Rotational Motion

JEE Main level, single correct, medium

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Question 188 of 280

A point P\mathrm{P} moves in counter-clockwise direction on a circular path as shown in the figure. The movement of ' P\mathrm{P} ' is such that it sweeps out a length s=t3+5s=t^3+5, where ss is in metres and tt is in seconds. The radius of the path is 20 m20 \mathrm{~m}. The acceleration of ' P\mathrm{P} ' when t=2 st=2 \mathrm{~s} is nearly

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Answer: (D) 14 m/s214 \mathrm{~m} / \mathrm{s}^2

S=t3+5S=t^3+5
∴\therefore \quad speed, v=dsdt=3t2\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=3 \mathrm{t}^2
and rate of change of speed =dvdt=6t=\frac{\mathrm{dv}}{\mathrm{dt}}=6 \mathrm{t}
∴\therefore \quad tangential acceleration at t=2 s,at=6×2=12 m/s2\mathrm{t}=2 \mathrm{~s}, \mathrm{a}_{\mathrm{t}}=6 \times 2=12 \mathrm{~m} / \mathrm{s}^2
at t=2 s,v=3(2)2=12 m/st=2 \mathrm{~s}, \mathrm{v}=3(2)^2=12 \mathrm{~m} / \mathrm{s}
∴\therefore \quad centripetal acceleration, ac=v2R=14420 m/s2\quad a_c=\frac{v^2}{R}=\frac{144}{20} \mathrm{~m} / \mathrm{s}^2
∴\therefore \quad net acceleration =at2+ai2=\sqrt{\mathrm{a}_t^2+\mathrm{a}_{\mathrm{i}}^2} ≈14 m/s2\approx 14 \mathrm{~m} / \mathrm{s}^2

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