Question

The position $x$ of a particle varies with time as $x = a{t^2} - b{t^3}$ . The acceleration of particle is zero at time $T$ equal to
\left( A \right)\dfrac{a}{b} \\\ \left( B \right)\dfrac{{2a}}{{3b}} \\\ \left( C \right)\dfrac{a}{{3b}} \\\ \left( D \right)zero \\\

Answer 1

Hint : In order to solve this question, we are going to first find the velocity of the particle from the relation for $x$ as given in the question, then by again differentiating the velocity, we find the acceleration and by equating the acceleration to zero, we get the desired value for the time $T$ .
The velocity of a particle in terms of the position is given by
$Velocity = \dfrac{{dx}}{{dt}}$
The acceleration of the particle in terms of position is given by
$Acceleration = \dfrac{{{d^2}x}}{{d{t^2}}}$

Complete Step By Step Answer:
Let us first write the equation for the position $x$ of the particle
$x = a{t^2} - b{t^3}$
On differentiating the above equation, we get the velocity of the particle as
$\Rightarrow Velocity = \dfrac{{dx}}{{dt}} = 2at - 3b{t^2}$
Now again differentiating the velocity equation, we get the acceleration equation of the particle
Acceleration = \dfrac{{{d^2}x}}{{d{t^2}}} = \dfrac{d}{{dt}}\left( {\dfrac{{dx}}{{dt}}} \right) \\\ \Rightarrow A = \dfrac{d}{{dt}}(2at - 3b{t^2}) = 2a - 6bt \\\
Thus, we can write that
$A = 2a - 6bt$
Now, when acceleration of the particle is zero,
We get,
A = 0 \\\ \Rightarrow 2a = 6bt \\\
Now, simplifying the above equation, in order to get the value of the time period,
$T = \dfrac{{2a}}{{6b}} = \dfrac{a}{{3b}}$
Thus, the acceleration of the particle is zero at time $T$ equal to $\dfrac{a}{{3b}}$
Hence, the option $\left( C \right)\dfrac{a}{{3b}}$ is correct.

Note :
It is important to know the relation between the distance, velocity and the acceleration of a particle in terms of the concept of the differentiation and also the basics of differentiation. The point at which the acceleration of the particle becomes zero is the time period for which that particle is moving with a constant velocity and the position is changing at a constant rate.