Question

A particle moves in a circle with a constant speed of $10m/s$. the magnitude of change in the velocity when the angular displacement of the particle is $60º$ is _____ $m/s$.

• A. $10m/s$
• B. $\sqrt[3]{10}m/s$
• C. $\sqrt{10}m/s$
• D. Zero

Answer 1

The correct option is A

$10m/s$

Step 1: Given Data

Uniform Circular Motion at a speed,

$v=10m/s$

$\theta =60º$

Step 2: Formula Used

$∆v=2v\times \sin (\theta /2)$

$∆v=finalvelocity-initialvelocity$

$∆v$is the change in velocity, $v$ is the constant speed, and $\theta$ is the angle made at the center.

Step. 3 Calculation of change in velocity,

$∆v=\sqrt{v²+v²+2v²\cos (120°)}$

$\Rightarrow ∆v=\sqrt{2v²(1+\cos 120°)}$

$∆v=\sqrt{2v²(1-½)}$

$∆v=\sqrt{2v²\times ½}$

$$\begin{gathered} ∆v=\sqrt{v²} \\ ∆v=10m/s \end{gathered}$$

Therefore, the change in velocity is $10m/s$.

Hence option A is the correct answer