Question

When a boy doubles his speed, his kinetic energy becomes ___times the initial kinetic energy.

Answer 1

K. E = $\frac{1}{2}m{v}^2$

Let initial speed be $v_1$

then final speed will be $2v_1$

Initial K. E = $\frac{1}{2}m{v}_1^2$

Final K. E = $\frac{1}{2}m(2v_1{)}^2=\frac{1}{2}m\times 4v_1^2=2m{v}_1^2$

$\frac{\text{Final K.E}}{\text{Initial K.E}}=\frac{2m{v}_1^2}{\frac{1}{2}m{v}_1^2}=4$

or, Final K. E = 4 $\times$ Initial K. E