Time, Velocity, and Speed

Read this text which explains what we mean when we talk about instantaneous versus average time, velocity, and speed. Note that we will explore the calculations you see in this text in detail in Section 2.4 below. For now, make sure you understand what these concepts mean before we begin discussing distance and displacement next.

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}