Question

A man runs 200 m in first 25 s and then turns back and run 100 m in next 15 s towards the starting point. Find out his
(a) average velocity and (b) his average speed.

Answer 1

Given,

In first event, displacement $S=200m$ in time $t_1=25\sec$

In second event, displacement $S=-100m$ in time $t_1=15\sec$

Average velocity$=\frac{Totaldisplacement}{totaltime}=\frac{S+(-S)}{t_1+t_2}=\frac{200+(-100)}{25+15}=2.5m{s}^{-1}$

Average speed$=\frac{Totaldistance}{totaltime}=\frac{S+S}{t_1+t_2}=\frac{200+100}{25+15}=10m{s}^{-1}$

Hence, Average velocity $2.5m{s}^{-1}$ and average speed is $10m{s}^{-1}$