Joseph jogs from one end A to the other end B | vedclass

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Question

(C) Total distance covered from $A$ to $C = AB + BC = 300 \, m + 100 \, m = 400 \, m$.Total time taken from $A$ to $C = 	ext{Time for } AB + 	ext{Ti

Vedclass · Accepted Answer

$1.904 \, m s^{-1}$ and $0.952 \, m s^{-1}$

Vedclass · Answer

(C) Total distance covered from $A$ to $C = AB + BC = 300 \, m + 100 \, m = 400 \, m$.Total time taken from $A$ to $C = 	ext{Time for } AB + 	ext{Time for } BC = (2 	imes 60 + 30) \, s + 60 \, s = 150 \, s + 60 \, s = 210 \, s$.Average speed = $	ext{Total distance} / 	ext{Total time} = 400 \, m / 210 \, s \approx 1.904 \, m s^{-1}$.Displacement from $A$ to $C = AB - BC = 300 \, m - 100 \, m = 200 \, m$.Average velocity = $	ext{Displacement} / 	ext{Total time} = 200 \, m / 210 \, s \approx 0.952 \, m s^{-1}$.

Joseph jogs from one end $A$ to the other end $B$ of a straight $300 \, m$ road in $2 \, min$ $30 \, sec$ and then turns around and jogs $100 \, m$ back to point $C$ in another $1 \, min$. What are Joseph's average speeds and velocities in jogging from $A$ to $C$?

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