Question

The speed of a particle moving in a circle of radius r = 2 m varies with time t as $v=t^2$ where t is in second and v in m/s. The net acceleration at t =2 s is

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Linear speed of particle at t = 2 s is
$v=(2{)}^2=4m/s$
$\therefore$ Radial acceleration $a_r=\frac{v^2}{r}=\frac{(4{)}^2}{2}=8m/s^2$
The tangential acceleration is $a_t=\frac{dv}{dt}=2t$
$\therefore$ Tangential aceleration at t = 2 s is $a_t=\left(2\right)\left(2\right)=4m/s^2$
$\therefore$ Net acceleration of particle at t = 2 s is
$a=\sqrt{(a_r{)}^2+(a_t{)}^2}=\sqrt{(8{)}^2+(4{)}^2}a=\sqrt{80}m/s^2$