Question

If the speed of a vehicle increases by 2 m/s, its kinetic energy is doubled. Then, the original speed of the vehicle is

• A. $(\sqrt{2}+1)\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$
• B. $\sqrt{2}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$
• C. $2(\sqrt{2}+1)\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$
• D. $\sqrt{2}(\sqrt{2}+1)\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Answer 1

The correct option is C

$2(\sqrt{2}+1)\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step 1. Given data:

Increase in speed of the vehicle = $2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Final kinetic energy = $2\times$Initial kinetic energy

Step 2. Formula used:

Kinetic energy = $\frac{1}{2}m{v}^2$

Where $m$= mass, $v$ = speed of the vehicle.

Step 3. Solution:

According to the question :

$v=v+2$

So kinetic energy according to this speed = $\frac{1}{2}m{\left(v+2\right)}^2=\frac{m}{2}\left(v^2+4v+4\right)$……….$\left(1\right)$

Also, Final kinetic energy = $2\left(\frac{1}{2}m{v}^2\right)=m{v}^2$…………..$\left(2\right)$

Equating $\left(1\right)$ and $\left(2\right)$, we get

$$\begin{gathered} m{v}^2=\frac{m}{2}\left(v^2+4v+4\right) \\ 2v^2=v^2+4v+4 \\ {v}^2-4v-4=0 \end{gathered}$$

Solving this quadratic equation and taking only positive root, we get

$v=2\left(\sqrt{2}+1\right)\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$
Hence the correct option is option C