Acceleration

Example 2.7 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?

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. $v_{0}=-20 \mathrm{~km} / \mathrm{h}, v_{\mathrm{f}}=0 \mathrm{~km} / \mathrm{h}, \Delta t=10.0 \mathrm{~s}$

2. Calculate $\Delta v$. The change in velocity here is actually positive, since

$\Delta v=v_{\mathrm{f}}-v_{0}=0-(-20 \mathrm{~km} / \mathrm{h})=+20 \mathrm{~km} / \mathrm{h}$

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

$\bar{a}=\frac{\Delta v}{\Delta t}=\frac{+20.0 \mathrm{~km} / \mathrm{h}}{10.0 \mathrm{~s}}$

4. Convert units.

$\bar{a}=\left(\frac{+20.0 \mathrm{~km} / \mathrm{h}}{10.0 \mathrm{~s}}\right)\left(\frac{10^{3} \mathrm{~m}}{1 \mathrm{~km}}\right)\left(\frac{1 \mathrm{~h}}{3600 \mathrm{~s}}\right)=+0.556 \mathrm{~m} / \mathrm{s}^{2}$

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.