Question

A particle describes a angle of 60° in a circular path with const speed of 5m/s . The change in velocity of the particle

Answer 1

Answer: The change in velocity of the particle is 5 m/s.

Explanation:

Given that,

Angle $\theta = 60^{\circ}$

Constant speed v = 5 m/s

Let us considered initial velocity $v_{1}$ and final velocity $v_{2}$

The change in velocity will be

$\Delta V= \sqrt{v_{1}^2+v_{2}^2-2v_{1}v_{2}\cos60^{\circ}}$

The particle moves with constant speed so, the initial velocity and final velocity is equal

$v_{1}=v_{2}= 5\ m/s$

Then, the change in velocity will be

$\Delta V = \sqrt{5^2+5^2-2\times5\times5\cos 60^{\circ}}$

$\Delta V = \sqrt{25+25-25}$

$\Delta V= 5 m/s$

Hence, The change in velocity of the particle is 5 m/s.