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

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Question

(B) The distance covered from $A$ to $B$ is $300 \, m$.The time taken to travel from $A$ to $B$ is $2 \, min \, 30 \, sec = (2 	imes 60) + 30 = 150 \

Vedclass · Accepted Answer

$2 \, m s^{-1}$ and $2 \, m s^{-1}$

Vedclass · Answer

(B) The distance covered from $A$ to $B$ is $300 \, m$.The time taken to travel from $A$ to $B$ is $2 \, min \, 30 \, sec = (2 	imes 60) + 30 = 150 \, s$.Average speed from $A$ to $B$ is defined as the total distance divided by the total time taken:$	ext{Average Speed} = \frac{	ext{Total Distance}}{	ext{Total Time}} = \frac{300 \, m}{150 \, s} = 2 \, m s^{-1}$.Velocity from $A$ to $B$ is defined as the displacement divided by the total time taken. Since the motion is in a straight line from $A$ to $B$,the displacement is equal to the distance $(300 \, m)$:$	ext{Velocity} = \frac{	ext{Displacement}}{	ext{Total Time}} = \frac{300 \, m}{150 \, s} = 2 \, m s^{-1}$.Thus,the average speed and velocity from $A$ to $B$ are both $2 \, m s^{-1}$.

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 $B$?

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