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12–2. The acceleration of a particle as it moves along a straight line is given by a = 14t 3 - 12 m>s2, where t is in seconds. If s = 2 m and v = 5 m>s when t = 0, determine the particle’s velocity and position when t = 5 s. Also, determine the total distance the particle travels during this time period.

SOLUTION v

dv =

(4 t - 1) dt

v = t4 - t + 5 s

ds = s =

(t4 - t + 5) dt

1 5 1 t - t2 + 5 t + 2 2 5

When t = 5 s, v = 625 m>s

Ans.

s = 639.5 m

Ans.

d = 639.5 - 2 = 637.5 m

Ans.

Since v Z 0 then

Ans: v = 625 m>s s = 639.5 m d = 637.5 m 2

12–3. The velocity of a particle traveling in a straight line is given by v = (6t - 3t2) m>s, where t is in seconds. If s = 0 when t = 0, determine the particle’s deceleration and position when t = 3 s. How far has the particle traveled during the 3-s time interval, and what is its average speed?

Solution v = 6t - 3t 2 a =

dv = 6 - 6t dt

At t = 3 s a = - 12 m>s2

Ans.

ds = v dt L0

ds =

(6t - 3t2)dt

s = 3t 2 - t 3 At t = 3 s Ans.

s = 0 Since v = 0 = 6t - 3t 2, when t = 0 and t = 2 s. when t = 2 s, s = 3(2)2 - (2)3 = 4 m

Ans.

sT = 4 + 4 = 8 m

( vsp ) avg =

sT 8 = = 2.67 m>s t 3

Ans.

Ans: a = -12 m>s2 s = 0 sT = 8 m (vsp)avg = 2.67 m>s 3

*12–4. A particle is moving along a straight line such that its position is defined by s = (10t2 + 20) mm, where t is in seconds. Determine (a) the displacement of the particle during the time interval from t = 1 s to t = 5 s, (b) the average velocity of the particle during this time interval, and (c) the acceleration when t = 1 s.

SOLUTION s = 10t2 + 20 (a) s|1 s = 10(1)2 + 20 = 30 mm s|5 s = 10(5)2 + 20 = 270 mm ¢s = 270 - 30 = 240 mm

Ans.

(b) ¢t = 5 - 1 = 4 s vavg = (c) a =

240 ¢s = = 60 mm>s ¢t 4

d2s = 20 mm s2 dt2

Ans.

(for all t)

Ans.

Ans: ∆s = 240 mm vavg = 60 mm>s a = 20 mm>s2 4

12–5. A particle moves along a straight line such that its position is defined by s = (t2 - 6t + 5) m. Determine the average velocity, the average speed, and the acceleration of the particle when t = 6 s.

Solution s = t2 - 6t + 5 v =

ds = 2t - 6 dt

a =

dv = 2 dt

v = 0 when t = 3 s t=0 = 5 s t = 3 = -4 s t=6 = 5 vavg =

∆s 0 = = 0 ∆t 6

( vsp ) avg =

Ans.

sT 9 + 9 = = 3 m>s ∆t 6

Ans.

a t = 6 = 2 m>s2

Ans.

Ans: vavg = 0 (vsp)avg = 3 m>s a t = 6 s = 2 m>s2 5

12–6. A stone A is dropped from rest down a well, and in 1 s another stone B is dropped from rest. Determine the distance between the stones another second later.

SOLUTION + T s = s1 + v1 t + sA = 0 + 0 +

1 2 a t 2 c

1 (9.81)(2)2 2

sA = 19.62 m sA = 0 + 0 +

1 (9.81)(1)2 2

sB = 4.91 m ¢s = 19.62 - 4.91 = 14.71 m

Ans.

Ans: s = 14.71 m 6

12–7. A bus starts from rest with a constant acceleration of 1 m>s2. Determine the time required for it to attain a speed of 25 m>s and the distance traveled.

SOLUTION Kinematics: v0 = 0, v = 25 m>s, s0 = 0, and ac = 1 m>s2. + B A:

v = v0 + act 25 = 0 + (1)t t = 25 s

+ B A:

Ans.

v2 = v02 + 2ac(s - s0) 252 = 0 + 2(1)(s - 0) s = 312.5 m

Ans.

Ans: t = 25 s s = 312.5 m 7

*12–8. A particle travels along a straight line with a velocity v = (12 - 3t 2) m>s, where t is in seconds. When t = 1 s, the particle is located 10 m to the left of the origin. Determine the acceleration when t = 4 s, the displacement from t = 0 to t = 10 s, and the distance the particle travels during this time period.

SOLUTION v = 12 - 3t 2

(1)

dv = - 6t t = 4 = -24 m>s2 dt

a = s

L-10

ds =

v dt =

Ans.

( 12 - 3t 2 ) dt

s + 10 = 12t - t 3 - 11 s = 12t - t 3 - 21 s t = 0 = - 21 s t = 10 = -901 ∆s = -901 - ( -21) = -880 m

Ans.

From Eq. (1): v = 0 when t = 2s s t = 2 = 12(2) - (2)3 - 21 = - 5 Ans.

sT = (21 - 5) + (901 - 5) = 912 m

Ans: a = - 24 m>s2 ∆s = - 880 m sT = 912 m 8

12–9. When two cars A and B are next to one another, they are traveling in the same direction with speeds vA and vB , respectively. If B maintains its constant speed, while A begins to decelerate at aA , determine the distance d between the cars at the instant A stops.

SOLUTION Motion of car A: v = v0 + act 0 = vA - aAt

t =

vA aA

v2 = v20 + 2ac(s - s0) 0 = v2A + 2( - aA)(sA - 0) sA =

v2A 2aA

Motion of car B: sB = vBt = vB a

vA vAvB b = aA aA

The distance between cars A and B is sBA = |sB - sA| = `

v2A vAvB 2vAvB - v2A ` = ` ` aA 2aA 2aA

Ans.

Ans: S BA = ` 9

2vA vB - v2A ` 2aA

12–10. A particle travels along a straight-line path such that in 4 s it moves from an initial position sA = - 8 m to a position sB = +3 m. Then in another 5 s it moves from sB to sC = -6 m. Determine the particle’s average velocity and average speed during the 9-s time interval.

SOLUTION Average Velocity: The displacement from A to C is ∆s = sC - SA = -6 - ( - 8) = 2 m. ∆s 2 vavg = = = 0.222 m>s Ans. ∆t 4 + 5 Average Speed: The distances traveled from A to B and B to C are sA S B = 8 + 3 = 11.0 m and sB S C = 3 + 6 = 9.00 m, respectively. Then, the total distance traveled is sTot = sA S B + sB S C = 11.0 + 9.00 = 20.0 m. (vsp)avg =

sTot 20.0 = = 2.22 m>s ∆t 4 + 5

Ans.

Ans: vavg = 0.222 m>s (vsp)avg = 2.22 m>s 10

12–11. Traveling with an initial speed of 70 km>h, a car accelerates at 6000 km>h2 along a straight road. How long will it take to reach a speed of 120 km>h? Also, through what distance does the car travel during this time?

SOLUTION v = v1 + ac t 120 = 70 + 6000(t) t = 8.33(10 - 3) hr = 30 s

Ans.

v2 = v21 + 2 ac(s - s1) (120)2 = 702 + 2(6000)(s - 0) s = 0.792 km = 792 m

Ans.

