Question

A boy starts towards the east with a uniform speed $5m/s$. After $t=2$ second he turns right and travels $40m$ with same speed. Again he turns right and travels for $8$ second with the same speed. Find out the displacement, average speed, average velocity, and total distance traveled.

Answer 1

REF.Image.

Distance covered in $25=v.t=10m$ $v=5m/s$

$85=v.t=40m$

His path :

Taking A are origin

$A\equiv (0,.0)$

$D=(-30,-40)$

$(t=\frac{d}{v}=\frac{40}{5}=8s)$

(1)Displacement $=d\left(AD\right)=\sqrt{{40}^2+{30}^2}=\sqrt{1600+900}=\sqrt{2500}=50m$

(ii) Total distance $=10+40+40=90m$

(iii) Avg. speed $=\frac{TotalDistance}{TotalTime}=\frac{90m}{2+8+8}=\frac{90}{18}=5m/s$

(iv)Avg. velocity $=\frac{TotalDisplacement}{Totaltime}=\frac{50}{18}=2.78m/s$ in $\overline{AD}direction$