Question

A car moves on a circular path of radius 5 m. At one instant the speed of car is $5m/s$ and it is decreasing at a rate of $5m/s^2$. The angle between acceleration vector and velocity vector at this instant is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{2\pi }{3}$
• D. $\frac{3\pi }{4}$

Answer 1

The correct option is D $\frac{3\pi }{4}$

Centripetal acceleration $a_c=\frac{v^2}{r}=\frac{25}{5}=5m/s^2$

the tangential acceleration is given as $a_T=5m/s^2$

so the net acceleration will be $\sqrt[2]{25^2+5^2}$ at angle $\pi /4$ to the $a_T$

so angle between this net acceleration and velocity vector will be $\pi -\frac{\pi }{4}=3\pi /4$