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A particle is moving in a straight line such that its distance at any time t is given by

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:

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

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

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

For maximum or minimum values of $v$, we must have

$\frac{d v}{d t}=0$

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

On solving the equation, we get

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

Now,

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

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

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

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

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

Again,

$\frac{d a}{d t}=6 t-12$

For maximum or minimum values of a, we must have

$\frac{d a}{d t}=0$

$\Rightarrow 6 t-12=0$

$\Rightarrow t=2$

Now,

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

So, acceleration is minimum at $t=2$.

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