Now suppose that at the end of its trip, the train in Figure 2.18(a) slows to a stop from a speed of 30.0 km/h in 8.00s. What is its average acceleration while stopping?

Data given in Figure 2.18

 

Strategy

In this case, the train is decelerating and its acceleration is negative because it is toward the left. As in the previous example, we must find the change in velocity and the change in time and then solve for acceleration.

Transcribed Image Text:

y 0 (b) Ax ¹x = 3.75 km Ax' = -1.50 km = 2.00 km 1x0 = 4.70 km ¹x6 = 5.25 km x = 6.70 km x (km) -x (km)