If angular displacement of a particle moving on a curved path be given as,θ=1.5t+2t2 where t is in sec, the angular velocity at t = 2 sec, will be- (in rad/sec.)

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If angular displacement of a particle moving on a curved path be given as, $θ = 1 .5 t + 2 t^{2}$ where t is in sec, the angular velocity at t = 2 sec, will be- (in rad/sec.)

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9.5

8.5

answer is C.

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Detailed Solution

$θ = 1 .5 t + 2 t^{2}, ∴ \frac{dθ}{dt} = 1 .5 + 4 t, Now, at t = 2 sec , (\frac{dθ}{dt}) = 1 .5 + 4 \times 2 = 1 .5 + 8 = 9 .5 rad / \sec$

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