Acceleration

Example 2.5 Calculate Acceleration: A Subway Train Slowing Down

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 8.00 s. What is its average acceleration while stopping?

Strategy

Figure 2.20

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.

Solution

1. Identify the knowns. $v_{0}=30.0 \mathrm{~km} / \mathrm{h}, v_{\mathrm{f}}=0 \mathrm{~km} / \mathrm{h}$ (the train is stopped, so its velocity is 0), and $\Delta t=8.00 \mathrm{~s}$

2. Solve for the change in velocity, $\Delta v$.

$\Delta v=v_{\mathrm{f}}-v_{0}=0-30.0 \mathrm{~km} / \mathrm{h}=-30.0 \mathrm{~km} / \mathrm{h}$

3. Plug in the knowns, $\Delta v$ and $\Delta t$, and solve for $\bar{a}$.

$\bar{a}=\frac{\Delta v}{\Delta t}=\frac{-30.0 \mathrm{~km} / \mathrm{h}}{8.00 \mathrm{~s}}$

4. Convert the units to meters and seconds.

$\bar{a}=\frac{\Delta v}{\Delta t}=\left(\frac{-30.0 \mathrm{~km} / \mathrm{h}}{8.00 \mathrm{~s}}\right)\left(\frac{10^{3} \mathrm{~m}}{1 \mathrm{~km}}\right)\left(\frac{1 \mathrm{~h}}{3600 \mathrm{~s}}\right)=-1.04 \mathrm{~m} / \mathrm{s}^{2}$

Discussion

The minus sign indicates that acceleration is to the left. This sign is reasonable because the train initially has a positive velocity in this problem, and a negative acceleration would oppose the motion. Again, acceleration is in the same direction as the change in velocity, which is negative here. This acceleration can be called a deceleration because it has a direction opposite to the velocity.