Ans: t = 30 s s = 792 m 11

*12–12. A particle moves along a straight line with an acceleration of a = 5>(3s1>3 + s 5>2) m>s2, where s is in meters. Determine the particle’s velocity when s = 2 m, if it starts from rest when s = 1 m . Use a numerical method to evaluate the integral.

SOLUTION a =

5 1 3

A 3s + s2 B

a ds = v dv 2

5 ds 1 3

5 2

L1 A 3s + s B 0.8351 =

v dv

1 2 v 2

v = 1.29 m>s

Ans.

Ans: v = 1.29 m>s 12

12–13. The acceleration of a particle as it moves along a straight line is given by a = (2t - 1) m>s2, where t is in seconds. If s = 1 m and v = 2 m>s when t = 0, determine the particle’s velocity and position when t = 6 s. Also, determine the total distance the particle travels during this time period.

Solution a = 2t - 1 dv = a dt L2

dv =

(2t - 1)dt

v = t2 - t + 2 dx = v dt Lt

ds =

s =

(t2 - t + 2)dt

1 3 1 t - t 2 + 2t + 1 3 2

When t = 6 s v = 32 m>s

Ans.

s = 67 m

Ans.

Since v ≠ 0 for 0 … t … 6 s, then Ans.

d = 67 - 1 = 66 m

Ans: v = 32 m>s s = 67 m d = 66 m 13

12–15. A particle is moving along a straight line such that its velocity is defined as v = ( - 4s2) m>s, where s is in meters. If s = 2 m when t = 0, determine the velocity and acceleration as functions of time.

SOLUTION v = - 4s2 ds = - 4s2 dt s

s - 2 ds =

- 4 dt

- s - 1| s2 = - 4t|t0 t =

1 -1 (s - 0.5) 4

s =

2 8t + 1

v = -4a a =

2 2 16 b = m>s 8t + 1 (8t + 1)2

Ans.

16(2)(8t + 1)(8) dv 256 = = m>s2 dt (8t + 1)4 (8t + 1)3

Ans.

Ans: 16 m>s (8t + 1)2 256 a= m>s2 (8t + 1)3 v =

*12–16. Determine the time required for a car to travel 1 km along a road if the car starts from rest, reaches a maximum speed at some intermediate point, and then stops at the end of the road. The car can accelerate at 1.5 m>s2 and decelerate at 2 m>s2.

SOLUTION Using formulas of constant acceleration: v2 = 1.5 t1 x =

1 (1.5)(t21) 2

0 = v2 - 2 t2 1 (2)(t22) 2

1000 - x = v2t2 -

Combining equations: t1 = 1.33 t2; v2 = 2 t2 x = 1.33 t22 1000 - 1.33 t22 = 2 t22 - t22 t2 = 20.702 s;

t1 = 27.603 s

t = t1 + t2 = 48.3 s

Ans.

Ans: t = 48.3 s 16

12–17. A particle is moving with a velocity of v0 when s = 0 and t = 0. If it is subjected to a deceleration of a = -kv3, where k is a constant, determine its velocity and position as functions of time.

SOLUTION dn = - kn3 dt

a =

n - 3 dn =

Ln0

- k dt

1 -2 1n - n0- 22 = - kt 2

-2 1 n = a 2kt + a 2 b b n0

Ans.

ds = n dt s

ds =

dt 1

2 1 a2kt + a 2 b b v0

t 2 1 2a 2kt + a 2 b b 3 n0 s = 2k 0

2 1 1 1 s = B ¢ 2kt + ¢ 2 ≤ ≤ - R n0 k n0 1

Ans.

Ans: v = a2kt + s=

1 - 1>2 b v20

1 1 1>2 1 c a2kt + 2 b d k v0 v0

12–18. A particle is moving along a straight line with an initial velocity of 6 m>s when it is subjected to a deceleration of a = (- 1.5v1>2) m>s2, where v is in m>s. Determine how far it travels before it stops. How much time does this take?

SOLUTION Distance Traveled: The distance traveled by the particle can be determined by applying Eq. 12–3. ds =

vdv a

ds =

v 1

L6 m>s - 1.5v2

s =

L6 m>s

- 0.6667 v2 dv 3

= a -0.4444v2 + 6.532 b m When v = 0,

s = - 0.4444a 0 2 b + 6.532 = 6.53 m

Ans.

Time: The time required for the particle to stop can be determined by applying Eq. 12–2. dt =

dv a

dt = 1

dv 1

L6 m>s 1.5v 2

t = - 1.333av2 b 6 m>s = a3.266 - 1.333v 2 b s When v = 0,

t = 3.266 - 1.333 a 0 2 b = 3.27 s

Ans.

Ans: s = 6.53 m t = 3.27 s 18

12–19. The acceleration of a rocket traveling upward is given by a = (6 + 0.02s) m>s2, where s is in meters. Determine the rocket’s velocity when s = 2 km and the time needed to reach this attitude. Initially, v = 0 and s = 0 when t = 0.

SOLUTION b

6 m>s2

b c sp

´ µ ¶

vp dvp

sp1

2000 m

dvp

dsp

b c sp dsp

b sp

dsp

´ µ ¶

0.02 s- 2

µ́ µ µ ¶

c 2 sp 2

2b sp c sp

1 2

2b sp c sp

dsp

2b sp1 c sp1

vp1

µ́ µ µ ¶

sp1

vp1

1 2b sp c sp

dsp

322.49 m>s

Ans.

19.27 s

Ans.

Ans: vp1 = 322.49 t1 = 19.27 s 19

m s

*12–20. The acceleration of a rocket traveling upward is given by a = 16 + 0.02s2 m>s2, where s is in meters. Determine the time needed for the rocket to reach an altitude of s = 100 m. Initially, v = 0 and s = 0 when t = 0.

SOLUTION a ds = n dv

16 + 0.02 s2 ds =

6 s + 0.01 s2 =

n dn

1 2 n 2

n = 212 s + 0.02 s2 ds = n dt 100

ds 212 s + 0.02 s2

1 20.02

1n B 212s + 0.02s2 + s20.02 +

t = 5.62 s

12 2 20.02

100

= t

Ans.

Ans: t = 5.62 s 20

12–21. When a train is traveling along a straight track at 2 m/s, it begins to accelerate at a = 160 v-42 m>s2, where v is in m/s. Determine its velocity v and the position 3 s after the acceleration.

SOLUTION a =

dv dt

dt =

dv a v

dt =

dv -4 L2 60v

3 =

1 (v5 - 32) 300

v = 3.925 m>s = 3.93 m>s

Ans.

ads = vdv ds = s

1 5 vdv = v dv a 60 3.925

ds =

1 60 L2

s =

1 v6 3.925 a b` 60 6 2

v5 dv

= 9.98 m

Ans.

Ans: v = 3.93 m>s s = 9.98 m 21

12–22. The acceleration of a particle along a straight line is defined by a = 12t - 92 m>s2, where t is in seconds. At t = 0, s = 1 m and v = 10 m>s. When t = 9 s, determine (a) the particle’s position, (b) the total distance traveled, and (c) the velocity.

