The answer to the question is 4 times i.e the kinetic energy of the car will increase four times when it will move with 60 mph as opposed to travelling with 30 mph.
CALCULATION:
Let us a consider a body having mass m which is moving with a velocity v.
The kinetic energy of the body is calculated as -
Kinetic energy K.E = [tex]\frac{1}{2}mv^2[/tex]
Let us consider the initial and final velocity of the car are denoted as [tex]v_{1}\ and\ v_{2}[/tex].
Here, [tex]v_{1}=\ 30 mph[/tex]
[tex]v_{2}=\ 60\ mph[/tex]
The initial kinetic energy of the car [tex]K.E_{1}=\ \frac{1}{2}mv_{1}^2[/tex]
The final kinetic energy of the car [tex]K.E_{2}=\ \frac{1}{2}mv_{2}^2[/tex]
Hence, the ratio of final kinetic energy to initial kinetic energy is calculated as -
[tex]\frac{K.E_{2}} {K.E_{1}} =\ \frac{1/2mv_{2}^2} {1/2mv_{1}^2}[/tex]
[tex]\frac{K.E_{2}} {K.E_{1}}=\ \frac{v_{2}^2} {v_{1}^2}[/tex]
[tex]=\ \frac{(60)^2}{(30)^2}[/tex]
[tex]=\ \frac{4}{1}[/tex]
Hence, the kinetic energy of the car will be increased to 4 times when it will move with speed from 30 mph to 60 mph.
