choices for the equal vectors by calculating the magnitude and direction and the scalar components, as needed. SOLUTION Vectors A and B cannot possibly be equal, because they have different x scalar components of Ax = 80.0 m and Bx = 60.0 m. Furthermore, vectors B and C cannot possibly be equal, because they have different magnitudes of B = 75.0 m and C = 100.0 m.
Chapter 1 Problems
23
Therefore, we conclude that vectors A and C are the equal vectors. To verify that this is indeed the case we have two choices. We can either calculate the magnitude and direction of A (and compare it to the given magnitude and direction of C) or determine the scalar components of C (and compare them to the given components of A). Either choice will do, although both are shown below. +y The magnitude and direction of A are A
A=
Ax2 + Ay2 =
(80.0 m ) + ( 60.0 m ) = 100.0 m 2
2
Ay −1 60.0 m = tan = 36.9 80.0 m Ax
= tan −1
These results are identical to those given for C. The scalar components of C are Cx = C cos 36.9 = (100.0 m ) cos 36.9 = 80.0 m
Ay +x
Ax +y C 36.9
Cx
Cy +x
C y = C sin 36.9 = (100.0 m ) sin 36.9 = 60.0 m
These results are identical to the components given for A.
42. REASONING Because both boats travel at Bwest W 101 km per hour, each one ends up 25.0° (0.500 h)(101 km/h) = 50.5 km from the dock after a half-hour. They travel along straight Bsouth B 53.0° paths, so this is the magnitude of both Gsouth displacement vectors: B = G = 50.5 km. Since G the displacement vector G makes an angle of 37° south of due west, its direction can also be expressed as 90° − 37° = 53° west of south. Gwest S With these angles and the magnitudes of both vectors in hand, we can consider the westward and southward components of each vector. SOLUTION a. First we calculate the magnitudes of the westward component of the displacement of each boat and then subtract them to find the difference:
Magnitude of Bwest = B cos 25.0 = ( 50.5 km ) cos 25.0 = 45.8 km Magnitude of Gwest = G sin 53.0 = ( 50.5 km ) sin 53.0 = 40.3 km
24
INTRODUCTION AND MATHEMATICAL CONCEPTS
The blue boat travels farther by the following amount:
45.8 km − 40.3 km = 5.5 km b. Similarly, we find for the magnitudes of the southward components that
Magnitude of Bsouth = B sin 25.0 = ( 50.5 km ) sin 25.0 = 21.3 km Magnitude of Gsouth = G cos53.0 = ( 50.5 km ) cos53.0 = 30.4 km The green boat travels farther by the following amount:
30.4 km − 21.3 km = 9.1 km
43. SSM REASONING The force F and its two components form a right triangle. The hypotenuse is 82.3 newtons, and the side parallel to the +x axis is Fx = 74.6 newtons. Therefore, we can use the trigonometric cosine and sine functions to determine the angle of F relative to the +x axis and the component Fy of F along the +y axis. SOLUTION a. The direction of F relative to the +x axis is specified by the angle as
74.6 newtons = 25.0 82.3 newtons
= cos−1
(1.5)
b. The component of F along the +y axis is
Fy = F sin 25.0 = ( newtons ) sin 25.0 = 34.8 newtons
(1.4)
______________________________________________________________________________ 44. REASONING AND SOLUTION The force F can be first resolved into two components; the z component Fz and the projection onto the x-y plane, Fp as shown on the left in the following figure. According to this figure, Fp = F sin 54.0° = (475 N) sin 54.0°= 384 N The projection onto the x-y plane, Fp, can then be resolved into x and y components.
Chapter 1 Problems
25
a. From the figure on the right, Fx = Fp cos 33.0° = (384 N) cos 33.0°=
322 N
b. Also from the figure on the right, Fy = Fp sin 33.0° = (384 N) sin 33.0°=
209 N
c. From the figure on the left, Fz = F cos 54.0° = (475 N) cos 54.0°=
279 N
______________________________________________________________________________ 45. SSM REASONING The individual displacements of the golf ball, A, B, and C are shown in the figure. Their resultant, R, is the displacement that would have been needed to "hole the ball" on the very first putt. We will use the component method to find R.
SOLUTION The components of each displacement vector are given in the table below. Vector
x Components
y Components
A B C
(5.0 m) cos 0° = 5.0 m (2.1 m) cos 20.0° = 2.0 m (0.50 m) cos 90.0° = 0
(5.0 m) sin 0° = 0 (2.1 m) sin 20.0° = 0.72 m (0.50 m) sin 90.0° = 0.50 m
R = A+B+C
7.0 m
1.22 m
26
INTRODUCTION AND MATHEMATICAL CONCEPTS
The resultant vector R has magnitude R = ( 7 .0 m) 2 + ( 1.22 m) 2 = 7 .1 m
and the angle is
1.22 m = 9.9 7.0 m
= tan −1
Thus, the required direction is 9.9 north of east . ______________________________________________________________________________ 46. REASONING To apply the component method for vector addition, we must first determine the x and y components of each vector. The algebraic sum of the three x components gives the x component of the resultant. The algebraic sum of the three y components gives the y component of the resultant. Knowing the x and y components of the resultant will allow us to use the Pythagorean theorem to determine the magnitude of the resultant. Finally, the directional angle of the resultant will be obtained using the trigonometric sine function. SOLUTION Referring to the drawing in the text, we see that the x and y components of the vectors are Ax = − A cos 20.0 = − ( 5.00 m ) cos 20.0 = −4.70 m Ay = A sin 20.0 = ( 5.00 m ) sin 20.0 = 1.71 m Bx = B cos 60.0 = ( 5.00 m ) cos 60.0 = 2.50 m
By = B sin 60.0 = ( 5.00 m ) sin 60.0 = 4.33 m
C x = 0.00 m
C y = −4.00 m
Rx = −4.70 m + 2.50 m + 0.00 m = −2.20 m
Ry = 1.71 m + 4.33 m − 4.00 m = 2.04 m
Note that the value for Ax is negative because this component points in the −x direction and that the value for Cx is zero because the vector C points along the −y axis. Note also that the value for Cy is negative because the vector C points along the −y axis. The x component Rx of the resultant vector, being negative, points in the −x direction. The y component Ry of the resultant vector, being positive, points in the +y direction. The drawing shows these two components and the resultant vector. Since the components are perpendicular, the magnitude R of the resultant can be obtained using the Pythagorean theorem.
+y R
Ry 90º
θ Rx
+x
Chapter 1 Problems
27
( −2.20 m )2 + ( 2.04 m )2 = 3.00 m
R = Rx2 + R y2 =
Referring to the drawing, we can see that sin = Rx / R , so that the directional angle θ is Ry −1 2.04 m = sin = 42.8 3.00 m R
= sin −1
Thus, the resultant vector points in a direction of 42.8º above the negative x axis . _____________________________________________________________________________________________
47. REASONING Using the component method for vector addition, we will find the x component of the resultant force vector by adding the x components of the individual vectors. Then we will find the y component of the resultant vector by adding the y components of the individual vectors. Once the x and y components of the resultant are known, we will use the Pythagorean theorem to find the magnitude of the resultant and trigonometry to find its direction. We will take east as the +x direction and north as the +y direction. SOLUTION The x component of the resultant force F is
b
b
g
b
g
g
Fx = 2240 N cos 34 .0 + 3160 N cos 90.0 = 2240 N cos 34 .0 FAx FBx The y component of the resultant force F is
b
g
b
g
Fy = − 2240 N sin 34 .0 + −3160 N FAy FBy
Using the Pythagorean theorem, we find that the magnitude of the resultant force is
F=
Fx2 + Fy2 =
b2240 N gcos 34.0 + − b2240 N gsin 34.0−3160 N = 4790 N 2
2
Using trigonometry, we find that the direction of the resultant force is
= tan −1
L b2240 N gsin 34.0+3160 N O = 67 .2 south of east M N b2240 N gcos 34.0 P Q
______________________________________________________________________________ 48. REASONING The resultant force in part a is FA, because that is the only force applied. The resultant force R in part b, where the two additional forces with identical magnitudes (FB = FC = F) are applied, is the sum of all three vectors: R = FA + FB + FC. Because the magnitude R of the resultant force R is k times larger than the magnitude FA of FA, we have
28
INTRODUCTION AND MATHEMATICAL CONCEPTS
R = kFA
(1)
To solve the problem, we need to find an expression for R in terms of FA and F. Let the +x direction be in the direction of FA and the +y direction be upward (see the following drawing), and consider the resultant of the two additional force vectors in part b. Because FB and FC are directed symmetrically about the x axis, and have the same magnitude F, their y components are equal and opposite. Therefore, they cancel out of the vector sum R, leaving only the x components of FB and FC. Now, since FA also has only an x component, the resultant R can be written as the sum of FA and the vector x components of FB and FC: R = FA + FBx + FCx. Because these vectors all point in the same direction, we can write down a first expression for the magnitude of the resultant:
R = FA + FBx + FCx
(2)
+y FB FA
20.0°
FBx
FC 20.0°
FCx
+x
R
The x components of the vectors FB and FC are adjacent to the 20.0°-angles, and so are
h related to the common magnitude F of both vectors by Equation 1.2 cos = a , with h ha = FBx and h = F: (3) FBx = F cos and FCx = F cos SOLUTION First, we use Equations (3) to replace FBx and FCx in Equation (2):
R = FA + F cos + F cos = FA + 2F cos
(4)
Then, we combine Equation (4) with Equation (1) (R = kFA) to eliminate R, and solve for the desired ratio F/FA:
Chapter 1 Problems
FA + 2 F cos = kFA
or
2 F cos = kFA − FA
29
2 F cos = ( k − 1) FA
or
F k −1 2.00 − 1 = = = 0.532 FA 2 cos 2 cos 20.0 49. REASONING Using the component method, we find the components of the resultant R that are due east and due north. The magnitude and direction of the resultant R can be determined from its components, the Pythagorean theorem, and the tangent function. SOLUTION The first four rows of the table below give the components of the vectors A, B, C, and D. Note that east and north have been taken as the positive directions; hence vectors pointing due west and due south will appear with a negative sign. Vector
East/West Component
North/South Component
A B C D
+ 2.00 km 0 – 2.50 km 0
0 + 3.75 km 0 –3.00 km
R=A+B+C+D
– 0.50 km
+ 0.75 km
The fifth row in the table gives the components of R. The magnitude of R is given by the Pythagorean theorem as R = (– 0.50 km) 2 + ( + 0.75 km) 2 = 0.90 km
The angle that R makes with the direction due west is
0.75 km = 56 north of west 0.50 km
= tan −1
R
RN
RE _____________________________________________________________________________________________
50. REASONING We will use the scalar x and y components of the resultant vector to obtain its magnitude and direction. To obtain the x component of the resultant we will add together the x components of each of the vectors. To obtain the y component of the resultant we will add together the y components of each of the vectors.
