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Question Number 15039 by Tinkutara last updated on 07/Jun/17

The acceleration of an object is given

by a (t) = \cos (π t) {ms}^{- 2} and its velocity

at time t = 0 is \frac{1}{2 π} m / s at origin . Its

velocity at t = \frac{3}{2} s is ?

The {object}^{'} s position at t = \frac{3}{2} s is ?

Answered by ajfour last updated on 07/Jun/17

v - \frac{1}{2 π} = \frac{\sin π t}{π}

v ∣_{t = 3 / 2} = \frac{1}{2 π} - \frac{1}{π} = - \frac{1}{2 π} m / s

x = \frac{t}{2 π} - \frac{(\cos π t - 1)}{π^{2}}

x ∣_{t = 3 / 2} = (\frac{3}{4 π} + \frac{1}{π^{2}}) m .

Commented by Tinkutara last updated on 07/Jun/17

But answer given is (i) : \frac{3}{2 π} m / s

(ii) : \frac{3}{4 π} + \frac{1}{π^{2}} m

Commented by mrW1 last updated on 07/Jun/17

the answer given in your book is wrong .

Commented by Tinkutara last updated on 07/Jun/17

Thanks Sir!

Answered by mrW1 last updated on 07/Jun/17

v (t) = \int a (t) dt = \int \cos (π t) dt = \frac{\sin (π t)}{π} + C

since v = \frac{1}{2 π} at t = 0 \Rightarrow C = \frac{1}{2 π}

\Rightarrow v (t) = \frac{\sin (π t)}{π} + \frac{1}{2 π}

x (t) = \int v (t) dt = - \frac{\cos (π t)}{π^{2}} + \frac{t}{2 π} + C

since x = 0 at t = 0 \Rightarrow C = \frac{1}{π^{2}}

\Rightarrow x (t) = \frac{1 - \cos (π t)}{π^{2}} + \frac{t}{2 π}

at t = \frac{3}{2}

v = - \frac{1}{π} + \frac{1}{2 π} = - \frac{1}{2 π}

x = \frac{1}{π^{2}} + \frac{3}{4 π}

Commented by Tinkutara last updated on 07/Jun/17

Thanks Sir!