Question

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?

Answer 1

a. Total Distance covered from AB = $300$ m
Total time taken = $2 \times 60 + 30$ s
=$150$ s

Therefore, Average Speed from AB = $\frac{Total Distance }{ Total Time}$
=$\frac{300 }{ 150}$ $m s ^{-1}$
=$2$ $m s^{-1}$
Therefore, Velocity from AB =$\frac{Displacement \;AB }{ Time}$ = $\frac{300 }{ 150}$ $m s^{-1}$
=$2 m s^{-1}$
Total Distance covered from AC =AB + BC
= $300 + 200$ m
Total time taken from A to C = Time taken for AB + Time taken for BC
= $(2 \times 60+30)+60$ $s$
= $210$ $s$
Therefore, Average Speed from AC = $\frac{Total\, Distance }{Total \,Time}$
=$\frac{ 400 }{210}$ $m s^{-1}$
= $1.904$ $m s^{-1}$

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b. Displacement (S) from A to C = AB - BC
= $300-100$ m = $200$ m
Time (t) taken for displacement from AC = $210$ $s$
Therefore, Velocity from AC = $\frac{Displacement (s) }{ Time(t)}$
= $\frac{200 }{ 210}$ $m s^{-1}$
= $0.952$ $m s^{-1}$