that are under the graph. From time t = 0 to t = 2, we need to find the area of a triangle with a
base of 2 and a height of 20. The area is 20. Next, from time t = 2 to t = 4, we need to find the
area of a rectangle with a base of 2 and a height of 20. The area is 40. Next, from time t = 4 to t
= 8, we need to find the area of a trapezoid with bases of 20 and 30 and a height of 4. The area
is 100. Next, from time t = 8 to t = 12, we need to find the area of a rectangle with a base of 4
and a height of 30. The area is 120. Finally, from time t = 12 to t = 16, we need to find the area
of a triangle with a base of 4 and a height of 30. The area is 60. Thus, the total distance that
train A traveled is 340 km.
Let’s repeat the process for train B. From time t = 0 to t = 4, we need to find the area of a
triangle with a base of 4 and a height of 40. The area is 80. Next, from time t = 4 to t = 10, we
need to find the area of a rectangle with a base of 6 and a height of 40. The area is 240. Finally,
from time t = 10 to t = 16, we need to find the area of a triangle with a base of 6 and a height of
- The area is 120. Thus, the total distance that train B traveled is 440 km.
Therefore, the positive difference between their distances is 100 km.
(d) Find the total distance that C traveled in 16 hours.
First, note that the graph of train C ’s velocity, v(t) = 8t − 0.25t^2 , is above the x-axis on the
entire interval. Therefore, in order to find the total distance traveled, we integrate v(t) over the
interval. We get
(8t − 0.25t^2 ) dt
Evaluate the integral: (8t − 0.25t^2 ) dt = .
5.