30
INTRODUCTION AND MATHEMATICAL CONCEPTS
SOLUTION The x and y components of the resultant vector R are Rx and Ry, respectively. In terms of these components, the magnitude R and the directional angle (with respect to the x axis) for the resultant are
R=
Rx2 + Ry2
and
Ry = tan Rx −1
(1)
+y
B A 20.0 35.0
+x
50.0
C The following table summarizes the components of the individual vectors shown in the drawing: Vector
x component
D y component
A
Ax = − (16.0 m ) cos 20.0 = −15.0 m
Ay = (16.0 m ) sin 20.0 = 5.47 m
B
Bx = 0 m
B y = 11.0 m
C
C x = − (12.0 m ) cos 35.0 = −9.83 m
C y = − (12.0 m ) sin 35.0 = −6.88 m
D
Dx = ( 26.0 m ) cos 50.0 = 16.7 m
Dy = − ( 26.0 m ) sin 50.0 = −19.9 m
R
Rx = −15.0 m + 0 m − 9.83 m + 16.7 m
Ry = 5.47 m + 11.0 m − 6.88 m − 19.9 m
= −8.1 m
= −10.3 m
Note that the component Bx is zero, because B points along the y axis. Note also that the components Cx and Cy are both negative, since C points between the −x and −y axes. Finally, note that the component Dy is negative since D points below the +x axis. Using equation (1), we find that +y Rx 2 2 +x R = Rx2 + R y2 = ( −8.1 m ) + ( −10.3 m ) = 13 m
Ry −10.3 m = tan = tan −1 = 52 −8.1 m Rx −1
Ry R
Since both Rx and Ry are negative, the resultant points between the −x and −y axes.
Chapter 1 Problems
31
51. REASONING If we let the directions due east and due north be the positive directions, then the desired displacement A has components
A E = (4.8 km ) cos 42 = 3.57 km A N = (4.8 km ) sin 42 = 3.21 km while the actual displacement B has components
B E = (2.4 km ) cos 22 = 2.23 km B N = (2.4 km ) sin 22 = 0.90 km Therefore, to reach the research station, the research team must go 3.57 km – 2.23 km = 1.34 km, eastward
and
3.21 km – 0.90 km = 2.31 km, northward
SOLUTION a. From the Pythagorean theorem, we find that the magnitude of the displacement vector required to bring the team to the research station is
N
R = (1.34 km) 2 + (2.31 km) 2 = 2.7 km
b. The angle is given by
R
2.31 km
1.34 km
2.31 km = 6.0 10 1 degrees, north of east 1.34 km
= tan –1
_____________________________________________________________________________________________
52. REASONING Let A be the vector from base camp to the first team, B the vector from base camp to the second team, and C the vector from the first team’s position to First team C the second team’s position. C is the vector whose magnitude and direction are given by the first team’s GPS unit. Since you can A get from the base camp to the second 19° team’s position either by traveling along W vector B alone, or by traveling first along Base camp A and then along C, we know that B is the vector sum of the other two: B = A + C.
N (y)
35°
Second team
B=A+C
E (x)
32
INTRODUCTION AND MATHEMATICAL CONCEPTS
The reading on the first team’s GPS unit is then C = B − A. The components of C are found from the components of A and B: Cx = Bx − Ax, Cy = By − Ay. Once we have these components, we can calculate the magnitude and direction of C, as shown on the GPS readout. Because the first team is northwest of camp, and the second team is northeast, we expect the vector C to be directed east and either north or south. SOLUTION Let east serve as the positive x direction and north as the positive y direction. We then calculate the components of C, noting that B’s components are both positive, and A’s x-component is negative: C x = (29 km) sin 35 − ( −38 km) cos 19 = 53 km B A x x
C y = (29 km) cos 35 − (38 km) sin 19 = 11 km B A y y
The second team is therefore 53 km east of the first team (since Cx is positive), and 11 km north (since Cy is positive). The straight-line distance between the teams can be calculated with the Pythagorean theorem (Equation 1.7):
C = Cx 2 + C y 2 =
(53 km )2 + (11 km )2 = 54 km
The direction of the vector C is to be measured relative to due east, so we apply the inverse tangent function (Equation 1.6) to get the angle θ:
C 11 km = tan −1 y = tan −1 = 12 53 km Cx
53.
SSM REASONING Since the finish line is coincident with the starting line, the net displacement of the sailboat is zero. Hence the sum of the components of the displacement vectors of the individual legs must be zero. In the drawing in the text, the directions to the right and upward are taken as positive. SOLUTION In the horizontal direction Rh = Ah + Bh + Ch + Dh = 0 Rh = (3.20 km) cos 40.0° – (5.10 km) cos 35.0° – (4.80 km) cos 23.0° + D cos = 0 D cos = 6.14 km
(1)
Chapter 1 Problems
33
In the vertical direction Rv = Av + Bv + Cv + Dv = 0. Rv = (3.20 km) sin 40.0° + (5.10 km) sin 35.0° – (4.80 km) sin 23.0° – D sin = 0. D sin = 3.11 km
(2)
Dividing (2) by (1) gives tan = (3.11 km)/(6.14 km)
= 26.9
or
Solving (1) gives D = (6.14 km)/cos 26.9° =
6.88 km
54. REASONING The following table shows the components of the individual displacements and the components of the resultant. The directions due east and due north are taken as the positive directions. East/West Component
Displacement (1) (2) (3) (4)
North/South Component
–27.0 cm 0 –(23.0 cm) cos 35.0° = –18.84 cm –(23.0 cm) sin 35.0° = –13.19 cm (28.0 cm) cos 55.0° = 16.06 cm –(28.0 cm) sin 55.0° = –22.94 cm (35.0 cm) cos 63.0° = 15.89 cm (35.0 cm) sin 63.0° = 31.19 cm
Resultant
–13.89 cm
–4.94 cm
SOLUTION a. From the Pythagorean theorem, we find that the magnitude of the resultant displacement vector is R = (13.89 cm) 2 + (4.94 cm) 2 = 14.7 cm
b. The angle is given by
= tan −1
4.94 cm I F G H13.89 cm J K= 19.6 , south of west
13.89 cm
4.94 cm
R
34
INTRODUCTION AND MATHEMATICAL CONCEPTS
55. REASONING The drawing shows the vectors A, B, and C. Since these vectors add to give a resultant that is zero, we can write that A + B + C = 0. This addition will be carried out by the component method. This means that the x-component of this equation must be zero (Ax + Bx + Cx = 0) and the y-component must be zero (Ay + By + Cy = 0). These two equations will allow us to find the magnitudes of B and C.
+y (north) A −x
145 units 35.0
B 65.0
+x (east)
15.0
C −y
SOLUTION The x- and y-components of A, B, and C are given in the table below. The plus and minus signs indicate whether the components point along the positive or negative axes. Vector
x Component
y Component
A
–(145 units) cos 35.0° = –119 units
+(145 units) sin 35.0° = +83.2 units
B C
+B sin 65.0° = +B (0.906) –C sin 15.0° = –C (0.259)
+B cos 65.0° = +B (0.423) –C cos 15.0° = –C (0.966)
A+B+C
–119 units + B (0.906) – C (0.259)
+83.2 units + B (0.423) – C (0.966)
Setting the separate x- and y- components of A + B + C equal to zero gives x-component
(–119 units) + B (0.906) – C (0.259) = 0
y-component
(+83.2 units) + B (0.423) – C (0.966) = 0
Solving these two equations simultaneously, we find that a. B = 178 units b. C = 164 units
56. REASONING We know that the three displacement vectors have a resultant of zero, so that A + B + C = 0. This means that the sum of the x components of the vectors and the sum of the y components of the vectors are separately equal to zero. From these two equations we will be able to determine the magnitudes of vectors B and C. The directions east and north are, respectively, the +x and +y directions. SOLUTION Setting the sum of the x components of the vectors and the sum of the y components of the vectors separately equal to zero, we have
Chapter 1 Problems
35
1550 m g cos 25.0 + B sin 41.0 + b − C cos 35.0 g =0 b Bx
Ax
Cx
1550 m g sin 25.0 + b − B cos 41.0 g + C sin 35.0 = 0 b Ay
By
Cy
These two equations contain two unknown variables, B and C. They can be solved simultaneously to show that a. B = 5550 m and b. C = 6160 m ______________________________________________________________________________ 57. REASONING The shortest distance between the tree and the termite mound is equal to the magnitude of the chimpanzee's displacement r. SOLUTION a. From the Pythagorean theorem, we have
51 m
r = (51 m) + (39 m) = 64 m 2
2
b. The angle is given by
= tan −1
39 m
r
39 m I F G H51 m J K= 37 south of east
_____________________________________________________________________________________________
58. REASONING When the monkey has climbed as far up the pole as it can, its leash is taut, making a straight line from the stake to the monkey, that is, L = 3.40 m long. The leash is the hypotenuse of a right triangle, and the other sides are a line drawn from the stake to the base of the pole (d = 3.00 m), and a line from the base of the pole to the monkey (height = h).
