Answer
$k= \frac{2}{3}s^{-2}$
Work Step by Step
We are given that
$v_x = kt^2 m/s$
At $ t=3.0s $
$ v = k(3.0 s)^2 m/s$
$ v = (9.0*k )m/s*(s^2)$
$ v= (9.0*k )ms$
we have
$ x_0 = -9.0m $ at $ t_0=0s$ and $ x_1 = 9.0m $ at $ t_1=3.0s$
$ \therefore v = \frac{x_1 -x_0}{t_1-t_0} m/s = \frac{18.0}{3.0}m/s$
$ v= 6.0m/s$
thus at $ t=3.0s, v=6.0m/s$
These two values must be equal
$ (9.0*k) ms= 6.0 m/s$
$ \implies k= \frac{2}{3}s^{-2}$