SOLUTION a = 2t - 9 v

L10

dv =

12t - 92 dt

v - 10 = t2 - 9 t v = t2 - 9 t + 10 s

ds =

s-1 = s =

1t2 - 9t + 102 dt

13 t - 4.5 t2 + 10 t 3

13 t - 4.5 t2 + 10 t + 1 3

Note when v = t2 - 9 t + 10 = 0: t = 1.298 s and t = 7.701 s When t = 1.298 s,

s = 7.13 m

When t = 7.701 s,

s = - 36.63 m

When t = 9 s,

s = -30.50 m

(a)

s = - 30.5 m

(b)

sTo t = (7.13 - 1) + 7.13 + 36.63 + (36.63 - 30.50)

(c)

Ans.

sTo t = 56.0 m

Ans.

v = 10 m>s

Ans.

Ans: (a) s = - 30.5 m (b) sTot = 56.0 m (c) v = 10 m>s 22

12–23. If the effects of atmospheric resistance are accounted for, a falling body has an acceleration defined by the equation a = 9.81[1 - v2(10-4)] m>s2, where v is in m>s and the positive direction is downward. If the body is released from rest at a very high altitude, determine (a) the velocity when t = 5 s, and (b) the body’s terminal or maximum attainable velocity (as t : q ).

SOLUTION Velocity: The velocity of the particle can be related to the time by applying Eq. 12–2. (+ T )

dt = t

dv a

dv 2 L0 9.81[1 - (0.01v) ]

dt = v

t =

1 dv dv c + d 9.81 L0 2(1 + 0.01v) L0 2(1 - 0.01v) 9.81t = 50ln a v =

1 + 0.01v b 1 - 0.01v

100(e0.1962t - 1)

(1)

e0.1962t + 1

a) When t = 5 s, then, from Eq. (1) v =

b) If t : q ,

e0.1962t - 1 e0.1962t + 1

100[e0.1962(5) - 1] e0.1962(5) + 1

= 45.5 m>s

Ans.

: 1. Then, from Eq. (1) vmax = 100 m>s

Ans.

Ans: (a) v = 45.5 m>s (b) v max = 100 m>s 23

*12–24. A sandbag is dropped from a balloon which is ascending vertically at a constant speed of 6 m>s. If the bag is released with the same upward velocity of 6 m>s when t = 0 and hits the ground when t = 8 s, determine the speed of the bag as it hits the ground and the altitude of the balloon at this instant.

SOLUTION (+ T )

s = s0 + v0 t +

1 a t2 2 c

h = 0 + ( -6)(8) +

1 (9.81)(8)2 2

= 265.92 m During t = 8 s, the balloon rises h¿ = vt = 6(8) = 48 m Altitude = h + h¿ = 265.92 + 48 = 314 m (+ T)

Ans.

v = v0 + ac t v = - 6 + 9.81(8) = 72.5 m s

Ans.

Ans: h = 314 m v = 72.5 m>s 24

12–25. A particle is moving along a straight line such that its acceleration is defined as a = (-2v) m>s2, where v is in meters per second. If v = 20 m>s when s = 0 and t = 0, determine the particle’s position, velocity, and acceleration as functions of time.

Solution a = - 2v dv = - 2v dt v

t dv = -2 dt L20 v L0

v = -2t 20

v = ( 20e -2t ) m>s a = L0

Ans.

dv = ( - 40e -2t ) m>s2 dt

ds = v dt =

Ans.

(20e-2t)dt

s = - 10e -2t t0 = - 10 ( e -2t - 1 ) s = 10 ( 1 - e -2t ) m

Ans.

Ans: v = ( 20e -2t ) m>s a = ( - 40e -2t ) m>s2 s = 10 ( 1 - e -2t ) m 25

12–26. The acceleration of a particle traveling along a straight line 1 1>2 is a = s m> s2, where s is in meters. If v = 0, s = 1 m 4 when t = 0, determine the particle’s velocity at s = 2 m.

SOLUTION Velocity: + B A:

v dv = a ds v

v dv = v

1 1>2 s ds L1 4

s v2 2 = 1 s3>2 ` 2 0 6 1

v =

1 23

1s3>2 - 121>2 m>s

When s = 2 m, v = 0.781 m>s.

Ans.

Ans: v = 0.781 m>s 26

12–27. When a particle falls through the air, its initial acceleration a = g diminishes until it is zero, and thereafter it falls at a constant or terminal velocity vf. If this variation of the acceleration can be expressed as a = 1g>v2f21v2f - v22, determine the time needed for the velocity to become v = vf>2 . Initially the particle falls from rest.

SOLUTION g dv = a = ¢ 2 ≤ A v2f - v2 B dt vf v

L0 v2f - v2¿

g v2f L0

vf + v y g 1 ln ¢ ≤` = 2t 2vf vf - v 0 vf t = t =

vf 2g vf 2g

ln ¢ ln ¢

t = 0.549 a

vf + v vf - v

≤

vf + vf> 2 vf - vf> 2 vf g

≤

Ans.

Ans: t = 0.549 a 27

vf g

*12–28. A sphere is fired downwards into a medium with an initial speed of 27 m>s. If it experiences a deceleration of a = ( -6t) m>s2, where t is in seconds, determine the distance traveled before it stops.

SOLUTION Velocity: v0 = 27 m>s at t0 = 0 s. Applying Eq. 12–2, we have

A+TB

dv = adt v

L27

dv =

-6tdt

v = A 27 - 3t2 B m>s

(1)

At v = 0, from Eq. (1) 0 = 27 - 3t2

t = 3.00 s

Distance Traveled: s0 = 0 m at t0 = 0 s. Using the result v = 27 - 3t2 and applying Eq. 12–1, we have

A+TB

ds = vdt s

ds =

A 27 - 3t2 B dt

s = A 27t - t3 B m

(2)

At t = 3.00 s, from Eq. (2) s = 27(3.00) - 3.003 = 54.0 m

Ans.

Ans: s = 54.0 m 28

12–29. A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m>s. At the same instant another ball B is thrown upward from the ground with an initial velocity of 20 m>s. Determine the height from the ground and the time at which they pass.

Solution Origin at roof: Ball A: 1 2

( + c ) s = s0 + v0t + act 2

- s = 0 + 5t -

1 (9.81)t 2 2

Ball B: 1 2

( + c ) s = s0 + v0t + act 2

- s = -30 + 20t -

1 (9.81)t 2 2

Solving, Ans.

t = 2 s s = 9.62 m Distance from ground,

Ans.

d = (30 - 9.62) = 20.4 m Also, origin at ground, s = s0 + v0t +

1 2 at 2 c

sA = 30 + 5t +

1 ( -9.81)t 2 2

sB = 0 + 20t +

1 ( - 9.81)t 2 2

Require sA = sB 30 + 5t +

1 1 ( -9.81)t 2 = 20t + ( -9.81)t 2 2 2

t = 2 s

Ans.

sB = 20.4 m

Ans.

Ans: h = 20.4 m t = 2s

12–30. A boy throws a ball straight up from the top of a 12-m high tower. If the ball falls past him 0.75 s later, determine the velocity at which it was thrown, the velocity of the ball when it strikes the ground, and the time of flight.

