Question

A jogger jogs from one end to the other of a straight track in 1.17 min and then back to the starting point in 1.67 min. What is the jogger’s average speed in jogging to the far end of the track (assuming the track is 100 m long) in m/s?

Answer 1

1.21 m/s

Explanation:

Given:

One way time = 1.17 min

Return time = 1.67 min

1 minute is 60 seconds, therefore:

One way time = 1.17 min = 1.17 * 60 = 70.2 sec

Return time = 1.67 min * 60 = 100.2 sec

We calculate the speed for each journey, knowing that the distance traveled is 100 meters, like this:

`\begin{gathered} v=(d)/(t) \\ \\ v_1=(d)/(t_1)=(100)/(70.2)=1.4245\text{ m/s} \\ \\ v_2=(d)/(t_2)=(100)/(100.2)=0.998\text{ m\/s} \end{gathered}`

Therefore, the average speed would be:

`\begin{gathered} v=(v_1+v_2)/(2)=(1.4245+0.998)/(2)=(2.4225)/(2) \\ \\ v=1.21\text{ m/s} \end{gathered}`

The average speed is 1.21 m/s