Question

A particle moving in uniform circular motion undergoes an angular displacement a (less than ${360}^0$) from point P to Q. Then angle between average velocity from P to Q and velocity at Q is

• A. $\frac{90-\alpha }{2}$
• B. $90-\alpha$
• C. $\alpha$
• D. $\frac{\alpha }{2}$

Answer 1

The correct option is D $\frac{\alpha }{2}$

(Refer to Image)

From the figure,

$\angle OBA={90}^o-\alpha /2$

$\therefore$ Velocity vector at $Q=\frac{\alpha }{2}$