Stake L h d
SOLUTION These three lengths are related by the Pythagorean theorem (Equation 1.7):
h 2 + d 2 = L2
or
h 2 = L2 − d 2
h = L2 − d 2 =
( 3.40 m )2 − ( 3.00 m )2 = 1.6 m
36
INTRODUCTION AND MATHEMATICAL CONCEPTS
59. SSM REASONING The ostrich's velocity vector v and the desired components are shown in the figure at the right. The components of the velocity in the directions due west and due north are v W and v N , respectively. The sine and cosine functions can be used to find the components. SOLUTION a. According to the definition of the sine function, we have for the vectors in the figure sin =
vN v
or
vN = v sin = (17.0 m/s) sin 68 = 15.8 m/s
or
vW = v cos = (17.0 m/s) cos 68.0 = 6.37 m/s
b. Similarly, cos =
vW v
_____________________________________________________________________________________________
60. REASONING In the expression for the volume flow rate, the dimensions on the left side of the equals sign are [L]3/[T]. If the expression is to be valid, the dimensions on the right side of the equals sign must also be [L]3/[T]. Thus, the dimensions for the various symbols on the right must combine algebraically to yield [L]3/[T]. We will substitute the dimensions for each symbol in the expression and treat the dimensions of [M], [L], and [T] as algebraic variables, solving the resulting equation for the value of the exponent n. SOLUTION We begin by noting that the symbol and the number 8 have no dimensions. It follows, then, that n M L 3 2 n n R n ( P2 − P1 ) L L T L T L Q= or = = = 2 8 L T LT M L T L L T
L3 = Ln T L T Thus, we find that n = 4 .
or
n L L = L 3
or
L 3 L = L4 = L n
Chapter 1 Problems
37
61. REASONING AND SOLUTION The east and north components are, respectively a.
Ae = A cos = (155 km) cos 18.0° =
147 km
b.
An = A sin = (155 km) sin 18.0° =
47.9 km
_____________________________________________________________________________________________
62.
REASONING According to the component method for vector addition, the x component of the resultant vector is the sum of the x component of A and the x component of B. Similarly, the y component of the resultant vector is the sum of the y component of A and the y component of B. The magnitude R of the resultant can be obtained from the x and y components of the resultant by using the Pythagorean theorem. The directional angle can be obtained using trigonometry.
North B R A
30.0
SOLUTION We find the following results:
Rx = ( 244 km ) cos 30.0 + ( −175 km ) = 36 km B
Ax
x
R y = ( 244 km ) sin 30.0 + ( 0 km ) = 122 km B
Ay R = Rx2 + R 2y =
y
( 36 km )2 + (122 km )2 = 127 km
Ry 122 km = tan −1 = 74 R 36 km x ______________________________________________________________________________
= tan −1
63. SSM REASONING The performer walks out on the wire a distance d, and the vertical distance to the net is h. Since these two distances are perpendicular, the magnitude of the displacement is given by the Pythagorean theorem as s = d 2 + h 2 . Values for s and h are given, so we can solve this expression for the distance d. The angle that the performer’s displacement makes below the horizontal can be found using trigonometry. SOLUTION a. Using the Pythagorean theorem, we find that
38
INTRODUCTION AND MATHEMATICAL CONCEPTS
s = d 2 + h2
or
d = s2 − h2 =
b26.7 ft g− b25.0 ft g = 9.4 ft 2
2
b. The angle that the performer’s displacement makes below the horizontal is given by
tan =
h d
or
= tan −1
hI F F25.0 ft IJ= 69 = tan G G J Hd K H9.4 ft K −1
64. REASONING The force vector F points at an angle of above the +x axis. Therefore, its x and y components are given by Fx = F cos and Fy = F sin . SOLUTION follows:
a. The magnitude of the vector can be obtained from the y component as Fy = F sin
or
F=
Fy sin
=
290 N = 370 N sin 52
b. Now that the magnitude of the vector is known, the x component of the vector can be calculated as follows: Fx = F cos = 370 N cos 52 = +230 N
b g
_____________________________________________________________________________________________
65.
SSM REASONING AND SOLUTION We take due north to be the direction of the +y axis. Vectors A and B are the components of the resultant, C. The angle that C makes with −1 the x axis is then = tan (B / A) . The symbol u denotes the units of the vectors. a. Solving for B gives B = A tan = (6.00 u) tan 60.0° =
10.4 u
b. The magnitude of C is C = A 2 + B 2 = ( 6.00 u) 2 + (10.4 u) 2 = 12 .0 u ______________________________________________________________________________
66. REASONING We are given that the vector sum of the three forces is zero, so F1 + F2 + F3 = 0 N. Since F1 and F2 are known, F3 can be found from the relation F3 = −(F1 + F2). We will use the x- and y-components of this equation to find the magnitude and direction of F3. SOLUTION The x- and y-components of the equation F3 = −( F1 + F2) are:
Chapter 1 Problems
39
x-component
F3x = − ( F1x + F2 x )
(1)
y-component
F3 y = − F1 y + F2 y
(
(2)
)
The table below gives the x- and y-components of F1 and F2: Vector
x component
y component
F1 F2
F1x = −(21.0 N) sin 30.0 = −10.5 N F2x = +15.0 N
F1y = −(21.0 N) cos 30.0 = +18.2 N F2y = 0 N
Substituting the values for F1x and F2x into Equation (1) gives
F3x = − ( F1x + F2 x ) = − ( −10.5 N + 15.2 N ) = −4.5 N Substituting F1y and F2y into Equation (2) gives
(
)
F3 y = − F1 y + F2 y = − ( +18.2 N + 0 N ) = −18.2 N The magnitude of F3 can now be obtained by employing the Pythagorean theorem:
F3 = F32x + F32y =
( −4.5 N )2 + ( −18.2 N )2 = 18.7 N
The angle that F3 makes with respect to the −x axis can be determined from the inverse tangent function (Equation 1.6),
F3 y −1 −18.2 N = tan = 76 F − 4.5 N 3x ______________________________________________________________________________
= tan −1
67. REASONING AND SOLUTION The following figures are scale diagrams of the forces drawn tail-to-head. The scale factor is shown in the figure. a. From the figure on the left, we see that FA − FB = 142 N, = 67 south of east . b. Similarly, from the figure on the right, FB − FA = 142 N, = 67 north of west .
40
INTRODUCTION AND MATHEMATICAL CONCEPTS
–FA = 90.0 N
FA = 90.0 N
75°
N W
E
–FB = 135 N
S
FA – FB
FB = 135 N FB – FA
Scale Factor:
20.0 N
75°
_____________________________________________________________________________________________
68. REASONING There are two right triangles in the drawing. Each H2 contains the common side that is 21 52 shown as a dashed line and is D labeled D, which is the distance between the buildings. The H1 hypotenuse of each triangle is one of the lines of sight to the top and base of the taller building. The remaining (vertical) sides of the triangles are labeled H1 and H2. Since the height of the taller building is H1 + H2 and the height of the shorter building is H1, the ratio that we seek is (H1 + H2)/H1. We will use the tangent function to express H1 in terms of the 52 angle and to express H2 in terms of the 21 angle. The unknown distance D will be eliminated algebraically when the ratio (H1 + H2)/H1 is calculated. SOLUTION The ratio of the building heights is
H + H2 Height of taller building = 1 Height of shorter building H1 Using the tangent function, we have that
Chapter 1 Problems
tan 52 =
H1 D
or
H1 = D tan 52
tan 21 =
H2 D
or
H 2 = D tan 21
41
Substituting these results into the expression for the ratio of the heights gives
H + H 2 D tan 52 + D tan 21 Height of taller building = 1 = Height of shorter building H1 D tan 52 = 1+
tan 21 = 1.30 tan 52
Since 1.30 is less than 1.50, your friend is wrong .
