Question

If the velocity of the body increases 3 times how does its K.E changes?

Answer 1

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

New velocity = $3v$

New K.E = $\frac{1}{2}$ $m(3v{)}^2$ = 9 ( $\frac{1}{2}$ $m{v}^2$ ) = 9 x K.E

Hence, new kinetic energy becomes 9 times.