Question

A particle is moving on a circular path of radius r with uniform speed V. The magnitude of change in velocity when the particle moves from P to Q is :- $(\angle PPOQ={40}^o)$

• A. $2vcos{40}^o$
• B. $2vsin{40}^o$
• C. $2vsin{20}^o$
• D. $2vcos{20}^o$

Answer 1

The correct option is C $2vsin{20}^o$

Let the velocity at point P

$\overrightarrow{v_1}=v\hat{i}$

So the velocity at point Q will be

$\overrightarrow{v_2}=vcos{40}^0\hat{i}-vsin{40}^0\hat{j}$

Change in velocity

$\overrightarrow{\Delta v}=\overrightarrow{v_2}-\overrightarrow{v_1}=(vcos{40}^0\hat{i}-vsin{40}^0\hat{j})-v\hat{i}$

Magnitude of change of velocity

$|\Delta v|=\sqrt{(vcos{40}^0-v{)}^2+(-vsin{40}^0{)}^2}$

On solving we get

$|\Delta v|=2vsin{20}^0$