Calculating Deceleration: The Subway Train

Finally, suppose the train in Figure 2.22 slows to a stop from a velocity of 20.0 km/h in 10.0 s. What is its average acceleration?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
Once again, let’s draw a sketch:

Figure 2.23
As before, we must find the change in velocity and the change in time to calculate average acceleration.
Solution
1. Identify the knowns. $v_{0} = - 20 km/h$ , $v_{f} = 0 km/h$ , $Δ t = 10 . 0 s$ .
2. Calculate $Δ v$ . The change in velocity here is actually positive, since
$Δ v = v_{f} - v_{0} = 0 - (- 20 km/h) = + 20 km/h .$
2.21

3. Solve for $\bar{a}$ .

$\bar{a} = \frac{Δ v}{Δ t} = \frac{+ 20 .0 km/h}{10 . 0 s}$
2.22

4. Convert units.

$\bar{a} = (\frac{+ 20 . 0 km/h}{10 . 0 s}) (\frac{10^{3} m}{1 km}) (\frac{1 h}{3600 s}) = + 0 .556 m {/s}^{2}$
2.23
Discussion
The plus sign means that acceleration is to the right. This is reasonable because the train initially has a negative velocity (to the left) in this problem and a positive acceleration opposes the motion (and so it is to the right). Again, acceleration is in the same direction as the change in velocity, which is positive here. As in Example 2.5, this acceleration can be called a deceleration since it is in the direction opposite to the velocity.

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