Question

A particle moves in the direction of east for 2s with velocity of $14m{s}^{-1}$. Then it moves towards north for 8s with a velocity of $5m{s}^{-1}$ . The average velocity of the particle is (in $m{s}^{-1}$)

• A. 1
• B. 5
• C. 7
• D. 10

Answer 1

The correct option is B 5

The average velocity is total displacement upon a total time.

$x_f=14\times 2=28m$

$y_f=8\times 5=40m$

$x_i=0,y_i=0$

So total displacement $=\sqrt{{28}^2+{40}^2}=48.8$

The total time is 10sec.

Average velocity $=\frac{48.8}{10}=4.88m/s\approx 5m/s$