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Question:

A point P moves in a counter-clockwise direction on a circular path. The movement of P is such that it sweeps out a length s = t³ + 5, where s is in meters and t is in seconds. The radius of the path is 20m. The acceleration of P when t = 2s is nearly:

12m/s²

7.2m/s²

13m/s²

14m/s²

Solution:

S = t³ + 5
•Speed, v = ds/dt = 3t² and rate of change of speed = dv/dt = 6t
• tangential acceleration at t = 2 s, at = 6 × 2 = 12 m/s²
at t = 2s, v = 3(2)² = 12 m/s
• Centripetal acceleration, ac = v²/R = 144/20 m/s²
•Net acceleration = √(at² + ac²) ≈ 14 m/s²