69. REASONING AND SOLUTION If D is the unknown vector, then A + B + C + D = 0 requires that DE = –(AE + BE + CE) or DE = (113 u) cos 60.0° – (222 u) cos 35.0° – (177 u) cos 23.0° =
–288 units
The minus sign indicates that DE has a direction of due west. Also, DN = –(AN + BN + CN) or DN = (113 u) sin 60.0° + (222 u) sin 35.0° – (177 u) sin 23.0° =
156 units
______________________________________________________________________________
CHAPTER 2 KINEMATICS IN
ONE DIMENSION ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. (b) Displacement, being a vector, conveys information about magnitude and direction. Distance conveys no information about direction and, hence, is not a vector. 2. (c) Since each runner starts at the same place and ends at the same place, the three displacement vectors are equal. 3. (c) The average speed is the distance of 16.0 km divided by the elapsed time of 2.0 h. The average velocity is the displacement of 0 km divided by the elapsed time. The displacement is 0 km, because the jogger begins and ends at the same place. 4. (a) Since the bicycle covers the same number of meters per second everywhere on the track, its speed is constant. 5. (e) The average velocity is the displacement (2.0 km due north) divided by the elapsed time (0.50 h), and the direction of the velocity is the same as the direction of the displacement. 6. (c) The average acceleration is the change in velocity (final velocity minus initial velocity) divided by the elapsed time. The change in velocity has a magnitude of 15.0 km/h. Since the change in velocity points due east, the direction of the average acceleration is also due east. 7. (d) This is always the situation when an object at rest begins to move. 8. (b) If neither the magnitude nor the direction of the velocity changes, then the velocity is constant, and the change in velocity is zero. Since the average acceleration is the change in velocity divided by the elapsed time, the average acceleration is also zero. 9. (a) The runners are always moving after the race starts and, therefore, have a non-zero average speed. The average velocity is the displacement divided by the elapsed time, and the displacement is zero, since the race starts and finishes at the same place. The average acceleration is the change in the velocity divided by the elapsed time, and the velocity changes, since the contestants start at rest and finish while running. 10. (c) The equations of kinematics can be used only when the acceleration remains constant and cannot be used when it changes from moment to moment. 11. (a) Velocity, not speed, appears as one of the variables in the equations of kinematics. Velocity is a vector. The magnitude of the instantaneous velocity is the speed.
Chapter 2 Answers to Focus on Concepts Questions
(
43
)
12. (b) According to one of the equation of kinematics v 2 = v02 + 2ax, with v0 = 0 m/s , the displacement is proportional to the square of the velocity.
(
)
13. (d) According to one of the equation of kinematics x = v0t + 12 at 2 , with v0 = 0 m/s , the displacement is proportional to the acceleration. 14. (b) For a single object each equation of kinematics contains four variables, one of which is the unknown variable. 15. (e) An equation of kinematics ( v = v0 + at ) gives the answer directly, since the initial velocity, the final velocity, and the time are known. 16. (c) An equation of kinematics x = 12 ( v0 + v ) t gives the answer directly, since the initial velocity, the final velocity, and the time are known.
(
)
17. (e) An equation of kinematics v 2 = v02 + 2ax gives the answer directly, since the initial velocity, the final velocity, and the acceleration are known. 18. (d) This statement is false. Near the earth’s surface the acceleration due to gravity has the approximate magnitude of 9.80 m/s2 and always points downward, toward the center of the earth. 19. (b) Free-fall is the motion that occurs while the acceleration is solely the acceleration due to gravity. While the rocket is picking up speed in the upward direction, the acceleration is not just due to gravity, but is due to the combined effect of gravity and the engines. In fact, the effect of the engines is greater than the effect of gravity. Only when the engines shut down does the free-fall motion begin.
(
)
20. (c) According to an equation of kinematics v = v0 + 2ax, with v = 0 m/s , the launch 2
2
speed v0 is proportional to the square root of the maximum height. 21. (a) An equation of kinematics ( v = v0 + at ) gives the answer directly. 22. (d) The acceleration due to gravity points downward, in the same direction as the initial velocity of the stone thrown from the top of the cliff. Therefore, this stone picks up speed as it approaches the nest. In contrast, the acceleration due to gravity points opposite to the initial velocity of the stone thrown from the ground, so that this stone loses speed as it approaches the nest. The result is that, on average, the stone thrown from the top of the cliff travels faster than the stone thrown from the ground and hits the nest first. 23. 1.13 s
44
KINEMATICS IN ONE DIMENSION
24. (a) The slope of the line in a position versus time graph gives the velocity of the motion. The slope for part A is positive. For part B the slope is negative. For part C the slope is positive. 25. (b) The slope of the line in a position versus time graph gives the velocity of the motion. Section A has the smallest slope and section B the largest slope. 26. (c) The slope of the line in a position versus time graph gives the velocity of the motion. Here the slope is positive at all times, but it decreases as time increases from left to right in the graph. This means that the positive velocity is decreasing as time increases, which is a condition of deceleration.
Chapter 2 Problems
45
CHAPTER 2 KINEMATICS IN
ONE DIMENSION PROBLEMS ______________________________________________________________________________ 1.
REASONING The distance traveled by the Space Shuttle is equal to its speed multiplied by the time. The number of football fields is equal to this distance divided by the length L of one football field. SOLUTION The number of football fields is
(
)(
−3
)
7.6 10 m / s 110 10 s x vt Number = = = = 9.1 L L 91.4 m ______________________________________________________________________________ 2.
3
REASONING The displacement is a vector that points from an object’s initial position to its final position. If the final position is greater than the initial position, the displacement is positive. On the other hand, if the final position is less than the initial position, the displacement is negative. (a) The final position is greater than the initial position, so the displacement will be positive. (b) The final position is less than the initial position, so the displacement will be negative. (c) The final position is greater than the initial position, so the displacement will be positive. SOLUTION The displacement is defined as Displacement = x – x0, where x is the final position and x0 is the initial position. The displacements for the three cases are: (a) Displacement = 6.0 m − 2.0 m = +4.0 m (b) Displacement = 2.0 m − 6.0 m = −4.0 m (c) Displacement = 7.0 m − (−3.0 m) = +10.0 m
3.
SSM REASONING The average speed is the distance traveled divided by the elapsed time (Equation 2.1). Since the average speed and distance are known, we can use this relation to find the time.
SOLUTION The time it takes for the continents to drift apart by 1500 m is Elapsed time =
Distance 1500 m = = 5 104 yr Average speed cm 1 m 3 yr 100 cm
46
KINEMATICS IN ONE DIMENSION
______________________________________________________________________________ 4.
REASONING Since the average speed of the impulse is equal to the distance it travels divided by the elapsed time (see Equation 2.1), the elapsed time is just the distance divided by the average speed. SOLUTION The time it takes for the impulse to travel from the foot to the brain is
Distance 1.8 m (2.1) = = 1.6 − s Average speed 1.1 m/s ______________________________________________________________________________ Time =
5.
x − x0 REASONING According to Equation 2.2 v = , the average velocity ( v ) is equal t − t 0 to the displacement ( x − x0 ) divided by the elapsed time ( t − t0 ) , and the direction of the average velocity is the same as that of the displacement. The displacement is equal to the difference between the final and initial positions. SOLUTION Equation 2.2 gives the average velocity as
v=
x − x0 t − t0
Therefore, the average velocities for the three cases are: (a) Average velocity = (6.0 m − 2.0 m)/(0.50 s) = +8.0 m/s (b) Average velocity = (2.0 m − 6.0 m)/(0.50 s) = −8.0 m/s (c) Average velocity = [7.0 m − (−3.0 m)]/(0.50 s) = +2.0 101 m/s The algebraic sign of the answer conveys the direction in each case. ______________________________________________________________________________ 6.
REASONING Distance and displacement are different physical quantities. Distance is a scalar, and displacement is a vector. Distance and the magnitude of the displacement, however, are both measured in units of length. SOLUTION a. The distance traveled is equal to three-fourths of the circumference of the circular lake. The circumference of a circle is 2πr, where r is the radius of the circle. Thus, the distance d that the couple travels is d=
3 4
( 2 r ) = 34 2 (1.50 km ) = 7.07 km
Chapter 2 Problems
47
b. The couple’s displacement is the hypotenuse of a right triangle with sides equal to the radius of the circle (see the drawing). The magnitude R of the displacement can be obtained with the aid of the Pythagorean theorem:
R = r 2 + r 2 = 2 (1.50 km ) = 2.12 km 2
The angle that the displacement makes with due east is r = tan −1 = tan −1 (1) = 45.0 north of east r ______________________________________________________________________________ 7.
REASONING AND SOLUTION In 12 minutes the sloth travels a distance of 60 s xs = vst = (0.037 m/s)(12 min) = 27 m 1 min
while the tortoise travels a distance of 60 s xt = vt t = (0.076 m/s)(12 min) = 55 m 1 min
The tortoise goes farther than the sloth by an amount that equals 55 m – 27 m = 28 m ______________________________________________________________________________ 8.
REASONING The younger (and faster) runner should start the race after the older runner, the delay being the difference between the time required for the older runner to complete the race and that for the younger runner. The time for each runner to complete the race is equal to the distance of the race divided by the average speed of that runner (see Equation 2.1). SOLUTION The difference between the times for the two runners to complete the race is t50 − t18 , where
t50 =
Distance ( Average Speed )50-yr-old
and t18 =
Distance ( Average Speed )18-yr-old
(2.1)
The difference between these two times (which is how much later the younger runner should start) is
48
KINEMATICS IN ONE DIMENSION
t50 − t18 =
Distance Distance − ( Average Speed )50-yr-old ( Average Speed )18-yr-old
10.0 103 m 10.0 103 m − = 64 s 4.27 m/s 4.39 m/s ______________________________________________________________________________ =
9.
REASONING In order for the bear to catch the tourist over the distance d, the bear must reach the car at the same time as the tourist. During the time t that it takes for the tourist to reach the car, the bear must travel a total distance of d + 26 m. From Equation 2.1, vtourist =
d t
(1)
and
vbear =
d + 26 m t
(2)
Equations (1) and (2) can be solved simultaneously to find d. SOLUTION Solving Equation (1) for t and substituting into Equation (2), we find
vbear =
d + 26 m (d + 26 m)vtourist = d / vtourist d
26 m vbear = 1 + v d tourist
Solving for d yields:
26 m
d=
=
26 m = 52 m 6.0 m/s −1 4.0 m/s
vbear −1 vtourist ______________________________________________________________________________ 10. REASONING AND SOLUTION Let west be the positive direction. The average velocity of the backpacker is
x +x e v= w t +t w e
where
t
x = w w v w
and
x t = e e v e
Combining these equations and solving for xe (suppressing the units) gives xe =
– (1 – v / vw ) xw
(1 – v / ve )
=
– 1 – (1.34 m/s ) / ( 2.68 m/s ) ( 6.44 km ) 1 – (1.34 m/s ) / ( 0.447 m/s )
= –0.81 km
Chapter 2 Problems
49
The distance traveled is the magnitude of xe, or 0.81 km . ______________________________________________________________________________ 11.