SOLUTION Kinematics: When the ball passes the boy, the displacement of the ball in equal to zero. Thus, s = 0. Also, s0 = 0, v0 = v1, t = 0.75 s, and ac = - 9.81 m>s2.

A+cB

s = s0 + v0t +

1 2 at 2 c

0 = 0 + v110.752 +

1 1-9.81210.7522 2

v1 = 3.679 m>s = 3.68 m>s

Ans.

When the ball strikes the ground, its displacement from the roof top is s = - 12 m. Also, v0 = v1 = 3.679 m>s, t = t2, v = v2, and ac = - 9.81 m>s2.

A+cB

s = s0 + v0t +

1 2 at 2 c

- 12 = 0 + 3.679t2 +

1 1- 9.812t22 2

4.905t22 - 3.679t2 - 12 = 0 t2 =

3.679 ; 21 - 3.67922 - 414.90521 -122 214.9052

Choosing the positive root, we have t2 = 1.983 s = 1.98 s

Ans.

Using this result,

A+cB

v = v0 + act v2 = 3.679 + 1 -9.81211.9832 = - 15.8 m>s = 15.8 m>s T

Ans.

Ans: v1 = 3.68 m>s t2 = 1.98 s v2 = 15.8 m>s T 30

12–31. The velocity of a particle traveling along a straight line is v = v0 - ks, where k is constant. If s = 0 when t = 0, determine the position and acceleration of the particle as a function of time.

SOLUTION Position: + B A:

ds y

dt = t

dt =

t t0 = t =

ds v L0 0 - ks

s 1 ln (v0 - ks) 2 k 0

v0 1 ln ¢ ≤ k v0 - ks

ekt =

v0 v0 - ks

s =

v0 A 1 - e - kt B k

v =

ds d v0 = c A 1 - e - kt B d dt dt k

Ans.

Velocity:

v = v0e - kt Acceleration: a =

dv d = A v e - kt B dt dt 0

a = -kv0e - kt

Ans.

Ans: v0 ( 1 - e - kt ) s = k a = - kv0e - kt 31

*12–32. Ball A is thrown vertically upwards with a velocity of v0. Ball B is thrown upwards from the same point with the same velocity t seconds later. Determine the elapsed time t 6 2v0>g from the instant ball A is thrown to when the balls pass each other, and find the velocity of each ball at this instant.

SOLUTION Kinematics: First, we will consider the motion of ball A with (vA)0 = v0, (sA)0 = 0, sA = h, tA = t¿ , and (ac)A = - g.

A+cB

h = 0 + v0t¿ + h = v0t¿ -

A+cB

1 (a ) t 2 2 cA A

sA = (sA)0 + (vA)0tA +

g 2 t¿ 2

1 (- g)(t¿)2 2

(1)

vA = (vA)0 + (ac)A tA vA = v0 + ( - g)(t¿) vA = v0 - gt¿

(2)

The motion of ball B requires (vB)0 = v0, (sB)0 = 0, sB = h, tB = t¿ - t , and (ac)B = - g.

A+cB

sB = (sB)0 + (vB)0tB + h = 0 + v0(t¿ - t) + h = v0(t¿ - t) -

A+cB

1 (a ) t 2 2 cBB

1 ( -g)(t¿ - t)2 2

g (t¿ - t)2 2

(3)

vB = (vB)0 + (ac)B tB vB = v0 + ( - g)(t¿ - t) vB = v0 - g(t¿ - t)

(4)

Solving Eqs. (1) and (3), g 2 g t¿ = v0(t¿ - t) - (t¿ - t)2 2 2 2v0 + gt t¿ = 2g v0t¿ -

Ans.

Substituting this result into Eqs. (2) and (4), vA = v0 - g a = -

2v0 + gt b 2g

1 1 gt = gtT 2 2

Ans.

2v0 + gt 2g 1 vA = gt T 2 1 vB = gt c 2

t= =

2v0 + gt - tb vB = v0 - g a 2g =

Ans:

1 gt c 2

Ans.

*12–44. The v–t graph for a particle moving through an electric field from one plate to another has the shape shown in the figure, where t¿ = 0.2 s and vmax = 10 m>s. Draw the s–t and a–t graphs for the particle. When t = t¿>2 the particle is at s = 0.5 m.

smax

v s

vmax

SOLUTION For 0 6 t 6 0.1 s,

t¿/ 2

v = 100 t

t¿

dv = 100 dt

a =

ds = v dt s

ds =

100 t dt

s = 50 t 2 When t = 0.1 s, s = 0.5 m For 0.1 s 6 t 6 0.2 s, v = -100 t + 20 a =

dv = - 100 dt

ds = v dt s

L0.5

ds =

1 -100t + 202dt L0.1

s - 0.5 = ( -50 t 2 + 20 t - 1.5) s = - 50 t 2 + 20 t - 1 When t = 0.2 s, s = 1m When t = 0.1 s, s = 0.5 m and a changes from 100 m/s2 to -100 m/s2. When t = 0.2 s, s = 1 m.

Ans: When t = 0.1 s, s = 0.5 m and a changes from 100 m>s2 to - 100 m>s2. When t = 0.2 s, s = 1 m. 44

12–45. The motion of a jet plane just after landing on a runway is described by the a–t graph. Determine the time t when the jet plane stops. Construct the v–t and s–t graphs for the motion. Here s = 0, and v = 150 m>s when t = 0.

a (m/s2) 10 10 20

Solution v–t Graph. The v9t function can be determined by integrating dv = a dt. For 0 ≤ t 6 10 s, a = 0. Using the initial condition v = 150 m>s at t = 0, v

L150 m/s

dv =

0 dt

v - 150 = 0 Ans.

v = 150 m/s

- 5 - 1 - 102 1 = , a = a t - 15bm>s2. Using the t - 10 20 - 10 2 initial condition v = 150 m>s at t = 10 s, For 10 s 6 t 6 20 s,

a - 1 -102 t

L150 m/s

dv =

1 a t - 15bdt L10 s 2

1 v - 150 = a t 2 - 15tb t10 s 4

At t = 20 s,

1 v = e t 2 - 15t + 275 fm/s 4 v t = 20 s =

Ans.

1 1202 2 - 151202 + 275 = 75 m/s 4

For 20 s 6 t ≤ t¿, a = –5 m/s. Using the initial condition v = 75 m>s at t = 20 s, v

L75 m/s

dv =

L20 s

- 5 dt

v - 75 = 1 - 5t2 t20 s

Ans.

v = { - 5t + 175} m/s

It is required that at t = t¿, v = 0. Thus 0 = -5 t′ + 175 Ans.

t′ = 35.0 s Using these results, the v9t graph shown in Fig. a can be plotted.

t¿

t (s)

12–45. Continued

s-t Graph. The s–t function can be determined by integrating ds = vdt. For 0 ≤ t 6 10 s, the initial condition is s = 0 at t = 0. L0

At t = 10 s,

ds =

150 dt Ans.

s = 5 150 t6 m s t = 10 s = 1501102 = 1500 m

For 10 s 6 t 6 20 s, the initial condition is s = 1500 m at t = 10 s. s

L1500 m

ds =

1 a t 2 - 15t + 275bdt L10 s 4

s - 1500 = a

At t = 20 s,

s = e

1 3 15 2 t t + 275tb t10 s 12 2

1 3 15 2 t t + 275t - 583 f m 12 2

s t = 20 s =

Ans.

