When a cars speed doubles, what happens to its kinetic energy?

The correct option is E It is quadrupled
Initial kinetic energy $K=\frac{1}{2}m{v}^2$

New speed of the car $v^{\prime }=2v$

Thus new kinetic energy $K^{\prime }=\frac{1}{2}m(v^{\prime }{)}^2=\frac{1}{2}m(2v{)}^2=4K$

Thus kinetic energy is quadrupled.