SSM REASONING AND SOLUTION a. The total displacement traveled by the bicyclist for the entire trip is equal to the sum of the displacements traveled during each part of the trip. The displacement traveled during each part of the trip is given by Equation 2.2: x = v t . Therefore,
60 s x1 = (7.2 m/s)(22 min) = 9500 m 1 min 60 s x2 = (5.1 m/s)(36 min) = 11 000 m 1 min 60 s x3 = (13 m/s)(8.0 min) = 6200 m 1 min The total displacement traveled by the bicyclist during the entire trip is then x = 9500 m + 11 000 m + 6200 m =
2.67 104 m
b. The average velocity can be found from Equation 2.2.
x 2.67 104 m 1min = = 6.74 m/s, due north ( ) t 22 min + 36 min + 8.0 min 60 s ______________________________________________________________________________ v=
12. REASONING The definition of average velocity is given by Equation 2.2 as Average velocity = Displacement/(Elapsed time). The displacement in this expression is the total displacement, which is the sum of the displacements for each part of the trip. Displacement is a vector quantity, and we must be careful to account for the fact that the displacement in the first part of the trip is north, while the displacement in the second part is south. SOLUTION According to Equation 2.2, the displacement for each part of the trip is the average velocity for that part times the corresponding elapsed time. Designating north as the positive direction, we find for the total displacement that Displacement =
( 27 m/s ) tNorth + ( −17 m/s ) tSouth Northward
Southward
where tNorth and tSouth denote, respectively, the times for each part of the trip. Note that the minus sign indicates a direction due south. Noting that the total elapsed time is
50
KINEMATICS IN ONE DIMENSION
tNorth + tSouth, we can use Equation 2.2 to find the average velocity for the entire trip as follows:
Average velocity =
Displacement ( 27 m/s ) t North + ( −17 m/s ) tSouth = Elapsed time t North + tSouth
t North tSouth = ( 27 m/s ) + ( −17 m/s ) t t North + tSouth North + tSouth 3 1 t North tSouth But = and = . Therefore, we have that t 4 t + t + t North South North South 4
3 1 Average velocity = ( 27 m/s ) + ( −17 m/s ) = +16 m/s 4 4
The plus sign indicates that the average velocity for the entire trip points north. 13. REASONING AND SOLUTION The upper edge of the wall will disappear after the train has traveled the distance d in the figure below.
0.90 m
B
B A
A
12°
b
2.0 m
12°
d The distance d is equal to the length of the window plus the base of the 12° right triangle of height 0.90 m. The base of the triangle is given by b =
0.90 m
0.90 m = 4.2 m tan12
12°
b
Thus, d = 2.0 m + 4.2 m = 6.2 m. The time required for the train to travel 6.2 m is, from the definition of average speed, t=
x 6.2 m = = 2.1 s v 3.0 m/s
Chapter 2 Problems
51
______________________________________________________________________________ 14. REASONING AND SOLUTION Since v = v0 + at, the acceleration is given by a = ( v – v0 ) / t . Since the direction of travel is in the negative direction throughout the
problem, all velocities will be negative. a.
a=
( − 29.0 m/s) − ( − 27.0 m/s) = −0.40 m/s 2 5.0 s
Since the acceleration is negative, it is in the same direction as the velocity and the car is speeding up. b.
a=
( − 23.0 m/s) − ( − 27.0 m/s) = +0.80 m/s 2 5.0 s
Since the acceleration is positive, it is in the opposite direction to the velocity and the car is slowing down or decelerating. ______________________________________________________________________________ v − v0 15. REASONING The average acceleration ( a ) is defined by Equation 2.4 a = as t − t0 the change in velocity ( v − v0 ) divided by the elapsed time ( t − t0 ) . The change in velocity
is equal to the final velocity minus the initial velocity. Therefore, the change in velocity, and hence the acceleration, is positive if the final velocity is greater than the initial velocity. The acceleration is negative if the final velocity is less than the initial velocity. The acceleration is zero if the final and initial velocities are the same. SOLUTION Equation 2.4 gives the average acceleration as
a=
v − v0 t − t0
a. The initial and final velocities are both +82 m/s, since the velocity is constant. The average acceleration is a = (82 m/s − 82 m/s)/(t – t0) = 0 m/s 2
b. The initial velocity is +82 m/s, and the final velocity is –82 m/s. The average acceleration is a = (−82 m/s − 82 m/s)/(12 s) = −14 m/s2 ______________________________________________________________________________
52
KINEMATICS IN ONE DIMENSION
16. REASONING Although the planet follows a curved, two-dimensional path through space, this causes no difficulty here because the initial and final velocities for this period are in opposite directions. Thus, the problem is effectively a problem in one dimension only. v Equation 2.4 a = relates the change v in the planet’s velocity to its average t acceleration and the elapsed time t = 2.16 years. It will be convenient to convert the elapsed time to seconds before calculating the average acceleration. SOLUTION a. The net change in the planet’s velocity is the final minus the initial velocity: v = v − v 0 = −18.5 km/s − 20.9 km/s = −39.4 km/s km 1000 m 4 v = −39.4 = −3.94 10 m/s s 1 km
b. Although the planet’s velocity changes by a large amount, the change occurs over a long time interval, so the average acceleration is likely to be small. Expressed in seconds, the interval is 365 d 24 h 60 min 60 s t = 2.16 yr = 6.81107 s 1 yr 1d 1h 1min
(
)
Then the average acceleration is a=
v −3.94 104 m/s = = −5.79 10−4 m/s 2 7 t 6.8110 s
17. REASONING Since the velocity and acceleration of the motorcycle point in the same direction, their numerical values will have the same algebraic sign. For convenience, we will choose them to be positive. The velocity, acceleration, and the time are related by Equation 2.4: v = v0 + at . SOLUTION a. Solving Equation 2.4 for t we have
t=
v − v0
t=
v − v0
b. Similarly,
a
a
=
(+31 m/s) – (+21 m/s) = 4.0 s +2.5 m/s 2
=
(+61 m/s) – (+51 m/s) = 4.0 s +2.5 m/s 2
Chapter 2 Problems
53
______________________________________________________________________________ 18. REASONING
We can use the definition of average acceleration a = ( v − v 0 ) / ( t − t0 )
(Equation 2.4) to find the sprinter’s final velocity v at the end of the acceleration phase, because her initial velocity ( v0 = 0 m/s , since she starts from rest), her average acceleration
a , and the time interval t − t0 are known. SOLUTION a. Since the sprinter has a constant acceleration, it is also equal to her average acceleration, so a = +2.3 m/s 2 Her velocity at the end of the 1.2-s period is
v = v0 + a ( t − t0 ) = ( 0 m/s ) + ( +2.3 m/s 2 ) (1.2 s ) = +2.8 m/s b. Since her acceleration is zero during the remainder of the race, her velocity remains constant at +2.8 m/s . ______________________________________________________________________________ 19. REASONING When the velocity and acceleration vectors are in the same direction, the speed of the object increases in time. When the velocity and acceleration vectors are in opposite directions, the speed of the object decreases in time. (a) The initial velocity and acceleration are in the same direction, so the speed is increasing. (b) The initial velocity and acceleration are in opposite directions, so the speed is decreasing. (c) The initial velocity and acceleration are in opposite directions, so the speed is decreasing. (d) The initial velocity and acceleration are in the same direction, so the speed is increasing. SOLUTION The final velocity v is related to the initial velocity v0, the acceleration a, and the elapsed time t through Equation 2.4 (v = v0 + a t). The final velocities and speeds for the four moving objects are: a. v = 12 m/s + (3.0 m/s2)(2.0 s) = 18 m/s. The final speed is 18 m/s . b. v = 12 m/s + (−3.0 m/s2)(2.0 s) = 6.0 m/s. The final speed is 6.0 m/s . c. v = −12 m/s + (3.0 m/s2)(2.0 s) = −6.0 m/s. The final speed is 6.0 m/s . d. v = −12 m/s + (−3.0 m/s2)(2.0 s) = −18 m/s. The final speed is 18 m/s . ______________________________________________________________________________ 20. REASONING The fact that the emu is slowing down tells us that the acceleration and the velocity have opposite directions. Furthermore, since the acceleration remains the same in both parts of the motion, we can determine its value from the first part of the motion and then use it in the second part to determine the bird’s final velocity at the end of the total 6.0-s time interval. SOLUTION
54
KINEMATICS IN ONE DIMENSION
a. The initial velocity of the emu is directed due north. Since the bird is slowing down, its acceleration must point in the opposite direction, or due south . b. We assume that due north is the positive direction. With the data given for the first part of the motion, Equation 2.4 shows that the average acceleration is a=
v − v0 t − t0
=
(10.6 m/s ) − (13.0 m/s ) = −0.60 m/s2 4.0 s − 0 s
The negative value for the acceleration indicates that it indeed points due south, which is the negative direction. Solving Equation 2.4 for the final velocity gives
(
)
v = v 0 + a ( t − t0 ) = +10.6 m/s + −0.60 m/s 2 ( 2.0 s − 0 s ) = +9.4 m/s Since this answer is positive, the bird’s velocity after an additional 2.0 s is in the positive direction and is 9.4 m/s, due north .