1 15 1203 2 1202 2 + 2751202 - 583 = 2583 m 12 2

12–45. Continued

For 20 s 6 t ≤ 35 s, the initial condition is s = 2583 m at t = 20 s. s

L2583 m

ds =

L20 s

1 - 5t + 1752 dt

5 s - 2583 = a - t 2 + 175tb t20 s 2

At t = 35 s,

5 s = e - t 2 + 175t + 83.3 f m 2

s t = 35 s = -

Ans.

5 1352 2 + 1751352 + 83.3 = 3146 m 2

using these results, the s-t graph shown in Fig. b can be plotted.

Ans: v = 150 m>s 1 v = e t2 - 15t + 275 f m>s 4 v = { - 5t + 175} m>s t = 35.0 s s = {150t} m 1 15 2 s = e t3 t + 275t - 583 f m 12 2 -5 2 s = e t + 175t + 83.3 f m 2

12–46. A two-stage rocket is fired vertically from rest at s = 0 with the acceleration as shown. After 30 s the first stage, A, burns out and the second stage, B, ignites. Plot the v–t and s–t graphs which describe the motion of the second stage for 0 … t … 60 s.

a (m/s2) B 24 A

Solution v9t Graph. The v - t function can be determined by integrating dv = a dt. For 0 … t 6 30 s, a = L0

12 2 t = a tb m>s2. Using the initial condition v = 0 at t = 0, 30 5

2 t dt L0 5

dv =

1 v = e t 2 f m>s 5

t (s)

Ans.

At t = 30 s, v`

t = 30 s

1 2 (30 ) = 180 m>s 5

For 30 6 t … 60 s, a = 24 m>s2. Using the initial condition v = 180 m>s at t = 30 s, L180 m>s v

dv =

L30 s

v - 180 = 24 t `

24 dt

t 30 s

Ans.

v = {24t - 540} m>s At t = 60 s, v`

t = 60 s

= 24(60) - 540 = 900 m>s

Using these results, v - t graph shown in Fig. a can be plotted. s -t Graph. The s - t function can be determined by integrating ds = v dt. For 0 … t 6 30 s, the initial condition is s = 0 at t = 0. L0

ds =

s = e

At t = 30 s, s`

1 2 t dt L0 5

1 3 t f m 15

t = 30 s

Ans.

1 (303) = 1800 m 15 Ans: 1 1 For 0 … t 6 30 s, v = e t2 f m>s, s = e t3 f m 5 15

For 30 … t … 60 s, v = {24t - 540} m>s, 48

12–47. The a–t graph of the bullet train is shown. If the train starts from rest, determine the elapsed time t¿ before it again comes to rest. What is the total distance traveled during this time interval? Construct the v–t and s–t graphs.

a(m/s2)

0.1t

SOLUTION v t Graph: For the time interval 0 … t 6 30 s, the initial condition is v = 0 when t = 0 s. + B A:

dv = adt v

dv =

0.1tdt

v = A 0.05t2 B m>s When t = 30 s, v t = 30 s = 0.05 A 302 B = 45 m>s or the time interval 30 s 6 t … t¿ , the initial condition is v = 45 m>s at t = 30 s. + B A:

dv = adt v

dv =

L30 s

L45 m>s

v = ¢-

¢-

1 t + 5 ≤ dt 15

1 2 t + 5t - 75 ≤ m>s 30

Thus, when v = 0, 0 = -

1 2 t¿ + 5t¿ - 75 30

Choosing the root t¿ 7 75 s, t¿ = 133.09 s = 133 s

Ans.

Also, the change in velocity is equal to the area under the a–t graph. Thus, ¢v = 0 =

adt

1 1 1 (3)(75) + B ¢ - t¿ + 5 ≤ (t¿ - 75) R 2 2 15

0 = -

1 2 t¿ + 5t¿ - 75 30

This equation is the same as the one obtained previously. The slope of the v–tgraph is zero when t = 75 s, which is the instant a = v t = 75 s = -

1 A 752 B + 5(75) - 75 = 112.5 m>s 30

dv = 0.Thus, dt

( 15 )t

5 t¿

t(s)

12–47. continued The v–t graph is shown in Fig. a. s t Graph: Using the result of v, the equation of the s–t graph can be obtained by integrating the kinematic equation ds = vdt. For the time interval 0 … t 6 30 s, the initial condition s = 0 at t = 0 s will be used as the integration limit. Thus, + B A:

ds = vdt s

ds =

s = a

0.05t2 dt

1 3 t bm 60

When t = 30 s, 1 A 303 B = 450 m 60

s t = 30 s =

For the time interval 30 s 6 t … t¿ = 133.09 s, the initial condition is s = 450 m when t = 30 s. + B A:

ds = vdt s

ds =

L450 m

s = a-

L30 s

1 2 t + 5t - 75 b dt 30

5 1 3 t + t2 - 75t + 750b m 90 2

When t = 75 s and t¿ = 133.09 s, s t = 75 s = -

1 5 A 753 B + A 752 B - 75(75) + 750 = 4500 m 90 2

s t = 133.09 s = -

5 1 A 133.093 B + A 133.092 B - 75(133.09) + 750 = 8857 m 90 2

Ans.

The s–t graph is shown in Fig. b. When t = 30 s, v = 45 m/s and s = 450 m. When t = 75 s, v = vmax = 112.5 m/s and s = 4500 m. When t = 133 s, v = 0 and s = 8857 m.

Ans: t = 133 s s` 50

t = 133.09 s

= 8857 m

*12–48. The v–t graph for a train has been experimentally determined. From the data, construct the s–t and a–t graphs for the motion for 0 … t … 180 s. When t = 0, s = 0.

v (m/s)

Solution s–t Graph. The s–t function can be determined by integrating ds = v dt. 1 6 For 0 … t 6 60 s, v = t = a t b m>s. Using the initial condition 60 10 s = 0 at t = 0,

ds =

1 a tbdt L0 10

1 2 t f m 20

s = e

Ans.

When t = 60 s,

s t - 60 s =

1 ( 602 ) = 180 m 20

For 60 s 6 t 6 120 s, v = 6 m>s. Using the initial condition s = 180 m at t = 60 s, s

L180 m

ds =

L60 s

s - 180 = 6t `

6 dt

t 60 s

Ans.

s = 56t - 1806 m

At t = 120 s,

s t - 120 s = 6(120) - 180 = 540 m For 120 s 6 t … 180 s,

v - 6 10 - 6 1 = ; v = e t - 2 f m>s. Using the initial t - 120 180 - 120 15

condition s = 540 m at t = 120 s, s

L540 m

ds =

1 t - 2b dt 15 L120 s a

s - 540 = a s = e

At t = 180 s,

t 1 2 t - 2tb ` 30 120 s

1 2 t - 2t + 300 f m 30

s t = 180 s =

Ans.

1 ( 1802 ) - 2(180) + 300 = 1020 m 30

Using these results, s–t graph shown in Fig. a can be plotted.

