Question

The equation of motion of a particle moving along a straight line is ${\text{s = 2}}{{\text{t}}^{\text{3}}}{\text{ - 9}}{{\text{t}}^{\text{2}}}{\text{ + 12t}}$ where the units of $s$ and ${\text{t}}$ are cm and sec. The acceleration of the particle will be zero after
$(A)\dfrac{3}{2}s$
$(B)\dfrac{2}{3}s$
$(C)\dfrac{1}{2}s$
$(D)1{\text{ }}s$

Answer 1

Hint:Since the equation of motion of a particle has been provided, we can determine the acceleration by differentiating the given equation twice. By differentiating the equation twice, the acceleration of motion of the particle can be determined. After which by substituting the acceleration to be equal to zero and then by rearranging the equation we can determine the time it takes for the particle to have zero acceleration.

Formula used:
$\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$

 Complete step by step solution:
The equation given to us is ${\text{s = 2}}{{\text{t}}^{\text{3}}}{\text{ - 9}}{{\text{t}}^{\text{2}}}{\text{ + 12t}}$
Where $s$ is the distance and ${\text{t}}$ is the time.
We know that the rate of change of distance with time gives us velocity and the rate of change of velocity with time gives us acceleration. Using this we can determine the time at which the acceleration will become zero.
Given: ${\text{s = 2}}{{\text{t}}^{\text{3}}}{\text{ - 9}}{{\text{t}}^{\text{2}}}{\text{ + 12t}}$
Now, find velocity that is the rate of change of distance ($s$) with time (${\text{t}}$)
$v = \dfrac{{ds}}{{dt}}{\text{ = (3}} \times {\text{2)}}{{\text{t}}^{{\text{(3 - 1)}}}}{\text{ - (9}} \times {\text{2)}}{{\text{t}}^{{\text{(2 - 1)}}}}{\text{ + (12}} \times {\text{1)}}$
$ \Rightarrow v = \dfrac{{ds}}{{dt}}{\text{ = 6}}{{\text{t}}^2}{\text{ - 18t + 12}}$
To find the acceleration, find the rate of change of velocity with time.
Therefore we get,
$a = \dfrac{{{d^2}s}}{{d{t^2}}}{\text{ = (6}} \times {\text{2)}}{{\text{t}}^{{\text{(2 - 1)}}}}{\text{ - (18}} \times {\text{1) + 0}}$
$ \Rightarrow a = \dfrac{{{d^2}s}}{{d{t^2}}}{\text{ = 12}}{{\text{t}}^1}{\text{ - 18}}$
Substitute acceleration is equal to zero To find the time.
Substituting acceleration equal to zero we get,
$ \Rightarrow {\text{0 = 12t - 18}}$
Rearranging we get,
$ \Rightarrow {\text{18 = 12t}}$
Simplifying the values we get,
${\text{t = }}\dfrac{{18}}{{12}} = \dfrac{3}{2}s$
From this, we have determined that the acceleration of the particle will become zero after ${\text{t = }}\dfrac{3}{2}s$
Hence option (A) is correct.

Note:While solving this question care must be taken to differentiate the equation twice to obtain the acceleration of the particle. If the equation of motion is differentiated only once velocity will be obtained. If this equation is then substituted to zero, we will obtain the wrong solution. So care must be taken to differentiate the equation twice to obtain acceleration.