21. REASONING AND SOLUTION from Equation 2.4 (v = v0 + at) as
v − v0
The magnitude of the car's acceleration can be found
26.8 m/s – 0 m/s = 8.18 m/s2 t 3.275 s ______________________________________________________________________________ a=
=
22. REASONING According to Equation 2.4, the average acceleration of the car for the first twelve seconds after the engine cuts out is
v −v a1 = 1f 10 t1
(1)
and the average acceleration of the car during the next six seconds is
a2 =
v2f − v20 t2
=
v2f − v1f t2
(2)
The velocity v1f of the car at the end of the initial twelve-second interval can be found by solving Equations (1) and (2) simultaneously.
Chapter 2 Problems
55
SOLUTION Dividing Equation (1) by Equation (2), we have
a1 (v1f − v10 ) / t1 (v1f − v10 )t2 = = a2 (v2f − v1f ) / t2 (v2f − v1f )t1 Solving for v1f , we obtain
v1f =
v1f =
a1t1v2f + a2 t2v10 a1t1 + a2 t2
=
(a1 / a2 )t1v2f + t2v10 (a1 / a2 )t1 + t2
1.50(12.0 s)(+28.0 m/s) + (6.0 s)( + 36.0 m/s) = +30.0 m/s 1.50(12.0 s) + 6.0 s
23. REASONING AND SOLUTION Both motorcycles have the same velocity v at the end of the four second interval. Now v = v0A + aAt for motorcycle A and v = v0B + aBt for motorcycle B. Subtraction of these equations and rearrangement gives v0A – v0B = (4.0 m/s2 – 2.0 m/s2)(4 s) =
+8.0 m/s
The positive result indicates that motorcycle A was initially traveling faster. ______________________________________________________________________________ 24. REASONING AND SOLUTION a = v / t , so
The average acceleration of the basketball player is
2 6.0 m/s x = 12 at 2 = 12 (1.5 s ) = 4.5 m 1.5 s ______________________________________________________________________________
25. SSM REASONING AND SOLUTION a. The magnitude of the acceleration can be found from Equation 2.4 (v = v0 + at) as
a=
v − v0 t
=
3.0 m/s – 0 m/s = 1.5 m/s 2 2.0 s
b. Similarly the magnitude of the acceleration of the car is
56
KINEMATICS IN ONE DIMENSION
a=
v − v0 t
=
41.0 m/s – 38.0 m/s = 1.5 m/s2 2.0 s
c. Assuming that the acceleration is constant, the displacement covered by the car can be found from Equation 2.9 (v2 = v02 + 2ax):
x=
v 2 − v02 2a
=
(41.0 m/s)2 − (38.0 m/s)2 = 79 m 2(1.5 m/s2 )
Similarly, the displacement traveled by the jogger is
x=
v 2 − v02 2a
=
(3.0 m/s)2 − (0 m/s) 2 = 3.0 m 2(1.5 m/s2 )
Therefore, the car travels 79 m – 3.0 m = 76 m further than the jogger. ______________________________________________________________________________ 26. REASONING The average acceleration is defined by Equation 2.4 as the change in velocity divided by the elapsed time. We can find the elapsed time from this relation because the acceleration and the change in velocity are given. SOLUTION a. The time t that it takes for the VW Beetle to change its velocity by an amount v = v – v0 is (and noting that 0.4470 m/s = 1 mi/h)
t =
v − v0 a
=
0.4470 m / s −0 m/s 1 mi / h
( 60.0 mi / h )
2.35 m / s 2
= 11.4 s
b. From Equation 2.4, the acceleration (in m/s2) of the dragster is 0.4470 m / s −0 m/s v − v0 1 mi / h a= = = 44.7 m / s 2 t − t0 0.600 s − 0 s ______________________________________________________________________________
( 60.0 mi / h )
27. REASONING We know the initial and final velocities of the blood, as well as its displacement. Therefore, Equation 2.9 v 2 = v02 + 2ax can be used to find the acceleration
(
)
Chapter 2 Problems
57
of the blood. The time it takes for the blood to reach it final velocity can be found by using x Equation 2.7 t = 1 . v + v ( ) 0 2 SOLUTION a. The acceleration of the blood is
a=
v 2 − v02 2x
2 2 26 cm / s ) − ( 0 cm / s ) ( = = 1.7 102 cm / s 2 2 ( 2.0 cm )
b. The time it takes for the blood, starting from 0 cm/s, to reach a final velocity of +26 cm/s is x 2.0 cm t= 1 =1 = 0.15 s v + v ) 2 ( 0 cm / s + 26 cm / s ) 2( 0 ______________________________________________________________________________ 28. REASONING AND SOLUTION a. From Equation 2.4, the definition of average acceleration, the magnitude of the average acceleration of the skier is
a=
v − v0 t − t0
=
8.0 m/s – 0 m/s = 1.6 m/s 2 5.0 s
b. With x representing the displacement traveled along the slope, Equation 2.7 gives: x = 12 (v0 + v)t = 12 (8.0 m/s + 0 m/s)(5.0 s) = 2.0 101 m
______________________________________________________________________________ 29.
SSM REASONING AND SOLUTION
(
The average acceleration of the plane can be
)
found by solving Equation 2.9 v 2 = v02 + 2ax for a. Taking the direction of motion as positive, we have a=
v 2 − v02 2x
=
(+6.1 m/s) 2 − (+69 m/s) 2 = – 3.1 m/s 2 2(+750 m)
The minus sign indicates that the direction of the acceleration is opposite to the direction of motion, and the plane is slowing down. ______________________________________________________________________________ 30. REASONING At a constant velocity the time required for Secretariat to run the final mile is given by Equation 2.2 as the displacement (+1609 m) divided by the velocity. The actual time required for Secretariat to run the final mile can be determined from Equation 2.8,
58
KINEMATICS IN ONE DIMENSION
since the initial velocity, the acceleration, and the displacement are given. difference between these two results for the time that we seek.
It is the
SOLUTION According to Equation 2.2, with the assumption that the initial time is t0 = 0 s, the run time at a constant velocity is t = t − t0 = t =
(
x +1609 m = = 97.04 s v +16.58 m/s
)
Solving Equation 2.8 x = v0t + 12 at 2 for the time shows that
t=
=
( ) 2 ( 12 a )
−v0 v02 − 4 12 a ( − x )
−16.58 m/s
( +16.58 m/s )2 − 4 ( 12 ) ( +0.0105 m/s2 ) ( −1609 m )
( )(
2 12 +0.0105 m/s 2
)
= 94.2 s
We have ignored the negative root as being unphysical. The acceleration allowed Secretariat to run the last mile in a time that was faster by 97.04 s − 94.2 s = 2.8 s
31.
SSM REASONING The cart has an initial velocity of v0 = +5.0 m/s, so initially it is moving to the right, which is the positive direction. It eventually reaches a point where the displacement is x = +12.5 m, and it begins to move to the left. This must mean that the cart comes to a momentary halt at this point (final velocity is v = 0 m/s), before beginning to move to the left. In other words, the cart is decelerating, and its acceleration must point opposite to the velocity, or to the left. Thus, the acceleration is negative. Since the initial
(
velocity, the final velocity, and the displacement are known, Equation 2.9 v 2 = v02 + 2ax
)
can be used to determine the acceleration. SOLUTION Solving Equation 2.9 for the acceleration a shows that
a=
v 2 − v02 2x
2 2 0 m/s ) − ( +5.0 m/s ) ( = = −1.0 m/s 2 2 ( +12.5 m )
32. REASONING At time t both rockets return to their starting points and have a displacement of zero. This occurs, because each rocket is decelerating during the first half of its journey.
Chapter 2 Problems
59
However, rocket A has a smaller initial velocity than rocket B. Therefore, in order for rocket B to decelerate and return to its point of origin in the same time as rocket A, rocket B must have a deceleration with a greater magnitude than that for rocket A. Since we know that the displacement of each rocket is zero at time t, since both initial velocities are given, and since we seek information about the acceleration, we begin our solution with Equation 2.8, for it contains just these variables. SOLUTION Applying Equation 2.8 to each rocket gives
xA = v0At + 12 aAt 2
xB = v0Bt + 12 aBt 2
0 = v0At + 12 aAt 2
0 = v0Bt + 12 aBt 2
0 = v0A + 12 aAt
0 = v0B + 12 aBt
t=
−2v0A aA
t=
−2v0B aB
The time for each rocket is the same, so that we can equate the two expressions for t, with the result that −2v0A −2v0B v0A v0B = or = aA aB aA aB Solving for aB gives
aB =
aA −15 m/s 2 v0B = (8600 m/s ) = −22 m/s2 v0A 5800 m/s
As expected, the magnitude of the acceleration for rocket B is greater than that for rocket A.
