Question

The displacement of a particle along a line at time t is given by: $x={{a}_{0}}+\dfrac{{{a}_{1}}t}{2}+\dfrac{{{a}_{2}}{{t}^{2}}}{3}$. The acceleration of the particle is
$\begin{align}
  & \text{A}\text{. }\dfrac{{{a}_{2}}}{3} \\
 & \text{B}\text{. }\dfrac{2{{a}_{2}}}{3} \\
 & \text{C}\text{. }\dfrac{{{a}_{1}}}{2} \\
 & \text{D}\text{. }{{\text{a}}_{0}}+\dfrac{{{a}_{2}}}{3} \\
\end{align}$

Answer 1

Hint:According to Newtonian mechanics, we know that acceleration is the second differentiation of displacement with respect to time. Acceleration is the rate at which a particle’s velocity is changing with respect to time.

Complete step-by-step answer:
Above given equation of displacement is a function of time. Given as:
$x={{a}_{0}}+\dfrac{{{a}_{1}}t}{2}+\dfrac{{{a}_{2}}{{t}^{2}}}{3}.......(1)$
Acceleration, a, is defined as the rate of change of velocity, v, with respect to time, t. The value of instantaneous acceleration can be given as:
$a=\dfrac{dv}{t}$
Where velocity is the rate of change in displacement, x, with respect to time. Mathematically given as:
$v=\dfrac{dx}{dt}$
Therefore, acceleration simplifies to:
$a=\dfrac{{{d}^{2}}x}{d{{t}^{2}}}$
Refer to the formula: $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$,
 First differentiation with respect to time of equation (1), will give velocity:
$\begin{align}
  & \dfrac{dx}{dt}=\dfrac{d}{dt}\left( {{a}_{0}} \right)+\dfrac{d}{dt}\left( \dfrac{{{a}_{1}}t}{2} \right)+\dfrac{d}{dt}\left( \dfrac{{{a}_{2}}{{t}^{2}}}{3} \right) \\
 & \Rightarrow \dfrac{dx}{dt}=\left( \dfrac{{{a}_{1}}}{2} \right)+\left( \dfrac{2{{a}_{2}}t}{3} \right) \\
\end{align}$
Second differentiation will give acceleration:
$\begin{align}
  & \dfrac{d}{dt}\left( \dfrac{dx}{dt} \right)=\dfrac{d}{dt}\left( \dfrac{{{a}_{1}}}{2} \right)+\dfrac{d}{dt}\left( \dfrac{2{{a}_{2}}t}{3} \right) \\
 & \Rightarrow \left( \dfrac{{{d}^{2}}x}{d{{t}^{2}}} \right)=\left( \dfrac{2{{a}_{2}}}{3} \right) \\
\end{align}$
So, we get acceleration $\overrightarrow{a}=\dfrac{2{{a}_{2}}}{3}$.
Correct option is B.
We can observe that a particle's acceleration is not a function of time, which means it remains constant with time.

Note: If we are given a displacement function, following Newtonian kinematics we can find any dynamic quantities related to it.
As we have seen in the above solution, we can find velocity and acceleration by doing first and second differentiation of the given displacement function respectively. Acceleration is vector quantity, which means it acts in a particular direction.
If we are given a displacement which is a function of time, we can find other dynamical quantities. By doing above calculations, we have got a positive acceleration which means that the particle's velocity is increasing with time.