Time, Velocity, and Speed

Now, let's return to this text which you read in Section 2.2, to study the calculations in more detail. Pay attention to Figure 2.10 and Figure 2.11 which show examples of how to calculate displacement, distance, average speed, and velocity.

Check Your Understanding

A commuter train travels from Baltimore to Washington, DC, and back in 1 hour and 45 minutes. The distance between the two stations is approximately 40 miles. What is (a) the average velocity of the train, and (b) the average speed of the train in m/s?

Solution

(a) The average velocity of the train is zero because x_{\mathrm{f}}=x_{0}; the train ends up at the same place it starts.

(b) The average speed of the train is calculated below. Note that the train travels 40 miles one way and 40 miles back, for a total distance of 80 miles.

\begin{gathered} \frac{\text { distance }}{\text { time }}=\frac{80 \text { miles }}{105 \text { minutes }} \\ \\ \frac{80 \text { miles }}{105 \text { minutes }} \times \frac{5280 \text { feet }}{1 \text { mile }} \times \frac{1 \text { meter }}{3.28 \text { feet }} \times \frac{1 \text { minute }}{60 \text { seconds }}=20 \mathrm{~m} / \mathrm{s} \end{gathered}