33. REASONING The stopping distance is the sum of two parts. First, there is the distance the car travels at 20.0 m/s before the brakes are applied. According to Equation 2.2, this distance is the magnitude of the displacement and is the magnitude of the velocity times the time. Second, there is the distance the car travels while it decelerates as the brakes are applied. This distance is given by Equation 2.9, since the initial velocity, the acceleration, and the final velocity (0 m/s when the car comes to a stop) are given. SOLUTION With the assumption that the initial position of the car is x0 = 0 m, Equation 2.2 gives the first contribution to the stopping distance as
x1 = x1 = vt1 = ( 20.0 m/s )( 0.530 s )
60
KINEMATICS IN ONE DIMENSION
(
Solving Equation 2.9 v 2 = v02 + 2ax
) for x shows that the second part of the stopping
distance is x2 =
v 2 − v02 2a
2 2 0 m/s ) − ( 20.0 m/s ) ( =
(
2 −7.00 m/s2
)
Here, the acceleration is assigned a negative value, because we have assumed that the car is traveling in the positive direction, and it is decelerating. Since it is decelerating, its acceleration points opposite to its velocity. The stopping distance, then, is 2 2 0 m/s ) − ( 20.0 m/s ) ( xStopping = x1 + x2 = ( 20.0 m/s )( 0.530 s ) + = 39.2 m 2
(
2 −7.00 m/s
)
34. REASONING The entering car maintains a constant acceleration of a1 = 6.0 m/s2 from the time it starts from rest in the pit area until it catches the other car, but it is convenient to separate its motion into two intervals. During the first interval, lasting t1 = 4.0 s, it accelerates from rest to the velocity v01 with which it enters the main speedway. This velocity is found from Equation 2.4 ( v = v0 + at ) , with v0 = 0 m/s, a = a1, t = t1, and v = v01: v10 = a1t1
(1)
The second interval begins when the entering car enters the main speedway with velocity v01, and ends when it catches up with the other car, which travels with a constant velocity v02 = 70.0 m/s. Since both cars begin and end the interval side-by-side, they both undergo the same displacement x during this interval. The displacement of each car is given by Equation 2.8 x = v0t + 12 at 2 . For the accelerating car, v0 = v10, and a = a1, so
(
)
x = v10t + 12 a1t 2
(2)
For the other car, v0 = v02 and a = 0 m/s2, and so Equation 2.8 yields x = v20t
(3)
SOLUTION The displacement during the second interval is not required, so equating the right hand sides of Equations (2) and (3) eliminates x, leaving an equation that may be solved for the elapsed time t, which is now the only unknown quantity:
Chapter 2 Problems
61
v10t + 12 a1t 2 = v20t v10 t + 12 a1t 2 = v20 t (4)
1 a t = v −v 20 10 2 1
t=
2 ( v20 − v10 ) a1
Substituting Equation (1) for v10 into Equation (4), we find that
t=
2 ( v20 − a1t1 ) a1
(
)
2 70.0 m/s − 6.0 m/s 2 ( 4.0 s ) = 15 s = 2 6.0 m/s
(
)
35. REASONING The drawing shows the two knights, initially separated by the displacement d, traveling toward each other. At any moment, Sir George’s displacement is xG and that of Sir Alfred is xA. When they meet, their displacements are the same, so xG = xA. Starting point for Sir George
xG
−
+
Starting point for Sir Alfred
xA d
According to Equation 2.8, Sir George's displacement as a function of time is xG = v0,Gt + 12 aGt 2 = ( 0 m/s ) t + 12 aGt 2 = 12 aGt 2
(1)
where we have used the fact that Sir George starts from rest ( v0,G = 0 m/s ) . Since Sir Alfred starts from rest at x = d at t = 0 s, we can write his displacement as (again, employing Equation 2.8) xA = d + v0,At + 12 aAt 2 = d + ( 0 m/s ) t + 12 aAt 2 = d + 12 aAt 2
(2)
62
KINEMATICS IN ONE DIMENSION
(
Solving Equation 1 for t2 t 2 = 2 xG / aG
) and substituting this expression into Equation 2
yields
2x x xA = d + 12 aA G = d + aA G aG aG
(3)
Noting that xA = xG when the two riders collide, we see that Equation 3 becomes
x xG = d + aA G aG Solving this equation for xG gives xG =
d . aA 1− aG
(
)
SOLUTION Sir George’s acceleration is positive aG = +0.300 m/s2 since he starts from rest and moves to the right (the positive direction). Sir Alfred’s acceleration is negative
( aA = −0.200 m/s2 ) since he starts from rest and moves to the left (the negative direction).
The displacement of Sir George is, then, xG =
d 88.0 m = = 52.8 m 2) aA ( − 0.200 m/s 1− 1− aG ( +0.300 m/s2 )
____________________________________________________________________________________________
36. REASONING The players collide when they have the same x coordinate relative to a common origin. For convenience, we will place the origin at the starting point of the first player. From Equation 2.8, the x coordinate of each player is given by
x1 = v01t1 + a1t12 = a1t12
(1)
1 2
(2)
1 2
1 2
1 2
x2 = d + v02t2 + a2t22 = d + a2t22
where d = +48 m is the initial position of the second player. When x1 = x2 , the players collide at time t = t1 = t2 . SOLUTION a. Equating Equations (1) and (2) when t = t1 = t2 , we have 1 1 2 a t = d + a t2 1 2 2 2
Chapter 2 Problems
63
We note that a1 = +0.50 m/s2, while a2 = −0.30 m/s2, since the first player accelerates in the +x direction and the second player in the –x direction. Solving for t, we have
t=
2d = a1 − a2
2 ( 48 m )
( 0.50 m/s2 ) − ( −0.30 m/s2 )
= 11 s
b. From Equation (1), 1 2
x1 = a1t12 =
1 2
( 0.50 m/s2 ) (11 s )2 = 3.0 101 m
______________________________________________________________________________ 37. REASONING At a constant velocity the time required for the first car to travel to the next exit is given by Equation 2.2 as the magnitude of the displacement (2.5 103 m) divided by the magnitude of the velocity. This is also the travel time for the second car to reach the next exit. The acceleration for the second car can be determined from Equation 2.8, since the initial velocity, the displacement, and the time are known. This equation applies, because the acceleration is constant. SOLUTION According to Equation 2.2, with the assumption that the initial time is t0 = 0 s, the time for the first car to reach the next exit at a constant velocity is x 2.5 103 m t = t − t0 = t = = = 76 s v 33 m/s
Remembering that the initial velocity v0 of the second car is zero, we can solve Equation 2.8
( x = v0t + 12 at 2 = 12 at 2 ) for the acceleration to show that a=
2x 2(2.5 103 m) = = 0.87 m/s 2 2 2 t (76 s)
Since the second car’s speed in the same direction as the velocity .
is
increasing,
this
acceleration
must
be
38. REASONING Let the total distance between the first and third sign be equal to 2d. Then, the time tA is given by
tA =
d ( v35 + v55 ) d d + = v55 v35 v55v35
(1)
64
KINEMATICS IN ONE DIMENSION
Equation 2.7 x = 12 ( v0 + v ) t can be written as t = 2 x / ( v0 + v ) , so that
tB =
2d ( v35 + v25 ) + ( v55 + v35 ) 2d 2d + = v55 + v35 v35 + v25 ( v55 + v35 )( v35 + v25 )
(2)
SOLUTION Dividing Equation (2) by Equation (1) and suppressing units for convenience, we obtain
tB 2v55v35 ( v35 + v25 ) + ( v55 + v35 ) 2 ( 55)( 35) ( 35 + 25 ) + ( 55 + 35 ) = = = 1.2 2 2 tA 55 + 35 35 + 25 ( ) ( ) v + v v + v ( )( ) 55
35
35
25
______________________________________________________________________________ 39. REASONING Because the car is traveling in the +x direction and decelerating, its acceleration is negative: a = −2.70 m/s2. The final velocity for the interval is given (v = +4.50 m/s), as well as the elapsed time (t = 3.00 s). Both the car’s displacement x and its initial velocity v0 at the instant braking begins are unknown. Compare the list of known kinematic quantities (v, a, t) to the equations of kinematics for constant acceleration: v = v0 + at (Equation 2.4), x = 12 ( v0 + v ) t (Equation 2.7), x = v0t + 12 at 2 (Equation 2.8), and v 2 = v02 + 2ax (Equation 2.9). None of these four
equations contains all three known quantities and the desired displacement x, and each of them contains the initial velocity v0. Since the initial velocity is neither known nor requested, we can combine two kinematic equations to eliminate it, leaving an equation in which x is the only unknown quantity. SOLUTION For the first step, solve Equation 2.4 ( v = v0 + at ) for v0: v0 = v − at
(1)
(
Substituting the expression for v0 in Equation (1) into Equation 2.8 x = v0t + 12 at 2
) yields
an expression for the car’s displacement solely in terms of the known quantities v, a, and t: x = ( v − at ) t + 12 at 2 = vt − at 2 + 12 at 2 x = vt − 12 at 2
Substitute the known values of v, a, and t into Equation (2):
(2)
Chapter 2 Problems
(
)
65
x = ( +4.50 m/s )( 3.00 s ) − 12 −2.70 m/s 2 ( 3.00 s ) = +25.7 m 2
Note: Equation (2) can also be obtained by combining Equation (1) with Equation 2.7 x = 1 ( v0 + v ) t , or, with more effort, by combining Equation (1) with Equation 2.9 2
( v2 = v02 + 2ax ) .
40. REASONING AND SOLUTION As the plane decelerates through the intersection, it covers a total distance equal to the length of the plane plus the width of the intersection, so x = 59.7 m + 25.0 m = 84.7 m The speed of the plane as it enters the intersection can be found from Equation 2.9. Solving Equation 2.9 for v0 gives
v0 = v 2 − 2ax = (45.0 m) 2 − 2(−5.70 m/s 2 )(84.7 m) = 54.7 m/s The time required to traverse the intersection can then be found from Equation 2.4. Solving Equation 2.4 for t gives
v − v0
45.0 m/s − 54.7 m/s = 1.7 s a −5.70 m/s 2 ______________________________________________________________________________ t=
41.
