A truck accelerates from speed v to 2v. Work done during this is
A. Three times as the work done in accelerating it from rest to v
B. Same as the work done in accelerating it from rest to v
C. Four times as the work done in accelerating it from rest to v
D. Lesser than the work done in accelerating it from rest to v

Hint: First we will find an expression for the work done on a body when its velocity changes and then we will try to convert it into a form such that it's value can be calculated from the quantities given in the question and then compare them.

Formula used:
F=ma
$2ad={{v}^{2}}-{{u}^{2}}$
W = F.d

Complete answer:
The work done on a body is given as the scalar product of the force acting on it and the displacement of the body during the time that force acts on it. Here we will assume a constant force acting to increase the velocity of the truck acting in the same direction as the direction of motion of the truck. The force can further be expressed as the product of the mass of the body and it's acceleration.

$\begin{align}
  & F=ma \\
 & W=Fd=mad \\
\end{align}$

Using the third law of motion given by Newton, we get

$\begin{align}
  & 2ad={{v}^{2}}-{{u}^{2}}\Rightarrow ad=\dfrac{{{v}^{2}}-{{u}^{2}}}{2} \\
 & W=m\left( \dfrac{{{v}^{2}}-{{u}^{2}}}{2} \right) \\
\end{align}$

As mass of the truck remains constant, the work done is proportional only to the difference of square of velocities.
\[W\propto {{v}^{2}}-{{u}^{2}}\]
The ratio of work done when the truck goes from v to 2v and the work done when the truck goes from zero to v is
$\dfrac{{{\left( 2v \right)}^{2}}-{{v}^{2}}}{{{v}^{2}}-0}=\dfrac{3{{v}^{2}}}{{{v}^{2}}}=3$

So, the correct answer is “Option A”.

Note:
We can also solve this question directly by using the work energy theorem which states that the work done on a body is equal to the change in its kinetic energy. So, we get
$W=\dfrac{m{{v}^{2}}}{2}-\dfrac{m{{u}^{2}}}{2}=m\left( \dfrac{{{v}^{2}}-{{u}^{2}}}{2} \right)$

This gives the same expression as we got earlier and the same result that the work done will be proportional to the difference of square of velocities.