Question

A particle starts from rest and moved in a straight line the velocity of the particle at time t after start is $vm{s^{ - 1}}$ where $v = 0.01{t^3} + 0.22{t^2} - 0.4t$. Find the time at which the acceleration of the particle is greatest?

Answer 1

Hint: We know the velocity of the particle from that we can get the acceleration of the particle. As the acceleration is defined as the change in velocity by the change in time. After getting the acceleration again we have to differentiate the acceleration with respect to time then equate it to zero to get the time for maximum acceleration.

Complete step by step answer:
As per the problem we have a particle starts from rest and moved in a straight line the velocity of the particle at time t after start is $vm{s^{ - 1}}$ where $v = 0.01{t^3} + 0.22{t^2} - 0.4t$.
We need to calculate the time at which the acceleration of the particle is greatest or we can say maximum.
We know the velocity function as,
$v = 0.01{t^3} + 0.22{t^2} - 0.4t$
Acceleration of a particle is defined as the ratio of change in velocity to change in time.
Hence on differentiating the velocity function we will get the acceleration of the particle as,
$a = \dfrac{{dv}}{{dt}}$
On putting v value we will get,
$a = \dfrac{{d\left( {0.01{t^3} + 0.22{t^2} - 0.4t} \right)}}{{dt}}$
$ \Rightarrow a = 3 \times 0.01{t^2} + 2 \times 0.22t - 0.4$
On further solving we will get,
$a = 0.03{t^2} + 0.44t - 0.4$
This is the function of acceleration.
Now we need to find the maximum value of this function. On again differentiating with respect to t and equation to zero we will identify the time,
$\dfrac{{da}}{{dt}} = \dfrac{{d\left( {0.03{t^2} + 0.44t - 0.4} \right)}}{t}$
$0 = 2 \times 0.06t + 0.44$
On rearranging and cancelling the common terms we will get,
$t = - 7.33s$
Here the time can’t be negative.
Hence there must be some error in the function given in the problem.

Note: Remember a few steps while calculating the maxima and minima of a given function $\left( {f\left( x \right)} \right)$. First differentiate the given function $\left( {f'\left( x \right)} \right)$ then let them equal to zero and find the critical points. Then find the second derivative to check if the function is maximum or minimum on that critical point.