Question

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

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

Answer 1

The correct option is D 14.14 m$s^{-1}$ in south-west direction

If the magnitude of vector remains same, only direction change by $\theta$ then
$\overrightarrow{△}v$ = ${\overrightarrow{v}}_2$ - ${\overrightarrow{v}}_1$, $\overrightarrow{△}v$ = ${\overrightarrow{v}}_2$ + (- ${\overrightarrow{v}}_1$)
Magnitude of change in vector |$\overrightarrow{△}v$| = 2 v sin ($\frac{\theta }{2}$)
|$\overrightarrow{△}v$| = 2 $\times$ 10 $\times$ sin($\frac{{90}^{\circ }}{2}$) = 10 $\sqrt{2}$ = 14.14 m/s
Direction is south - west as shown in figure.