A particle is moving in a straight line such that its distance at any time t is given by S-t4-4-2t3-4t2-7 then its acceleration is minimum at t -

The correct option is A 2Distance S=t^4/4 2t^3+4t^2+7 #160;Acceleration ⇒ a=3t^2 12t+8... Double derivative of S a=f(t)=3t^2 12t+8 For ma ...

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Standard XII

Physics

Differentiation from 1st Principles

A particle is...

Question

A particle is moving in a straight line such that its distance at any time t is given by $S = \frac{t^{4}}{4} - 2 t^{3} + 4 t^{2} + 7$ then its acceleration is minimum at t =

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Solution

The correct option is A 2
Distance $S = \frac{t^{4}}{4} - 2 t^{3} + 4 t^{2} + 7$

Acceleration $\Rightarrow a = 3 t^{2} - 12 t + 8...$ Double derivative of $S$

$a = f (t) = 3 t^{2} - 12 t + 8$

For maxima or minima,

$f^{'} (t) = 0$

$\Rightarrow 6 t - 12 = 0$

$\Rightarrow t = 2$

$f^{''} (t) = 6 > 0$

So, the minimum value of acceleration is at $t = 2$

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