Question

A particle is performing circular motion of radius $1$m. Its speed is $v=\left(2t^2\right)m/s$. What will be magnitude of its acceleration at $t=1$s.

Answer 1

Radial acceleration $a_n=v^2/r$

at t = 1s, v= 2 m/s $a_n=2^2/1=4m/s^2$

Tangential acceleration $a_t=dv/dt=4t$

at t = 1 s, $a_t=4m/s^2$

$so,a=\sqrt{a_n^2+a_t^2}$

$a=\sqrt{4^2+4^2}=4\sqrt{2}m/s^2$