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Solution for A Particle is Moving in a Straight Line Such that Its Distance at Any Time T is Given by S = T 4 4 − 2 T 3 + 4 T 2 − 7 . Find When Its Velocity is Maximum and Acceleration Minimum. - CBSE (Commerce) Class 12 - Mathematics

Question

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

Solution

$\text { Given }: \hspace{0.167em} s = \frac{t^4}{4} - 2 t^3 + 4 t^2 - 7$

$\Rightarrow v = \frac{ds}{dt} = t^3 - 6 t^2 + 8t$

$\Rightarrow a = \frac{dv}{dt} = 3 t^2 - 12t + 8$

$\text { For maximum or minimum values of v, we must have }$

$\frac{dv}{dt} = 0$

$\Rightarrow 3 t^2 - 12t + 8 = 0$

$\text { On solving the equation, we get }$

$t = 2 \pm \frac{2}{\sqrt{3}}$

$\text { Now },$

$\frac{d^2 v}{d t^2} = 6t - 12$

$\text {At t } = 2 - \frac{2}{\sqrt{3}}:$

$\frac{d^2 v}{d t^2} = 6\left( 2 - \frac{2}{\sqrt{3}} \right) - 12$

$\Rightarrow \frac{- 12}{\sqrt{3}} < 0$

$\text { So, velocity is maximum at t } = \left( 2 - \frac{2}{\sqrt{3}} \right) .$

$\text { Again },$

$\frac{da}{dt} = 6t - 12$

$\text { For maximum or minimum values of a, we must have }$

$\frac{da}{dt} = 0$

$\Rightarrow 6t - 12 = 0$

$\Rightarrow t = 2$

$\text { Now,}$

$\frac{d^2 a}{d t^2} = 6 > 0$

$\text { So, acceleration is minimum at t }=2.$

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Solution A Particle is Moving in a Straight Line Such that Its Distance at Any Time T is Given by S = T 4 4 − 2 T 3 + 4 T 2 − 7 . Find When Its Velocity is Maximum and Acceleration Minimum. Concept: Graph of Maxima and Minima.
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