120

180

t (s)

*12–48. Continued

a–t Graph. The a–t function can be determined using a = For 0 … t 6 60 s, a =

1 t2 d 1 10

For 60 s 6 t 6 120 s, a = For 120 s 6 t … 180 s, a =

= 0.1 m>s2

d(6) dt

dv . dt Ans. Ans.

= 0

1 t - 22 d 1 15

= 0.0667 m>s2

Ans.

Using these results, a–t graph shown in Fig. b can be plotted.

Ans: For 0 … t 6 60 s, 1 s = e t 2 f m, 20 a = 0.1 m>s2. For 60 s 6 t 6 120 s, s = {6t - 180} m, a = 0. For 120 s 6 t … 180 s, 1 s = e t 2 - 2t + 300 f m, 30 a = 0.0667 m>s2.

12–49. The v–s graph for a go-cart traveling on a straight road is shown. Determine the acceleration of the go-cart at s = 50 m and s = 150 m. Draw the a–s graph.

v (m/s)

SOLUTION For 0 … s 6 100 v = 0.08 s,

100

200

s (m)

dv = 0.08 ds

a ds = (0.08 s)(0.08 ds) a = 6.4(10 - 3) s At s = 50 m,

a = 0.32 m>s2

Ans.

For 100 6 s 6 200 v = - 0.08 s + 16, dv = - 0.08 ds a ds = (-0.08 s + 16)( -0.08 ds) a = 0.08(0.08 s - 16) At s = 150 m,

a = - 0.32 m>s2

Ans.

Also, v dv = a ds a = v(

dv ) ds

At s = 50 m, a = 4(

8 ) = 0.32 m>s2 100

Ans.

At s = 150 m, a = 4(

-8 ) = -0.32 m>s2 100

Ans.

At s = 100 m, a changes from amax = 0.64 m>s2 to amin = -0.64 m>s2 .

Ans: a = 0.32 m>s2, a = - 0.32 m>s2 53

12–50. The jet car is originally traveling at a velocity of 10 m>s when it is subjected to the acceleration shown. Determine the car’s maximum velocity and the time t′ when it stops. When t = 0, s = 0.

a (m/s2)

t¿

Solution v9t Function. The v-t function can be determined by integrating dv = a dt . For 4 0 … t 6 15 s, a = 6 m>s2. Using the initial condition v = 10 m>s at t = 0, v

L10 m>s

dv =

t (s)

6dt

v - 10 = 6t v = {6t + 10} m>s The maximum velocity occurs when t = 15 s. Then

vmax = 6(15) + 10 = 100 m>s

Ans.

For 15 s 6 t … t′, a = - 4 m>s, Using the initial condition v = 100 m>s at t = 15 s, v

L100 m>s

dv =

L15 s

v - 100 = ( - 4t) `

- 4dt t 15 s

v = { - 4t + 160} m>s It is required that v = 0 at t = t′. Then

Ans.

0 = - 4t′ + 160 t′ = 40 s

Ans: vmax = 100 m>s t′ = 40 s 54

12–51. v (m/s)

The race car starts from rest and travels along a straight road until it reaches a speed of 26 m>s in 8 s as shown on the v–t graph. The flat part of the graph is caused by shifting gears. Draw the a–t graph and determine the maximum acceleration of the car.

26 v 4t 6 14 v 3.5t

Solution For 0 … t 6 4 s a =

t (s)

∆v 14 = = 3.5 m>s2 ∆t 4

For 4 s … t 6 5 s a =

∆v = 0 ∆t

For 5 s … t 6 8 s a =

∆v 26 - 14 = = 4 m>s2 ∆t 8 - 5

a max = 4.00 m>s2

Ans.

Ans: amax = 4.00 m>s2 55

v(m/s)

*12–5 2. A car starts from rest and travels along a straight road with a velocity described by the graph. Determine the total distance traveled until the car stops. Construct the T–U and B–U graphs.

30 v

Solution

t 30

0.5t

45 t(s)

U V Graph: For the time interval 0 U 30 s, the initial condition is T 0 when U 0 s.

ET vEU T

UEU

U2 T A Im 2 When U 30 s, T

302 450 m 2

For the time interval 30 s U 90 s, the initial condition is T 450 m when U 30 s.

ET vEU T

(450 m

(30s

1 T 3 U2 4

( 0.5U

45)EU

45U 675 4 m

When U 90 s, T U 90 s

1 902 4

45(90) 675 1350 m

Ans.

The T U graph shown is in Fig. a. C V Graph: For the time interval 0 U 30 s, B

Ev E (U) 1 m s2 EU EU

For the time interval 30 s U 90 s, B

Ev E ( 0.5U EU EU

45) 0.5 m s2

The B U graph is shown in Fig. b. Note: Since the change in position of the car is equal to the area under the v U graph, the total distance traveled by the car is cT

(

vEU

T U 90 s 0

1 (90)(30) 2

T U 90 s 1350 m

Ans: T U 90 s 1350 m

12–53. The v–s graph for an airplane traveling on a straight runway is shown. Determine the acceleration of the plane at s = 100 m and s = 150 m. Draw the a–s graph.

v (m/s)

Solution a

dv ds

50 40

v v

100 m

150 m

200 m

40 m/s

50 m

50 m/s

0 s3 s1

§ s3 · § v1 · ¨ ¸ v1¨ ¸ © s1 ¹ © s1 ¹

s1 s4 s2

s4 s1 ª º v2 v1 v2 v1 » «v1 s2 s1 ¬ ¼ s2 s1

0.1s

0.4s

100

200

8.00 m/s2

Ans.

4.50 m/s2

Ans.

s (m)

The graph 2

0 0.01s1 s1

a1 V 1

s1 1.01s1 s2

a2 V 2

V 1 v1

s2 / m

s1 s1

V 2 s1 º v2 v1 s2 /m ª v2 v1 » «v1 s2 s1 ¬ ¼ s2 s1

Acceleration in m/s^2

20 15 a1 V 1 a2 V 2

10 5 0

100

150

200

V1 V2

Distance in m Ans: a3 = 8.00 m/s2 a4 = 4.50 m/s2 57

12–54. a (m/s2)

The a–s graph for a jeep traveling along a straight road is given for the first 300 m of its motion. Construct the v –s graph. At s = 0, v = 0.

SOLUTION a s Graph: The function of acceleration a in terms of s for the interval 0 m … s 6 200 m is a - 0 2 - 0 = s - 0 200 - 0

a = (0.01s) m>s2

For the interval 200 m 6 s … 300 m, a - 2 0 - 2 = s - 200 300 - 200

a = ( - 0.02s + 6) m>s2

v s Graph: The function of velocity v in terms of s can be obtained by applying vdv = ads. For the interval 0 m ◊ s<200 m, vdv = ads v

vdv =

0.01sds

v = (0.1s) m>s At s = 200 m,

v = 0.100(200) = 20.0 m>s

For the interval 200 m 6 s … 300 m, vdv = ads v

L20.0 m>s

vdv =

L200 m

( - 0.02s + 6)ds

v = A 2-0.02s2 + 12s - 1200 B m>s At s = 300 m,

v = 2 - 0.02(3002) + 12(300) - 1200 = 24.5 m>s

200

300

s (m)

12–55. The v–t graph for the motion of a car as it moves along a straight road is shown. Draw the s–t and a–t graphs. Also determine the average speed and the distance traveled for the 15-s time interval. When t = 0, s = 0.

v (m/s) 15 v 0.6t 2

Solution s–t Graph. The s–t function can be determined by integrating ds = v dt . For 0 … t 6 5 s, v = 0.6t 2. Using the initial condition s = 0 at t = 0, L0

ds =

t (s)

0.6t 2dt

s = 50.2t 3 6 m

Ans.

