Question

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

Answer 1

Hint: For answering this question we will use the concept when $f\left( x \right)$ is maximum or minimum $f'\left( x \right)$ is zero and for maximum value, $f''\left( x \right)<0$ and for minimum value $f''\left( x \right)>0$. And we also know that the velocity is the derivative of distance travelled by the particle with respect to time. And acceleration similarly is the derivative of velocity with respect to time.

Complete step by step answer:
In the question it is given that a particle is moving in a straight line such that its distance $s$ at the time $t$ is given by $s=\dfrac{{{t}^{4}}}{4}-2{{t}^{3}}+4{{t}^{2}}-7$ .
Then the velocity that is the measure of distance travelled per second can be given as $s'$ .
Then the acceleration that is the measure of rate of velocity can be given as $s''$ .
We can derive velocity as $s'=\dfrac{4{{t}^{3}}}{4}-2\left( 3{{t}^{2}} \right)+4\left( 2t \right)$ .
This can be simplified as $s'={{t}^{3}}-6{{t}^{2}}+8t$.
Let us assume the velocity as $v$ then we can write it as $v={{t}^{3}}-6{{t}^{2}}+8t$ .
We can derive acceleration as $s''=3{{t}^{2}}-6\left( 2t \right)+8$ .
This can be simplified as $s''=3{{t}^{2}}-12t+8$ .
Let us assume the acceleration as $a$ then we can write it as $a=3{{t}^{2}}-12t+8$ .
When velocity is maximum, $v'$ is zero and $v''$ is negative.
By applying this we can say $v'=3{{t}^{2}}-12t+8=0$ and $v''=6t-12<0$ .
Here we have a quadratic equation as we know that the roots of a quadratic equation $a{{x}^{2}}+bx+c=0$ are given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
So here for $v'=3{{t}^{2}}-12t+8=0$ we will have $t=2\pm \dfrac{2}{\sqrt{3}}$ .
Now by verifying for $v''=6t-12<0$ for $t=2+\dfrac{2}{\sqrt{3}}$, we will have $v''=6\left( 2+\dfrac{2}{\sqrt{3}} \right)-12=\dfrac{12}{\sqrt{3}}=4\sqrt{3}>0$ and for $t=2-\dfrac{2}{\sqrt{3}}$, we will have $v''=6\left( 2-\dfrac{2}{\sqrt{3}} \right)-12=\dfrac{-12}{\sqrt{3}}=-4\sqrt{3}<0$ .
When acceleration is minimum, $a'$ is zero and $a''$ is positive.
By applying this we can say $a'=6t-12$ and $a''=-12<0$ .
So here we will have only minimum value and it will be at $t=2$.

Hence we can conclude that when a particle is moving in a straight line such that its distance $s$ at any time $t$ is given by $s=\dfrac{{{t}^{4}}}{4}-2{{t}^{3}}+4{{t}^{2}}-7$ . Its velocity will be maximum when $t=2-\dfrac{2}{\sqrt{3}}$ and acceleration will be minimum when $t=2$ .

Note: While answering questions of this type we should remember that we use the concept when $f\left( x \right)$ is maximum or minimum $f'\left( x \right)$ is zero and for maximum value, $f''\left( x \right)<0$ and for minimum value $f''\left( x \right)>0$. If we confuse and take for maximum, $f''\left( x \right)>0$ we will end up having a complete wrong answer that is its velocity will be maximum when $t=2+\dfrac{2}{\sqrt{3}}$ and the point of time at which its acceleration will be minimum is uncertain.