Answer to Question #236131 in Physics for Cobra

Question

A cross-country runner runs 5 km east along a course then turns around and runs 5 km west along the same path. She returns to her starting point in 36 minutes. What is her average speed in m/s? What is the magnitude of her average velocity in m/s?

Answer 1

(a) The average speed of the runner is defined as the total distance traveled by the runner divided by the total time:

$$v_{avg}=\dfrac{d_{tot}}{t_{tot}}=\dfrac{5000\ m+5000\ m}{36\ min\cdot\dfrac{60\ s}{1\ min}}=4.63\ \dfrac{m}{s}.$$

(b) The average velocity of the runner is defined as the displacement per unit time:

$$\vec{v}_{avg}=\dfrac{Displacement}{Time\ taken}.$$

Because the runner returns to her starting point, the displacement is equal to zero and

we get:

$$\vec{v}_{avg}=\dfrac{Displacement}{Time\ taken}=\dfrac{0\ m}{36\ min\cdot\dfrac{60\ s}{1\ min}}=0\ \dfrac{m}{s}.$$