SSM REASONING
=
As the train passes through the crossing, its motion is described by
1 Equations 2.4 (v = v0 + at) and 2.7 x = (v + v0 )t , which can be rearranged to give 2
v − v0 = at
and
v + v0 =
2x t
These can be solved simultaneously to obtain the speed v when the train reaches the end of the crossing. Once v is known, Equation 2.4 can be used to find the time required for the train to reach a speed of 32 m/s. SOLUTION Adding the above equations and solving for v, we obtain 2x 2(20.0 m) v = 12 at + = 12 (1.6 m/s 2 )(2.4 s) + = 1.0 101 m/s t 2.4 s
66
KINEMATICS IN ONE DIMENSION
The motion from the end of the crossing until the locomotive reaches a speed of 32 m/s requires a time v − v0 32 m/s − 1.0 101 m/s t= = = 14 s a 1.6 m/s2 ______________________________________________________________________________ 42. REASONING Since the car is moving with a constant velocity, the displacement of the car in a time t can be found from Equation 2.8 with a = 0 m/s2 and v0 equal to the velocity of the car: xcar = vcar t . Since the train starts from rest with a constant acceleration, the displacement of the train in a time t is given by Equation 2.8 with v0 = 0 m/s: xtrain = 12 atrain t 2
At a time t1 , when the car just reaches the front of the train, xcar = Ltrain + xtrain , where Ltrain is the length of the train. Thus, at time t1 ,
vcar t1 = Ltrain + 12 atrain t12
(1)
At a time t2, when the car is again at the rear of the train, xcar = xtrain . Thus, at time t2 vcar t2 = 12 atrain t22
(2)
Equations (1) and (2) can be solved simultaneously for the speed of the car vcar and the acceleration of the train atrain . SOLUTION a. Solving Equation (2) for atrain we have atrain =
2vcar t2
Substituting this expression for atrain into Equation (1) and solving for vcar , we have vcar =
Ltrain t t1 1 − 1 t2
=
92 m = 13 m/s 14 s (14 s) 1 − 28 s
b. Direct substitution into Equation (3) gives the acceleration of the train:
(3)
Chapter 2 Problems
atrain =
2vcar
=
67
2 (13 m/s) = 0.93 m/s 2 28 s
t2 ______________________________________________________________________________
43.
SSM REASONING AND SOLUTION When air resistance is neglected, free fall conditions are applicable. The final speed can be found from Equation 2.9;
v 2 = v02 + 2ay
where v0 is zero since the stunt man falls from rest. If the origin is chosen at the top of the hotel and the upward direction is positive, then the displacement is y = –99.4 m. Solving for v, we have
v = − 2ay = – 2(–9.80 m/s2 )(–99.4 m) = – 44.1 m/s The speed at impact is the magnitude of this result or 44.1 m/s . ______________________________________________________________________________ 44. REASONING Because there is no effect due to air resistance, the rock is in free fall from its launch until it hits the ground, so that the acceleration of the rock is always −9.8 m/s2, assuming upward to be the positive direction. In (a), we will consider the interval beginning at launch and ending 2.0 s later. In (b), we will consider the interval beginning at launch and ending 5.0 s later. Since the displacement isn’t required, Equation 2.4 ( v = v0 + at ) suffices to solve both parts of the problem. The stone slows down as it rises, so we expect the speed in (a) to be larger than 15 m/s. The speed in (b) could be smaller than 15 m/s (the rock does not reach its maximum height) or larger than 15 m/s (the rock reaches its maximum height and falls back down below its height at the 2.0-s point). SOLUTION a. For the interval from launch to t = 2.0 s, the final velocity is v = 15 m/s, the acceleration is a = −9.8 m/s2, and the initial velocity is to be found. Solving Equation 2.4 ( v = v0 + at ) for v0 gives v0 = v − at = 15 m/s − (−9.8 m/s2 )(2.0 s) = 35 m/s
Therefore, at launch, Speed = 35 m/s
b. Now we consider the interval from launch to t = 5.0 s. The initial velocity is that found in part (a), v0 = 35 m/s. The final velocity is
68
KINEMATICS IN ONE DIMENSION
v = v0 + at = 35 m/s + (−9.8 m/s2 )(5.0 s) = −14 m/s
(2.4)
Instantaneous speed is the magnitude of the instantaneous velocity, so we drop the minus sign and find that Speed = 14 m/s
45. REASONING AND SOLUTION In a time t the card will undergo a vertical displacement y given by y = 12 at2 where a = –9.80 m/s2. When t = 60.0 ms = 6.0 10–2 s, the displacement of the card is 0.018 m, and the distance is the magnitude of this value or d1 = 0.018 m . Similarly, when t = 120 ms, d2 = 0.071 m , and when t = 180 ms, d3 = 0.16 m . ______________________________________________________________________________ 46. REASONING Assuming that air resistance can be neglected, the acceleration is the same for both the upward and downward parts, namely −9.80 m/s2 (upward is the positive direction). Moreover, the displacement is y = 0 m, since the final and initial positions of the ball are the same. The time is given as t = 8.0 s. Therefore, we may use Equation 2.8 2 1 y = v0 t + 2 a t to find the initial velocity v0 of the ball.
(
)
(
)
SOLUTION Solving Equation 2.8 y = v0 t + 12 a t 2 for the initial velocity v0 gives
v0 =
y − 12 a t 2
(
)
0 m − 12 −9.80 m/s 2 (8.0 s )
2
= = +39 m/s t 8.0 s ______________________________________________________________________________ 47. REASONING AND SOLUTION The figure at the right shows the paths taken by the pellets fired from gun A and gun B. The two paths differ by the extra distance covered by the pellet from gun A as it rises and falls back to the edge of the cliff. When it falls back to the edge of the cliff, the pellet from gun A will have the same speed as the pellet fired from gun B, as Conceptual Example 15 discusses. Therefore, the flight time of pellet A will be greater than that of B by the amount of time that it takes for pellet A to cover the extra distance.
A ____ B
Chapter 2 Problems
69
The time required for pellet A to return to the cliff edge after being fired can be found from Equation 2.4: v = v0 + a t. If "up" is taken as the positive direction then v0 = +30.0 m/s and v = –30.0 m/s. Solving Equation 2.4 for t gives t =
v − v0 a
(−30.0 m/s) − (+30.0 m/s) = −9.80 m/s 2
=
6.12 s
Notice that this result is independent of the height of the cliff. ______________________________________________________________________________ 48. REASONING The initial velocity and the elapsed time are given in the problem. Since the rock returns to the same place from which it was thrown, its displacement is zero (y = 0 m). Using this information, we can employ Equation 2.8 y = v0t + 12 a t 2 to determine the
(
)
acceleration a due to gravity. SOLUTION Solving Equation 2.8 for the acceleration yields
2 ( y − v0 t )
2 0 m − ( +15 m / s )( 20.0 s ) = = −1.5 m / s 2 2 t2 ( 20.0 s ) ______________________________________________________________________________ a=
49.
SSM REASONING The initial velocity of the compass is +2.50 m/s. The initial position of the compass is 3.00 m and its final position is 0 m when it strikes the ground. The displacement of the compass is the final position minus the initial position, or y = –3.00 m. As the compass falls to the ground, its acceleration is the acceleration due to gravity, a = –9.80 m/s2. Equation 2.8 y = v0t + 1 a t 2 can be used to find how much time elapses
(
2
)
before the compass hits the ground. SOLUTION Starting with Equation 2.8, we use the quadratic equation to find the elapsed time.
t=
( ) 2 ( 12 a )
−v0 v02 − 4 12 a ( − y )
=
− ( 2.50 m/s )
( 2.50 m/s )2 − 4 ( −4.90 m/s 2 ) − ( −3.00 m )
(
2 −4.90 m/s 2
)
There are two solutions to this quadratic equation, t1 = 1.08 s and t2 = –0.568 s. The second solution, being a negative time, is discarded. ______________________________________________________________________________
70
KINEMATICS IN ONE DIMENSION
50. REASONING The initial speed of the ball can be determined from Equation 2.9 v 2 = v02 + 2ay . Once the initial speed of the ball is known, Equation 2.9 can be used a
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second time to determine the height above the launch point when the speed of the ball has decreased to one half of its initial value. SOLUTION When the ball has reached its maximum height, its velocity is zero. If we take upward as the positive direction, we have from Equation 2.9 that
( 0 m/s )2 − 2(–9.80 m/s2 )(16 m) = 18 m/s
v0 = v 2 − 2ay =
When the speed of the ball has decreased to one half of its initial value, v = 12 v0 , and Equation 2.9 gives
v 2 − v02
( 12 v0 )2 − v02
v02 1 (18 m/s)2 1 y= = = − 1 = − 1 = 12 m 2a 2a 2a 4 2(–9.80 m/s2 ) 4 ______________________________________________________________________________ 51. REASONING AND SOLUTION a.
v 2 = v02 + 2ay
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v = (1.8 m/s ) + 2 –9.80 m/s 2 ( –3.0 m ) = 7.9 m/s 2
The minus is chosen, since the diver is now moving down. Hence, v = −7.9 m/s . b. The diver's velocity is zero at his highest point. The position of the diver relative to the board is y=–
v02 2a
=–
(1.8 m/s )2
(
2 –9.80 m/s 2
)
= 0.17 m
The position above the water is 3.0 m + 0.17 m = 3.2 m . ______________________________________________________________________________
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52. REASONING Equation 2.9 v 2 = v02 + 2ay can be used to determine the maximum height above the launch point, where the final speed is v = 0 m / s . However, we will need to know 1 2
the initial speed v0, which can be determined via Equation 2.9 and the fact that v = v0 when y = 4.00 m (assuming upward to be the positive direction). SOLUTION When the ball has reached its maximum height, we have v = 0 m / s and y = ymax , so that Equation 2.9 becomes