Question

A particle moves along a straight line. Its position at any instant is given by $ x = 32t - \dfrac{{8{t^3}}}{3} $ where $ x $ is in metre and $ t $ in second. Find the acceleration of the particle at the instant when the particle is at rest.
(A) $ - 16m/{s^2} $
(B) $ - 32m/{s^2} $
(C) $ 32m/{s^2} $
(D) $ 16m/{s^2} $

Answer 1

Hint : The instantaneous velocity is generally given as the derivative of the distance at a particular time. The acceleration is the derivative of the velocity. So from the question by taking the double derivative of the position, we will get the acceleration.

Formula used: In this solution we will be using the following formula;
 $\Rightarrow v = \dfrac{{dx}}{{dt}} $ , where $ v $ is the instantaneous velocity, $ x $ is the position at any particular time $ t $ .
 $\Rightarrow a = \dfrac{{dv}}{{dt}} $ where $ a $ is the acceleration, $ v $ is the velocity at a particular time.

Complete step by step answer
To find the acceleration at a particular time, first we must find the expression of the velocity at any given time from the expression of the position given. This is called the instantaneous velocity, and it is given as
 $\Rightarrow v = \dfrac{{dx}}{{dt}} $
Since, $ x = 32t - \dfrac{{8{t^3}}}{3} $ , then $ v $ is given as
 $\Rightarrow v = \dfrac{{d\left( {32t - \dfrac{{8{t^3}}}{3}} \right)}}{{dt}} = 32 - 8{t^2} $
Similarly, acceleration is given as
 $\Rightarrow a = \dfrac{{dv}}{{dt}} $ hence we get,
 $\Rightarrow a = \dfrac{{d\left( {32 - 8{t^2}} \right)}}{{dt}} = - 16t $
Now to find time when a particle is at rest, we must solve for $ t $ from the equation $ v = 32 - 8{t^2} $ for when $ v $ is zero. Hence,
 $\Rightarrow 0 = 32 - 8{t^2} $
 $ \Rightarrow 8{t^2} = 32 $
Thus, by dividing both side by 8 we have
 $\Rightarrow {t^2} = 4 $
 $ \Rightarrow t = 2 $
Hence, acceleration when particle is at rest is given as
 $\Rightarrow a = - 16t = - 16\left( 2 \right) = - 32 $
 $ \therefore a = - 32m/{s^2} $
Hence, the correct option is B.

Note
In general, velocity is defined as the time-rate of change of position. This can be considered to be given as $ \bar v = \dfrac{{{x_2} - {x_1}}}{{{t_2} - {t_1}}} = \dfrac{{\Delta x}}{{\Delta t}} $ . However, in many circumstances, velocity is not constant, hence, to get a velocity at any particular time, it would require us to pick the two times (with their corresponding position) very close to each other such that the difference is almost zero. This is as though we’re measuring the velocity at a point (since the two points are so close together).
This statement is given based on mathematical principles that
 $\Rightarrow v = \mathop {\lim }\limits_{t \to 0} \dfrac{{\Delta x}}{{\Delta t}} $ , and this expression is written as
 $\Rightarrow v = \dfrac{{dx}}{{dt}} $
Similarly, we can derive
 $\Rightarrow a = \dfrac{{dv}}{{dt}} $.