# A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of P is such that it sweeps out a length $\dpi{100} s=t^{3}+5,$ where $\dpi{100} s$ is in metres and $\dpi{100} t$ is in seconds. The radius of the path is 20 m , The acceleration of $\dpi{100} P$ When $\dpi{100} t= 2s$ is nearly Option 1) $14 m\: s^{-2}$ Option 2) $13\: m\: s^{-2}$ Option 3) $12\: m\: s^{-2}$ Option 4) $7.2 m\: s^{-2}$

As we learnt in

Centripetal acceleration -

When a body is moving in a uniform circular motion, a force is responsible to change direction of its velocity.This force acts towards the centre of circle and is called centripetal force.Acceleration produced by this force is centripetal acceleration.

$a= \frac{v^{2}}{r}$

- wherein

Figure Shows Centripetal acceleration

$S = t^{3}+3$

$V= \frac{ds}{dt} = 3{t^2}+0$

$\Rightarrow v = 3t^2$

tangential acceleration

$= a_{t} = \frac{dv}{dt} = \frac{d\left ( 3t^2 \right )}{dt}$

$a_{t} = 6t$

At $t = 2 sec$

$v= 3 \left ( 2 \right )^2 = 12ms^{-1}$

$a_{t} = 6\times 2 = 12ms^{-2}$

$\therefore\ \; centripetal\ acceleration = \vec a_{{c}} = \frac{{v^2}}{r}=\frac{(12)^{2}}{20}=\frac{144}{20}$

$a_{c}= 7.2 ms ^{^-2}$

$\therefore Net \:acceleration$

$a= \sqrt{a_{c}^{2}+a_{t}^{2}}=\sqrt{7.2^{2}+12^{2}}$

$a_{_{c}} \simeq14 ms^{-2}$

Option 1)

$14 m\: s^{-2}$

Correct

Option 2)

$13\: m\: s^{-2}$

Incorrect

Option 3)

$12\: m\: s^{-2}$

Incorrect

Option 4)

$7.2 m\: s^{-2}$

Incorrect

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