Question

In a situation shown below, if the speed of the tip of the seconds hand is $v\text{m/s}$, find the magnitude of change in its velocity.

• A. $1.732v\text{m/s}$
• B. $1.414v\text{m/s}$
• C. $0.707v\text{m/s}$
• D. $v\text{m/s}$

Answer 1

The correct option is A $1.732v\text{m/s}$

For one complete revolution of second's hand, $\theta =2\pi \text{radians}$.
$\therefore \theta$ for $20\text{s}=\frac{2\pi }{60}\times 20=\frac{2\pi }{3}\text{radians}$.

Thus, the magnitude of change in velocity is given by
$|\overrightarrow{v_2}-\overrightarrow{v_1}|=\sqrt{v_2^2+v_1^2-2v_2{v}_1\cos \theta }$
Given $v_1=v_2=v$
So, $|\overrightarrow{v_2}-\overrightarrow{v_1}|=\sqrt{v^2+v^2-2v^2\cos \frac{2\pi }{3}}=\sqrt{3}v=1.732v\text{m/s}$