Question

A person walks a distance of $30m$ along west with a speed of $2m/s$ and $40m$ along north with a speed of $4m/s$. Find his average speed and average velocity

• A. $2.8m/s,2m/s$
• B. $3m/s,4m/s$
• C. $5m/s,2.8m/s$
• D. $1.8m/s,2.4m/s$

Answer 1

The correct option is A $2.8m/s,2m/s$

Time during motion in west is $t_1=\frac{30m}{2m/s}=15s$

Time during motion in north is $t_2=\frac{40m}{4m/s}=10s$

net $distance$ is $30m+40m=70m$

so average speed will be $v=\frac{70m}{(15+10)s}=2.8m/s$

Displacement is a vector quantity so the net displacement will be the $vector$ sum.

North and West are $perpendicular$ to each other so the net displacement will be $x=\sqrt{{30}^2+{40}^2}=50m$

so magnitude of the average velocity will be $V=\frac{50m}{25s}=2m/s$