Question

A body starts from origin and moves along x-axis such that at any instant velocity is $(4{t^3} - 2t)$, where t is in second and velocity is in $m{s^{ - 1}}$. The acceleration of the particle when it is 2m from the origin is
A. $28m{s^{ - 2}}$
B. $22m{s^{ - 2}}$
C. $12m{s^{ - 2}}$
D. $10m{s^{ - 2}}$

Answer 1

Hint: The meaning of velocity of an object is defined as the rate of change of the object’s position with respect to a position and time. It might sound complicated but velocity is basically speeding in a specific direction and it is a vector quantity.
Acceleration is the rate of change of velocity with respect to time. We are going to use these two definitions and calculate the required parameters and finally arrive at the result which is the acceleration of a body at 2m distance from the origin.

Complete Answer:
Given that velocity of the body at any instant of time is,
${v_t} = 4{t^3} - 2t$
We know that velocity is the rate of change of displacement. Thus, by using this definition we shall calculate the time required to cover 2m.
$\therefore {v_t} = \dfrac{{ds}}{{dt}}$
$ \Rightarrow \dfrac{{ds}}{{dt}} = 4{t^3} - 2t$
$ \Rightarrow ds = \left( {4{t^3} - 2t} \right)dt$
Integrating on both sides,
$ \Rightarrow \int {ds} = \int {\left( {4{t^3} - 2t} \right)dt} $
\[ \Rightarrow s = {t^4} - {t^2}\]
It is given that displacement is 2m, therefore we get:
\[ \Rightarrow {t^4} - {t^2} - 2 = 0\]
\[ \Rightarrow {t^2}({t^2} - 2) + ({t^2} - 2) = 0\]
\[ \Rightarrow ({t^2} + 1)({t^2} - 2) = 0\]
On solving above equation, we find the value of ‘t’ and is given by,
\[ \Rightarrow t= \pm \sqrt 2 \sec = \sqrt 2 \sec \]
Now acceleration of the body is, again by definition we know that, acceleration is rate of change of velocity. Therefore,
${a_t} = \dfrac{{d{v_t}}}{{dt}} = 12{t^2} - 2$
Substituting for the value of “t”, we get:
${a_t} = \left( {12 \times 2} \right) - 2 = 22m/{s^2}$
Therefore, acceleration of the particle when it is 2m away from origin is $22m{s^{ - 2}}$.
Hence, the correct option is B.

Note: While calculating time we get two values of “t” , one is $t = \pm i\sec $ and $t = \pm \sqrt 2 \sec $. We cannot consider the value of time as an imaginary number, thus we must consider a positive real value which is $t = \sqrt 2 \sec $.