## Essential University Physics: Volume 1 (3rd Edition)

$v=\sqrt{\frac{4ax^{1.5}}{3m}}$
We know that the potential energy is the integral of the force with respect to distance, so we find: $U = \int (a\sqrt x)dx=\frac{2}{3}ax^{1.5}$ We know that the kinetic energy is equal to the change in potential energy. Thus: $\frac{1}{2}mv^2 = \frac{2}{3}ax^{1.5}$ $v=\sqrt{\frac{4ax^{1.5}}{3m}}$