At t = 5 s,

s t = 5 s = 0.2 ( 53 ) = 25 m v - 15 0 - 15 1 = ; v = (45 - 3t). Using the initial condition t - 5 15 - 5 2

For 5 s 6 t … 15 s, s = 25 m at t = 5 s, s

ds =

L25 m

1 (45 - 3t)dt 2 L5 s

s - 25 =

45 3 t - t 2 - 93.75 2 4

1 s = e (90t - 3t 2 - 275) f m 4

Ans.

At t = 15 s, s =

1 3 90(15) - 3 ( 152 ) - 275 4 = 100 m 4

Ans.

Thus the average speed is vavg =

sT 100 m = = 6.67 m>s t 15 s

Ans.

using these results, the s–t graph shown in Fig. a can be plotted.

Ans: s = 5 0.2t3 6 m 1 s = e ( 90t - 3t2 - 275 ) f m 4 s = 100 m

vavg = 6.67 m>s 59

*12–56. A motorcycle starts from rest at s = 0 and travels along a straight road with the speed shown by the v–t graph. Determine the total distance the motorcycle travels until it stops when t = 15 s. Also plot the a–t and s–t graphs.

v (m/s)

v 1.25t

v 5 v t 15

Solution

t (s)

For t 6 4 s a = L0

dv = 1.25 dt

ds =

1.25 t dt

s = 0.625 t 2 When t = 4 s, s = 10 m For 4 s 6 t 6 10 s a =

dv = 0 dt

L10

ds =

5 dt

s = 5 t - 10 When t = 10 s, s = 40 m For 10 s 6 t 6 15 s a =

dv = -1 dt

L40

ds =

(15 - t) dt L10

s = 15 t - 0.5 t 2 - 60 Ans.

When t = 15 s, s = 52.5 m

Ans: When t = 15 s, s = 52.5 m 60

12–57. A motorcycle starts from rest at s = 0 and travels along a straight road with the speed shown by the v–t graph. Determine the motorcycle’s acceleration and position when t = 8 s and t = 12 s.

v (m/s)

v 1.25t

v 5 v t 15

Solution

t (s)

At t = 8 s a =

dv = 0 dt

Ans.

∆ s = 1 v dt s - 0 =

1 (4)(5) + (8 - 4)(5) = 30 2 Ans.

s = 30 m At t = 12 s a =

dv -5 = = -1 m>s2 dt 5

Ans.

∆ s = 1 v dt s - 0 =

1 1 1 3 3 (4)(5) + (10 - 4)(5) + (15 - 10)(5) - a b(5)a b(5) 2 2 2 5 5

s = 48 m

Ans.

Ans: At t = 8 s, a = 0 and s = 30 m. At t = 12 s, a = - 1 m>s2 and s = 48 m. 61

12–58. Two cars start from rest side by side and travel along a straight road. Car A accelerates at 4 m>s2 for 10 s and then maintains a constant speed. Car B accelerates at 5 m>s2 until reaching a constant speed of 25 m/s and then maintains this speed. Construct the a–t, v–t, and s–t graphs for each car until t = 15 s. What is the distance between the two cars when t = 15 s?

SOLUTION Car A: v = v0 + ac t vA = 0 + 4t At t = 10 s,

vA = 40 m>s

s = s0 + v0t +

1 2 at 2 c

sA = 0 + 0 +

1 (4)t2 = 2t2 2

At t = 10 s,

sA = 200 m

t 7 10 s,

ds = v dt sA

L200

ds =

L10

40 dt

sA = 40t - 200 At t = 15 s,

sA = 400 m

Car B: v = v0 + a c t vB = 0 + 5t When vB = 25 m/s,

t =

25 = 5s 5

s = s0 + v0t + sB = 0 + 0 +

1 2 at 2 c

1 (5)t2 = 2.5t2 2

When t = 10 s, vA = (vA)max = 40 m/s and sA = 200 m. When t = 5 s, sB = 62.5 m. When t = 15 s, sA = 400 m and sB = 312.5 m.

12–58.

Continued

At t = 5 s,

sB = 62.5 m

t 7 5 s,

ds = v dt sB

L62.5

ds =

25 dt

sB - 62.5 = 25t - 125 sB = 25t - 62.5 When t = 15 s,

sB = 312.5

Distance between the cars is ¢s = sA - sB = 400 - 312.5 = 87.5 m

Ans.

Car A is ahead of car B.

Ans: When t = 5 s, sB = 62.5 m. When t = 10 s, vA = (vA)max = 40 m>s and sA = 200 m. When t = 15 s, sA = 400 m and sB = 312.5 m. ∆s = sA - sB = 87.5 m 63

12–59. A particle travels along a curve defined by the equation S (T3 3T2 2T) m. where T is in seconds. Draw the S T, V T, and A T graphs for the particle for 0 T 3 s.

SOLUTION S T3

3T2

DS 3T2 DT

Dv 6T dt

v 0 at 0 3T2

T 1.577 s, and T 0.4226 s, S|T 1.577

0.386 m

S|T 0.4226 0.385 m

*12–60. The speed of a train during the first minute has been recorded as follows: t (s) 0 20 40 60 0 16 21 24 v (m>s) Plot the v-t graph, approximating the curve as straight-line segments between the given points. Determine the total distance traveled.

SOLUTION The total distance traveled is equal to the area under the graph. sT =

1 1 1 (20)(16) + (40 - 20)(16 + 21) + (60 - 40)(21 + 24) = 980 m 2 2 2

Ans.

Ans: sT = 980 m 65

12–61. A two-stage rocket is fired vertically from rest with the acceleration shown. After 15 s the first stage A burns out and the second stage B ignites. Plot the v–t and s–t graphs which describe the motion of the second stage for 0 … t … 40 s.

a (m/s2) B A 20 15

Solution

t (s)

For 0 … t 6 15 a = t L0

dv =

v =

1 2 t 2

t dt

v = 112.5 when t = 15 s L0

ds =

L0 2

1 2 t dt

1 3 t 6 s = 562.5 when t = 15 s s =

For 15 6 t 6 40 a = 20 v

L112.5

dv =

L1.5

20 dt

v = 20t - 187.5 v = 612.5 when t = 40 s s

L562.5

ds =

(20 t - 187.5) dt L15

s = 10 t 2 - 187.5 t + 1125 s = 9625 when t = 40 s

Ans: For 0 … t 6 15 s, 1 v = e t 2 f m>s 2 1 s = e t3 f m 6 For 15 s 6 t … 40 s, v = {20t - 187.5} m>s s = {10t 2 - 187.5t + 1125} m 66

12–62. The motion of a train is described by the a–s graph shown. Draw the v–s graph if v = 0 at s = 0.

a (m/s2)

300

600

s (m)

Solution v–s Graph. The v- s function can be determined by integrating v dv = a ds. 3 1 For 0 … s 6 300 m, a = a bs = a s b m>s2. Using the initial condition 300 100 v = 0 at s = 0, L0

v dv =

1 2 v2 = s 2 200

1 s b ds 100

1 s f m>s 10

v = e

At s = 300 m,

Ans.

