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 s=t^{3}+5, where s is in metres and t is in seconds. The radius of the path is 20 m , The acceleration of P When 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}

 

Answers (1)

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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