INTEGRAL CALCULUS: AN INTRODUCTION THE DISTANCE PROBLEM In physics, recall that in a velocity-time graph, the area under the graph represents the distance traveled by an object. Why is this so? Recall that the velocity of a body is expressed as a ratio of distance to velocity:
VELOCITY
DISTANCE TIME
=
(1)
Thus, if the velocity is constant, then the distance traveled is given by:
DISTANCE = VELOCITY ×
(2)
TIME
However, if an object has varying velocities, then equation (2) will no longer be applicable. Instead, we find the distance traveled by such object by computing the distance traveled within each time interval (assuming that the velocity attained is approximately equal in each interval). Then we sum up all the individual distances. Consider this illustration:
t (s)
0
0.5
1.0
1.5
2.0
2.5
3.0
v (ft/s)
0
6.2
10.8
14.9
18.1
19.4
20.2
The speed of a runner increased steadily during the first three seconds of a race. Her speed at half-second intervals is given in the table above. Let's assume we want to find the lower and upper estimates for the distance that she traveled during these three seconds. First, we have to ensure that the time and speed have consistent units. In this case, the time is in seconds (s), and the velocity is in feet per second (ft/s). So, we’re okay. Here, the time intervals are equal (0.5s). Thus we say that
∆t = 0.5 and the speeds will be represented by f(tn). The task is to find the distance traveled by the runner in three seconds. On the following page is a graph of the runner’s motion in those three seconds. Therefore, a = t0 = 0 and b = tn = 3. So, we have
t0
=
0.0
t4
=
2.0
t1
=
0.5
t5
=
2.5
t2
=
1.0
t6
=
3.0
t3
=
1.5