Question

A scooter going due east at $10m{s}^{-1}$ turns right through an angle of 90°. If the speed of the scooter remains unchanged in taking turn, the change is the velocity of the scooter is

• A. $20.0m{s}^{-1}$ south eastern direction
• B. Zero
• C. $10.0m{s}^{-1}$ in southern direction
• D. $14.14m{s}^{-1}$ in south-west direction

Answer 1

The correct option is D

$14.14m{s}^{-1}$ in south-west direction

If the magnitude of vector remains same, only direction change by $\theta$ then
$\overrightarrow{\Delta v}=\overrightarrow{v_2}-\overrightarrow{v_1},\overrightarrow{\Delta v}=\overrightarrow{v_2}+\left(\overrightarrow{-v_1}\right)$
Magnitude of change in vector $|\overrightarrow{\Delta v}|=2vsin\left(\frac{\theta }{2}\right)$
$|\overrightarrow{\Delta v}|=2\times 10\times sin\left(\frac{{90}^{\circ }}{2}\right)=10\sqrt{2}=14.14m/s$
Direction is south-west as shown in figure.