Question

A scooter going due east at $10m{s}^{-1}$ 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. $10.0m{s}^{-1}$ southeastern direction
• B. $14.14m{s}^{-1}$ in south-west direction
• C. $20.0m{s}^{-1}$ southeastern direction
• D. Zero

Answer 1

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

Draw a labelled diagram.

Find the change is the velocity.
Given,
Velocity of the scooter is $10m/s$
Angle is ${90}^{\circ }$
If the magnitude of vector remains same, only direction change by $\theta$.
Change in velocity of the scooter,
$\overrightarrow{\Delta v}={\overrightarrow{v}}_2-{\overrightarrow{v}}_1$
$\overrightarrow{\Delta v}={\overrightarrow{v}}_2+(-{\overrightarrow{v}}_1)$
$\overrightarrow{\Delta v}=-10\hat{j}+(-10\hat{i})$
Magnitude of change in vector,
$|\left(\overrightarrow{\Delta v}\right)|=\sqrt{v_1^2+v_2^2}=10\sqrt{2}m/s=14.14m/s$
Direction is south-west.
Final Answer: (D)