Question #236131

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?


1
Expert's answer
2021-09-12T19:00:13-0400

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


vavg=dtotttot=5000 m+5000 m36 min60 s1 min=4.63 ms.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:


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

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

we get:


vavg=DisplacementTime taken=0 m36 min60 s1 min=0 ms.\vec{v}_{avg}=\dfrac{Displacement}{Time\ taken}=\dfrac{0\ m}{36\ min\cdot\dfrac{60\ s}{1\ min}}=0\ \dfrac{m}{s}.

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