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. $20.0m{s}^{-1}$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

Step 1. Given data:

The velocity of the scooter given is $10m{s}^{-1}$.

The angle between $\overrightarrow{v_1}$ and $\overrightarrow{v_2}$ is ${90}^{\circ }$.

$v_1$and $v_2$ are initial and final velocities.

Step 2. Formula

$\begin{array}{rcl}\overrightarrow{∆v}& =& \overrightarrow{v_2}-\overrightarrow{v_1}\\ & =& \overrightarrow{v_2}+\left(-\overrightarrow{v_1}\right)\\ \overrightarrow{∆v}& =& \sqrt{{v_1}^2+{v_2}^2}\end{array}$

Step 3. Find the change in velocity of the scooter:

$\begin{array}{rcl}\overrightarrow{∆v}& =& \overrightarrow{v_2}-\overrightarrow{v_1}\\ & =& \overrightarrow{v_2}+\left(-\overrightarrow{v_1}\right)\\ \overrightarrow{∆v}& =& \sqrt{{v_1}^2+{v_2}^2}\\ \overrightarrow{∆v}& =& \sqrt{{\left(10\right)}^2+{\left(10\right)}^2}\\ \overrightarrow{∆v}& =& 10\sqrt{2}\\ \overrightarrow{∆v}& =& 14.14m/s\end{array}$

Hence, the (D) option is correct.