Question
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 meters and t is in seconds. The radius of the path is 20 m. The acceleration of 'P' when t = 2 s is nearly
Answer: Option D
:
D
S=t3+5
Speed,v=dsdt=3t2 and rate of change of speed =dvdt=6t
∴ tangential acceleration at t=2S,at = 6 × 2=12m/s2
Att=2s,v=3(2)2=12m/s2
∴ centripetal acceleration,
=v2R=14420m/s2
∴ net acceleration=√a2t+a2i≈14m/s2
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:
D
S=t3+5
Speed,v=dsdt=3t2 and rate of change of speed =dvdt=6t
∴ tangential acceleration at t=2S,at = 6 × 2=12m/s2
Att=2s,v=3(2)2=12m/s2
∴ centripetal acceleration,
=v2R=14420m/s2
∴ net acceleration=√a2t+a2i≈14m/s2
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