Question

A body starts from the origin and moves along the $x - axis$ such that the velocity at any instant is given by $\left( {4{t^3} - 2t} \right)$ where $t$ is in sec and velocity in m/s. what is the acceleration of the particle when it is $2m$ from the origin?
$\left( a \right)$$28m/{s^2}$
$\left( b \right)$$22m/{s^2}$
$\left( c \right)$$12m/{s^2}$
$\left( d \right)$$10m/{s^2}$

Answer 1

Hint: Here we have to find the acceleration of the particle and we know the definition of velocity which rate change of the displacement concerning the time. After that on integrating we will get the time and then we will put the value in the velocity formula and get the required result.
Formula used:
Velocity,
$ \Rightarrow V = \dfrac{{dx}}{{dt}}$
Where,$V$ is the velocity, and $\dfrac{{dx}}{{dt}}$ is the change in the rate of displacement concerning the time.

Complete step by step solution: As in the question, we have velocity given which will be equal to the
$ \Rightarrow V = 4{t^3} - 2t$
Also, it is given that we have to find the acceleration when the particle is $2m$ from the origin distance.
Now velocity can be written as
\[ \Rightarrow \dfrac{{dx}}{{dt}} = 4{t^3} - 2t\]
Taking time derivative on the other side, we get
$ \Rightarrow dx = \left( {4{t^3} - 2t} \right)dt$
Now we will integrate the equation
$ \Rightarrow \int\limits_0^2 {dx} = \int\limits_0^t {\left( {4{t^3} - 2t} \right)} dt$
On integrating, we get
$ \Rightarrow 2 = 4\left[ {\dfrac{{{t^4}}}{4}} \right] - {t^2}$
After solving the above equation
$ \Rightarrow 2 = {t^4} - {t^2}$
Taking time common
$ \Rightarrow {t^2}\left( {{t^2} - 1} \right) = 2$
And then solving the equation,
$ \Rightarrow {t^4} - {t^2} - 2 = 0$
After expanding the equation to get the value to time,
$ \Rightarrow {t^4} - 2{t^2} + {t^2} - 2 = 0$
Now taking time common
\[ \Rightarrow {t^2}\left( {{t^2} - 2} \right) + 1\left( {{t^2} - 2} \right) = 0\]
We get two values of time but since time cannot be negative
$ \Rightarrow {t^2} = 2,{t^2} + 1 = 0$
Therefore,
$ \Rightarrow t = \sqrt 2 s$
Since the velocity equation is being given in the question and we have to find the acceleration.
$ \Rightarrow V = 4{t^3} - 2t$
So on differentiating w.r.t time, we get
\[ \Rightarrow a = \dfrac{{dv}}{{dt}} = 12{t^2} - 2\]
Now we will put the values
$ \Rightarrow 12 \times 2 - 2$
$ \Rightarrow 22m/{s^2}$

Therefore the acceleration will be $22m/{s^2}$.

Notes Acceleration depends on a modification in velocity, not speed. Velocity depends on speed and direction. If you mechanically change your velocity so you alter your acceleration. Even if you’re traveling at a continuing speed however your ever-changing direction (going in a circle for example), your velocity is ever-changing so your acceleration is ever-changing.