Question

A car is moving on a circular path with a constant speed of $10m/s$. What is the magnitude of change in velocity when it moves through ${120}^{\circ }$?

• A. $0$
• B. $10\sqrt{3}m/s$
• C. $10m/s$
• D. $10\sqrt{2}m/s$

Answer 1

The correct option is B $10\sqrt{3}m/s$

Let the initial position and final position be A and B respectively.
Velocity at initial point A = $v_A$
Velocity at final point B = $v_B$
Angle between $v_A$ and $v_B$ is $\theta ={120}^0$
Change in velocity
$\Delta v=|v_B-v_A|=\sqrt{v_B^2+v_A^2-2v_B{v}_A\cos \theta }$
Since speed is constant $\therefore v_A=v_B=v=10m/s$
$\Delta v=\sqrt{2v^2}\times \sqrt{1-\cos \theta }=v\sqrt{2}\times \sqrt{2{\sin }^2\frac{\theta }{2}}$$\Delta v=2v\sin \frac{\theta }{2}=2\times 10\times \sin {60}^0\Delta v=10\sqrt{3}m/s$