Question

A particle moves along the x-axis and its displacement at any time is given as \[x\left( t \right) = 2{t^3} - 3{t^2} + 4t\] in SI units. The velocity of the particle when its acceleration is zero, is:
$
  {\text{A}}{\text{. 2}}{\text{.5}}m{s^{ - 1}} \\
  {\text{B}}{\text{. 3}}{\text{.5}}m{s^{ - 1}} \\
  {\text{C}}{\text{. 4}}{\text{.54}}m{s^{ - 1}} \\
  {\text{D}}{\text{. 8}}{\text{.5}}m{s^{ - 1}} \\
$

Answer 1

Hint: We are given the displacement of the particle which is a function of time. The velocity of the particle is equal to the first order time derivative of displacement and acceleration is the second order time derivative of displacement.

Formula used:
If the displacement of a particle is a function of time and given as x(t), then we can calculate the velocity of the particle by the following formula.
\[v = \dfrac{{dx}}{{dt}}{\text{ }}...\left( i \right)\]
The acceleration of the particle is given as
$a = \dfrac{{dv}}{{dt}} = \dfrac{{{d^2}x}}{{d{t^2}}}{\text{ }}...\left( {ii} \right)$

Detailed step by step solution:
We are given a particle which is moving along the x-direction. Its displacement changes with time and is given in terms of time t by the following expression.
\[x\left( t \right) = 2{t^3} - 3{t^2} + 4t\]
We need to calculate the velocity of the given particle when its acceleration is zero. We know how the velocity and acceleration can be calculated from the displacement of the particle as given by equations (i) and (ii). Let us first calculate these quantities.
The velocity of the particle is given as
$
  v = \dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left( {2{t^3} - 3{t^2} + 4t} \right) \\
   \Rightarrow v = 6{t^2} - 6t + 4{\text{ }}...\left( {iii} \right) \\
$
Now the acceleration can be calculated as follows:
$
  a = \dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}\left( {6{t^2} - 6t + 4} \right) \\
   \Rightarrow a = 12t - 6{\text{ }}...\left( {iv} \right) \\
$
Now we will calculate the velocity of the given particle when its acceleration is zero. First taking the acceleration to be zero in equation (iv), we get
$
  12t - 6 = 0 \\
  12t = 6 \\
  t = \dfrac{6}{{12}} = \dfrac{1}{2}s \\
$
This means that acceleration of the particle is zero at t = 0.5 s. Now we will put this value in equation (iii) and will get the desired answer.
 $
  v = 6{t^2} - 6t + 4{\text{ }} \\
   = 6{\left( {\dfrac{1}{2}} \right)^2} - 6 \times \dfrac{1}{2} + 4 \\
   = \dfrac{3}{2} - 3 + 4 = 2.5m{s^{ - 1}} \\
$
Hence, the correct answer is option A.

Note: When the acceleration of a particle is zero then velocity of the particle is zero according to equation (i). This means that the particle is executing uniform motion where it moves with constant velocity equal to 2.5 m/s.