Question

A particle moves along a straight line such that its displacement s at any time t is given by $s={{t}^{3}}-6{{t}^{2}}+3t+4m$, t being in seconds. Find the velocity of the particle when the acceleration is zero.

Answer 1

Hint: We need to form the concept of velocity and acceleration from displacement. We use the differential forms to find the acceleration of the particle. Equating with 0 we get the point. Then we use that to find the velocity of the particle at that point.

Complete step-by-step answer:
We have been given the equation of a particle’s displacement s at any time t by $s={{t}^{3}}-6{{t}^{2}}+3t+4m$.
We know the velocity of a moving particle at any instant of time is its rate of displacement.
So, the instantaneous displacement at a fixed point is considered as the velocity of the particle at that point.
We take the formula $v=\dfrac{ds}{dt}$ as the velocity of that particle.
Again, acceleration of a moving particle at any instant of any time is its rate of change of velocity. Instantaneous velocity at a fixed point is considered as the acceleration of the particle at that point.
We take the formula $a=\dfrac{dv}{dt}$ as the acceleration of that particle.
We convert the formula of a with respect to the displacement s.
\[a=\dfrac{dv}{dt}=\dfrac{d}{dt}\left( \dfrac{ds}{dt} \right)=\dfrac{{{d}^{2}}s}{d{{t}^{2}}}\].
Now for our given displacement $s={{t}^{3}}-6{{t}^{2}}+3t+4m$, we need to find the velocity of the particle when the acceleration is zero.
$v=\dfrac{ds}{dt}=\dfrac{d}{dt}\left( {{t}^{3}}-6{{t}^{2}}+3t+4m \right)$, \[a=\dfrac{{{d}^{2}}s}{d{{t}^{2}}}=\dfrac{{{d}^{2}}}{d{{t}^{2}}}\left( {{t}^{3}}-6{{t}^{2}}+3t+4m \right)\].
We differentiate twice to get the velocity first and acceleration after that.
So, $v=\dfrac{d}{dt}\left( {{t}^{3}}-6{{t}^{2}}+3t+4m \right)=3{{t}^{2}}-12t+3$ and \[a=\dfrac{{{d}^{2}}}{d{{t}^{2}}}\left( {{t}^{3}}-6{{t}^{2}}+3t+4m \right)=\dfrac{d}{dt}\left( 3{{t}^{2}}-12t+3 \right)=6t-12\].
The displacement is 0 which means \[6t-12=0\].
Solving this we get $t=2$.
The velocity of the particle at $t=2$ will be $v=3{{\left( 2 \right)}^{2}}-12\left( 2 \right)+3=-9$.
So, the velocity of the particle when the acceleration is zero is -9.

Note: We are finding the velocity and the acceleration of the particle at a fixed point which is instantaneous. So, the differential gives us the slope of the curve at that point. With the slight change of the value the velocity and the acceleration will also change.