Question

A particle is moving in a circle in a radius $r$ with a constant speed $v$ . The change in velocity after the particle has traveled a distance equal to $(\frac{1}{8}{)}^{th}$of the circumference of the circle is:

• A. zero
• B. $0.5V$
• C. $0.765V$
• D. $0.125V$

Answer 1

The correct option is C $0.765V$

Let initial position of particle be A velocity be
${\overrightarrow{v}}_A=v\hat{i}+0\hat{j}$
After it has transversed ${\frac{1}{8}}^{th}$ of circle $\Rightarrow (\frac{{360}^0}{8}={45}^0)$
So, $\theta ={45}^0$
So, now velocity is,${\overrightarrow{v}}_B=vcos{45}^0\hat{i}-vsin{45}^0\hat{j}$
So, change in velocity in x direction$=vcos{45}^0-v$
$=-0.292v$
Change in velocity in y direction $=-vsin{45}^0-0$
$=-0.707v$
Net change in velocity$=-0.292v\hat{i}-0.707v\hat{j}$
$=\sqrt{(0.292{)}^2+(0.707{)}^2}=0.765V$