Acceleration

Example 2.6 Calculating Average Velocity: The Subway Train

What is the average velocity of the train in part b of Example 2.2, and shown again below, if it takes 5.00 min to make its trip?

Figure 2.22

Strategy

Average velocity is displacement divided by time. It will be negative here, since the train moves to the left and has a negative displacement.

Solution

1. Identify the knowns. \(x_{\mathrm{f}}^{\prime}=3.75 \mathrm{~km}, x_{0}^{\prime}=5.25 \mathrm{~km}, \Delta t=5.00 \mathrm{~min}\).

2. Determine displacement, \(\Delta x^{\prime}\) '. We found \(\Delta x\) ' to be \(-1.5 \mathrm{~km}\) in Example 2.2.

3. Solve for average velocity.

\(\bar{v}=\frac{\Delta x^{\prime}}{\Delta t}=\frac{-1.50 \mathrm{~km}}{5.00 \mathrm{~min}}\)

4. Convert units.

\(\bar{v}=\frac{\Delta x^{\prime}}{\Delta t}=\left(\frac{-1.50 \mathrm{~km}}{5.00 \mathrm{~min}}\right)\left(\frac{60 \mathrm{~min}}{1 \mathrm{~h}}\right)=-18.0 \mathrm{~km} / \mathrm{h}\)

Discussion

The negative velocity indicates motion to the left.