Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging

             (a) from A to B and

              (b) from A to C?

Answers (1)

Given,

(a) Distance between A and B = 300\ m

Time taken to reach from A to B = 2\ min\ 30\ s = 150\ s 

\therefore Average speed from A to B = \frac{Distance}{Time} = \frac{300}{150} = 2\ ms^{-1}

And, Average velocity from A to B = \frac{Displacement}{Time} = \frac{300}{150} = 2\ ms^{-1}

(In this case, average speed is equal to the average velocity)

(b) Distance travelled from A to reach C = AB + BC = 300 + 100 = 400\ m 

And, Displacement from A to C = AC= 300-100 = 200\ m

Also, time taken to reach C from A = 2\ min\ 30\ s + 1\ min= (150+60)\ s =210\ s

\therefore Average speed from A to C = \frac{Distance}{Time} = \frac{400}{210} = 1.90\ ms^{-1}

And, Average velocity from A to C = \frac{Displacement}{Time} = \frac{200}{210} = 0.95\ ms^{-1}

(In this case, average speed is not equal to the average velocity)

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