Question

A particle is moving on a circular path of radius $1m$ and its speed is changing as $v=2t$. The magnitude of the acceleration of particle at time $t=1s$ is

• A. 4 $m/s^2$
• B. 2 $m/s^2$
• C. 2$\sqrt{3}$ $m/s^2$
• D. 2$\sqrt{5}$ $m/s^2$

Answer 1

The correct option is D 2$\sqrt{5}$ $m/s^2$

$a_t=\frac{dv}{dt}=2m/s^2$
$a_c(att=1s)=\frac{v^2}{r}=\frac{4(1{)}^2}{1}=4m/s^2$
$a=\sqrt{a_t^2+a_c^2}=\sqrt{20}=2\sqrt{5}m/s^2$