A body starts from the origin and moves along the x-axis such that velocity at any instant is given by (4t³-2t) , where t is in second and velocity is in m/s. What is the acceleration of the particle, when it is 2m from the origin?

The correct option is 22 m/s².

To find the acceleration of the particle when it is at a distance of 2 m from the origin, we can follow these steps:

Step 1: Find the position function, $x\left(t\right)$
The velocity $v\left(t\right)$ of the particle at any instant $t$ is given by:
$v\left(t\right)=4t^3-2t$
Since velocity is the rate of change of position, we can integrate the velocity function with respect to time to get the position $x\left(t\right)$:
$x\left(t\right)=\int v\left(t\right)dt=\int (4t^3-2t)dt$
$x\left(t\right)=t^4-t^2+C$
where $C$ is the constant of integration.

Since the particle starts from the origin, at $t=0$, the position $x\left(0\right)=0$:
$0=0^4-0^2+C\Rightarrow C=0$
Therefore, the position equation is:
$x\left(t\right)=t^4-t^2$

Step 2: Find the time when the particle is 2 m from the origin
We set the position $x\left(t\right)=2$:
$t^4-t^2=2$
Letting $y=t^2$, the equation becomes a quadratic equation:
$y^2-y-2=0$
Factoring the equation:
$(y-2)(y+1)=0$
This gives $y=2$ or $y=-1$.
Since $y=t^2$ must be non-negative, we reject $y=-1$. Thus:
$t^2=2$

Step 3: Calculate the acceleration
Acceleration $a\left(t\right)$ is the derivative of velocity $v\left(t\right)$ with respect to time:
$a\left(t\right)=\frac{dv}{dt}=\frac{d}{dt}(4t^3-2t)=12t^2-2$
Substitute $t^2=2$ into the acceleration equation:
$a=12\left(2\right)-2=24-2=22{\text{m/s}}^2$

The acceleration of the particle when it is 2 m from the origin is 22 m/s².