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 20m. The acceleration of ‘P’ when t = 2s is nearly.

Question

Infinity Learn · Accepted Answer

$$s=t3+5$$ ∴  Speed $$v=dsdt=3t2$$         Tangential acceleration $$(at)=dvdt=6tAt$$ t $$=$$ $$2s$$,  $$v=3$$×$$(2)2=12ms$$−$$1$$ and $$at=6$$×$$2=12ms$$−$$2Centripetal$$ acceleration $$(ac)=v2R=(12)220=14420ms$$−$$2$$∴  Acceleration of P is                              $$a=at2+ac2Substituting$$ the values of at  and ac , we get $$a=14ms$$−$$2$$ ,

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 20m. The acceleration of ‘P’ when t = 2s is nearly.

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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=t3+5 , where s is in metres and t is in seconds. The radius of the path is 20m. The acceleration of ‘P’ when t = 2s is nearly.

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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 20m. The acceleration of ‘P’ when t = 2s is nearly.