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 length $s = t^{3} + 5,$ where $s$ is in metre and t is in second. The radius of the path is 20 m. The acceleration of P when t =2s is nearly [image:4_image25.emf]

Question

Dr. Abubacker · Accepted Answer

(d) $Given,\ \ s = t^{3} + 5$
$Speed,\ \ v = \frac{ds}{dt} = {3t}^{2}$
$and\ rate\ of\ change\ of\ speed,\ a_{t} = \frac{dv}{dt} = 6t$
$	herefore Tangential\ acceleration\ at\ t = 2\ s,$
$a_{t} = 6 	imes 2 = 12\ {ms}^{- 2}$
$and\ at\ t = 2s,\ v = 3({2)}^{2} = 12{ms}^{- 1}$
$	herefore Centripetal\ acceleration,\ a_{c} = \frac{v^{2}}{R} = \frac{144}{20}{ms}^{- 2}$
$	herefore Net\ acceleration = a_{t}^{2} + a_{i}^{2} \approx 14{ms}^{- 2}$

Q. 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 length $s = t^{3} + 5,$ where $s$ is in metre and t is in second. The radius of the path is 20 m. The acceleration of P when t =2s is nearly

Explanation

Reviewed by Dr. Abubacker

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