An Accelerating Subway Train

Again, make a simple sketch.

Figure 3.5

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

1. Identify the knowns: v₀ = 30.0 km/h; v_f = 0; and Δt = 8.00 s.
2. Convert the units. From the first problem, we know that 30.0 km/h = 8.333 m/s.
3. Calculate change in velocity, $\Delta v=v_{\text{f}}-v_0=0-8.333\text{m/s}=-8.333\text{m/s,}$ where the minus sign means that the change in velocity points to the left.

4. We know Δt = 8.00 s, so all we have to do is insert the known values into the equation for average acceleration.

   $$\overline{a}=\frac{\Delta v}{\Delta t}=\frac{-8.333\text{m/s}}{8.00\text{s}}=-1.04\frac{\text{m}}{{\text{s}}^{\text{2}}}$$

   3.3