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 = \(\frac{1}{2}mv^2\)
Let us consider the initial and final velocity of the car are denoted as \(v_{1}\ and\ v_{2}\).
Here, \(v_{1}=\ 30 mph\)
\(v_{2}=\ 60\ mph\)
The initial kinetic energy of the car \(K.E_{1}=\ \frac{1}{2}mv_{1}^2\)
The final kinetic energy of the car \(K.E_{2}=\ \frac{1}{2}mv_{2}^2\)
Hence, the ratio of final kinetic energy to initial kinetic energy is calculated as -
\(\frac{K.E_{2}} {K.E_{1}} =\ \frac{1/2mv_{2}^2} {1/2mv_{1}^2}\)
\(\frac{K.E_{2}} {K.E_{1}}=\ \frac{v_{2}^2} {v_{1}^2}\)
\(=\ \frac{(60)^2}{(30)^2}\)
\(=\ \frac{4}{1}\)
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.
