Question

The equation of motion of a particle moving along a straight line is $s = 2{t^3} - 9{t^2} + 12t$ where the units of \[s\] and \[t\] are cm and sec. the acceleration of the particle will be zero after
$(A)\dfrac{3}{2}\sec $
$(B)\dfrac{2}{3}\sec $
$(C)\dfrac{1}{2}\sec $
$(D)$ Never.

Answer 1

Hint: Acceleration is the rate at which the velocity changes in time. In terms of speed and direction.
The motion in the circle gets accelerated even if the speed is constant. Because in the circle the direction is changing continuously.
The acceleration is also the vector quantity.


Complete step-by-step solution:
The object moving in a straight line gets accelerated when it gets speed up or it gets slowed down.
The graph of acceleration a versus displacement s is a horizontal line without change and how a point on the graph is related to any other point.
$\sum a.ds = as$
Now the question can be solved as follows,
Let us consider as follows,
 \[s{\text{ }} = {\text{ }}2{t^3}\;-{\text{ }}9{t^2}\; + {\text{ }}12t\]
\[a = \dfrac{{{d^2}s}}{{d{t^2}}}\]
Now let's integrate the above equation we get,
\[\dfrac{{ds}}{{dt}} = 6{t^2}-18t + 12\]
Differentiate the above equation again,
\[\dfrac{{{d^2}s{\text{ }}}}{{d{t^2}\;}} = 12t-18\]
Now,
\[a = 0,{\text{ }}\dfrac{{{d^2}s}}{{d{t^2}}} = 0\]
By substituting the above conditions we obtain the equation as follows,
\[12t-18 = 0\]
By the further simplification, the above equation becomes,
\[ \Rightarrow t = \dfrac{{18}}{{12}} = \dfrac{3}{2}\]
From this, we can find the time.
Hence the correct answer is an option $(A)$.

Note:The displacement is defined as the particle moving in a straight line with the change in position.
The displacement of a particle when it is moving in a straight line the vector is defined as the change in its position.
The rate of change of the position of a particle concerning time is called the velocity of the particle.