Question

A particle is moving along a circular path of radius $3$, in which a way that the distance travelled along the circumference is given by $s=\frac{t^2}{2}+\frac{t^3}{3}$. If the acceleration of the particle when $t=2sec$ is?

• A. $1.3m/s^2$
• B. $13m/s^2$
• C. $3m/s^2$
• D. $10m/s^2$

Answer 1

Answer: (B)

$13m/s^2$

Given,

$r=3m$

$t=2sec$

$s=\frac{t^2}{2}+\frac{t^3}{3}$

Tangential acceleration, $a_t=\frac{d^{2}s}{d{t}^2}$

$a_t=\frac{d\left(\frac{d(t^2/2+t^3/3)}{dt}\right)}{dt}$

$a_t=\frac{d(t+t^2)}{dt}$

$a_t=1+2t$

$a_t{|}_{t=2sec}=1+2\times 2$

$a_t=5m/s^2$

Linear speed, $v=\frac{ds}{dt}$

$v=t+t^2$

$v_{t=2sec}=2+2\times 2=6m/s$

Radial acceleration, $a_r=\frac{v^2}{r}$

$a_r=\frac{6\times 6}{3}$

$a_r=12m/s^2$

Acceleration, $a=\sqrt{a_t^2+a_r^2}$

$a=\sqrt{(5{)}^2+(12{)}^2}=\sqrt{169}$

$a=13m/s^2$

The correct option is B.