Question

A scooter going due east at 10 ms^–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. 20.0 ms^–1 south eastern direction
• B. Zero
• C. 10.0 ms^–1 in southern direction
• D. 14.14 ms^–1 in south-west direction

Answer 1

Answer: (D)

14.14 ms^–1 in south-west direction

Answer 2

If the magnitude of vector remains same, only direction change by $\theta$ then

$\overset{\rightarrow}{\Delta v} = \overset{\rightarrow}{v_{2}} - \overset{\rightarrow}{v_{1}}$, $\overset{\rightarrow}{\Delta v} = \overset{\rightarrow}{v_{2}} + ( - \overset{\rightarrow}{v_{1})}$

Magnitude of change in vector $|\overset{\rightarrow}{\Delta v}| = 2v\sin\left( \frac{\theta}{2} \right)$

$|\overset{\rightarrow}{\Delta v}| = 2 \times 10 \times \sin\left( \frac{90{^\circ}}{2} \right)$= $10\sqrt{2}$= $14.14m/s$

Direction is south-west as shown in figure.