Answer
The particle's kinetic energy increases by a factor of 9
Work Step by Step
We can write an expression for the kinetic energy when the speed is $v$:
$KE_1 = \frac{1}{2}mv^2$
We can write an expression for the kinetic energy when the speed is $3v$:
$KE_2 = \frac{1}{2}m(3v)^2$
$KE_2 = 9\times ~\frac{1}{2}mv^2$
$KE_2 = 9\times ~KE_1$
The particle's kinetic energy increases by a factor of 9.
