Question

If the speed of a vehicle increases by $$2m/s$$, its kinetic energy is doubled. Then, the original speed of the vehicle is:
A.$$(\sqrt{2}+1)m/s$$
B.$$\sqrt{2}m/s$$
C.$$2(\sqrt{2}+1)m/s$$
D.$$\sqrt{2}(\sqrt{2}+1)m/s$$

Answer 1

Hint: Speed is defined as distance divided by time. Speed consists only of magnitude. Whereas velocity is equal to displacement divided by time. Velocity is the vector quantity. It consists of both direction and magnitude. Velocity can also be said as speed in a particular direction. Therefore we can consider speed and velocity as the same thing for this case. Therefore it is given the question that when speed increases the kinetic energy is doubled. From this relation, we can calculate the initial speed of the vehicle.

Complete answer:
Let the initial kinetic energy be $$K={\displaystyle \frac{1}{2}}m{v}^2$$ ……… (1)
After increasing the speed by $$2m/s$$,the final velocity (speed) will be $$v+2$$. Therefore
Final kinetic energy be $$2K={\displaystyle \frac{1}{2}}m(v+2{)}^2$$ ……… (2)
Substituting$$K$$from equation (1) in the equation (2) we get,
$$2({\displaystyle \frac{1}{2}}\times m{v}^2)={\displaystyle \frac{1}{2}}m(v+2{)}^2$$
$$\Rightarrow v^2={\displaystyle \frac{1}{2}}(v+2{)}^2$$
$$\Rightarrow 2v^2=(v+2{)}^2$$
Squaring on both sides,
$$\sqrt{2}v=(v+2)$$
$$\Rightarrow (\sqrt{2}-1)v=2$$
$$\Rightarrow v={\displaystyle \frac{2}{(\sqrt{2}-1)}}$$
Multiplying by its conjugate we get,
$$v={\displaystyle \frac{2}{(\sqrt{2}-1)}}\times {\displaystyle \frac{(\sqrt{2}+1)}{(\sqrt{2}+1)}}$$
$$\Rightarrow v={\displaystyle \frac{2(\sqrt{2}+1)}{{(\sqrt{2})}^2-1^2}}$$
$$\Rightarrow v=2(\sqrt{2}+1)m/s$$

Therefore the correct option is C.

Note:
When we need to accelerate an object we need to apply a certain amount of force. To apply this force, we need to do work. When the work is done on an object energy is transferred and then the object will start moving at a new speed. This energy that is transferred to an object is known as kinetic energy. Also, kinetic energy is a scalar quantity and therefore it contains the only magnitude. The SI unit of kinetic energy is said to be Joules which is equal to $$kg-m^2{s}^{-2}$$. The CGS unit of kinetic energy is said to be erg. One of the best examples of kinetic energy is vibrational motion.