A point P moves in counter-clockwise direction on a circular path of radius 20 m . The movement of ' P ' is such that it sweeps out a length s = t^3 + - Physics MCQ AIEEE 2026 | ExamTest.live

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A point P moves in counter-clockwise direction on a circular path of radius 20 m . 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 acceleration of ' P' when t = 2 s is nearly:

Solution & Step-by-step Explanation

1. Tangential velocity v = \frac{d s}{d t} = 3 t^{2} .At t = 2 s, v = 3 (2^{2}) = 12 m/s .2. Tangential acceleration a_{t} = \frac{d v}{d t} = 6 t .At t = 2 s, a_{t} = 6 (2) = 12 m/s^{2} .3. Centripetal acceleration a_{c} = \frac{v ^{2}}{R} = \frac{1 2 ^{2}}{20} = \frac{144}{20} = 7.2 m/s^{2} .4. Net acceleration a = a_{t}^{2} + a_{c}^{2} = 1 2^{2} + 7. 2^{2} = 144 + 51.84 = 195.84 \approx 14 m/s^{2} .

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A point P moves in counter-clockwise direction on a circular path of radius 20 m . 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 acceleration of ' P' when t = 2 s is nearly:

13 m/s^{2}

12 m/s^{2}

7.2 m/s^{2}

14 m/s^{2}

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