1 (300) = 30 m>s 10 a - 3 0 - 3 1 For 300 m 6 s … 600 m, = ;a = es + 6 f m>s2, using the s - 300 600 - 300 100 initial condition v = 30 m>s at s = 300 m, v s = 300 m =

L30 m>s

v dv =

L300 m

2 v

1 s + 6b ds 100

s 1 2 v ` = as + 6 sb ` 2 30 m>s 200 300 m

v2 1 2 - 450 = 6 s s - 1350 2 200

1 2 s - 1800 f m>s 100

v = e

v =

12 (600) -

At s = 600 m,

12s -

Ans.

1 ( 6002 ) - 1800 = 42.43 m>s 100

Using these results, the v - s graph shown in Fig. a can be plotted.

Ans: v = e 67

1 s f m>s 10

12–63. If the position of a particle is defined as s = (5t - 3t2) m, where t is in seconds, construct the s–t, v–t, and a–t graphs for 0 … t … 2.5 s.

Solution s–t Graph. When s = 0, 0 = 5t - 3t 2

0 = t15 - 3t2

t = 0

and

t = 1.675

At t = 10 s, s t = 2 s = 512.52 - 312.52 2 = - 6.25 m

ds v–t Graph. The velocity function can be determined using v = , Thus dt ds v = = 55 - 6t6 m/s dt

At t = 0, v t = 0 = 5 - 6102 = 5 m/s

At t = 2.5 s, v t = 2.5 s = 5 - 612.52 = - 10 m/s when v = 0, 0 = 5 - 6t

t = 0.833 s

Then, at t = 0.833 s, s t = 0.833 s = 510.8332 - 310.8332 2 = 2.08 m a–t Graph. The acceleration function can be determined using a = a =

dv = - 6 m/s2 dt

dv . Thus dt

Ans.

s–t, v–t and a–t graph shown in Fig. a, b, and c can be plotted using these results.

Ans: v = {5 - 6t} m>s a = - 6 m>s 2 68

*12–64. From experimental data, the motion of a jet plane while traveling along a runway is defined by the v–t graph. Construct the s–t and a–t graphs for the motion. When t = 0, s = 0.

v (m/s) 60

SOLUTION s - t Graph: The position in terms of time t can be obtained by applying ds 20 . For time interval 0 s … t 6 5 s, y = y = t = (4t) m>s. dt 5

t (s)

ds = ydt s

ds =

4tdt

s = A 2t2 B m s = 2 A 52 B = 50 m,

When t = 5 s, For time interval 5 s 6 t 6 20 s,

ds = ydt s

L50 m

ds =

L5 a

20dt

s = (20t - 50) m When t = 20 s,

s = 20(20) - 50 = 350 m

For time interval 20 s 6 t … 30 s,

60 - 20 y - 20 , y = (4t - 60) m>s. = t - 20 30 - 20 ds = ydt

L350 m

ds =

L20 a

(4t - 60) dt

s = A 2t2 - 60t + 750 B m When t = 30 s,

s = 2 A 302 B - 60(30) + 750 = 750 m

a - t Graph: The acceleration function in terms of time t can be obtained by dy applying a = . For time interval 0 s … t 6 5 s, 5 s 6 t 6 20 s and dt dy dy dy 20 s 6 t … 30 s, a = = 4.00 m s2, a = = 0 and a = = 4.00 m s2, dt dt dt respectively.

Ans: For 0 … t 6 5 s, s = 5 2t 2 6 m and a = 4 m>s2. For 5 s 6 t 6 20 s, s = {20t - 50} m and a = 0. For 20 s 6 t … 30 s, s = 5 2t 2 - 60t + 750 6 m and a = 4 m>s2.

12–65. The v–s graph for a test vehicle is shown. Determine its acceleration when s = 100 m and when s = 175 m.

v (m/s)

SOLUTION

150

0 … s … 150 m:

n =

1 s 3

dn =

1 ds 3

200

s (m)

n dn = a ds 1 1 s a dsb = a ds 3 3 a =

a =

At s = 100 m, 150 m … s … 200 m;

1 s 9

1 (100) = 11.1 m>s2 9

Ans.

n = 200 - s, dn = - ds n dn = a ds

(200 - s)( - ds) = a ds a = s - 200 At s = 175 m,

a = 175 - 200 = - 25 m>s2

Ans.

Ans: At s = 100 m, a = 11.1 m>s2 At s = 175 m, a = -25 m>s2 70

12–66. The a–t graph for a car is shown. Construct the v–t and s–t graphs if the car starts from rest at t = 0. At what time t' does the car stop?

a (m/s2)

SOLUTION a1

5 m>s2

2 m>s2

10 s

t¿ 10

t1 2

t 2

vp1 ( t)

ap2 ( t)

vp2 ( t)

vp1 t1 a2 t t1

sp2 ( t)

sp1 t1 vp1 t1 t t1

Guess

12 s

Given

0 0.01t1 t1

sp1 ( t)

vp2 ( t' )

t 6

ap1 ( t)

1 2 a2 t t1 2

Find ( t' )

22.5 s

t1 1.01t1 t'

Velocity (m/s)

vp1 W 1 20 vp2 W 2

W 1 W 2

Time (s)

Ans.

t (s)

Continued

300

Distance (m)

12–66.

sp1 W 1 200 sp2 W 2

100

W1 W2

Time (s)

Ans: t = 22.5 s 72

12–67. The boat travels along a straight line with the speed described by the graph. Construct the s–t and a-s graphs. Also, determine the time required for the boat to travel a distance s = 400 m if s = 0 when t = 0.

v (m/s)

SOLUTION s t Graph: For 0 … s 6 100 m, the initial condition is s = 0 when t = 0 s. + B A:

ds v

dt = t

v2 s

dt = L0 2s1>2 L0 1>2 t = s

0.2s

20 s (m) 100

s = A t2 B m

400

When s = 100 m, 100 = t2

t = 10 s

Ans.

For 100 m 6 s … 400 m, the initial condition is s = 100 m when t = 10 s. + B A:

ds v

dt = t

ds 0.2s L10 s L100 m s t - 10 = 5ln 100 s t - 2 = ln 5 100 s et>5 - 2 = 100 s et>5 = 100 e2 dt =

s = A 13.53et>5 B m When s = 400 m, 400 = 13.53et>5 t = 16.93 s = 16.9 s

Ans.

The s–t graph is shown in Fig. a. a s Graph: For 0 m … s 6 100 m, a = v

dv = A 2s1>2 B A s - 1>2 B = 2 m>s2 ds

For 100 m 6 s … 400 m, a = v

dv = (0.2s)(0.2) = 0.04s ds Ans: When s = 100 m, t = 10 s. When s = 400 m, t = 16.9 s. a s = 100 m = 4 m>s2 a s = 400 m = 16 m>s2

When s = 100 m and 400 m, a s = 100 m = 0.04(100) = 4 m>s2 a s = 400 m = 0.04(400) = 16 m>s2 The a–s graph is shown in Fig. b. 73