Question

A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by $(4t^3-2t)$, where t is in sec and velocity in m/s. What is the acceleration of the particle, when it is 2 m from the origin

• A.
• B.
• C.
• D.

Answer 1

The correct option is (B)

Step 1, Given data

Velocity, v = $4t^3-2t$ .

Step 2, Finding the equation

We know,

Acceleration $a=\frac{dv}{dt}$

So, $a=\frac{d(4t^3-2t)}{dt}$

Or, $\therefore a=\frac{dv}{dt}=12t^2-2$

We also know that

$v=\frac{dx}{dt}$

$Or,\int dx=\int vdt$

Now we can write

and $x=\int vdt=\int (4t^3-2t)dt=t^4-t^2$

According to the question when particle is at 2 m from the origin,

We can write

$t^4-t^2=2$

Or,

$\Rightarrow t^4-t^2-2=0$

$\Rightarrow t^4+t^2-2t^2-2=0$

$\Rightarrow t^2(t^2+1)-2(t^2+1)=0$

After solving this polynomial

$(t^2-2)(t^2+1)=0\Rightarrow t=\sqrt{2}$sec (only possible physical root)

Now this value of t will be substituted in the equation of acceleration

Acceleration at $t=\sqrt{2}$ sec given by,

$a=12t^2-2=12\times 2-2=22m/s^2$

Hence the acceleration is 22 m/$s^2$ .