#### Answer

a) This never happens.
b) This happens at $t = \frac{2c}{3d}$.

#### Work Step by Step

a) The particle is moving in the x-direction when the y-component is 0. Thus, we find:
$4ct -3dt^2 =0\\4ct = 3dt^2 \\ 4c = 3dt \\ t = \frac{4c}{3d}$
The particle is moving in the y-direction when the x-component is 0. Thus, we find:
$2ct - 6dt^2 = 0 \\ 2ct = 6dt^2 \\ t = \frac{c}{3d}$
The particle is at rest when both the x and y-components are 0. However, seeing the values above, we see that this will never happen.
b) Since the second derivative of the x-acceleration is not 0, we see that there is a possibility that the acceleration will be just in the x-direction. The particle is accelerating in the x-direction when the y-acceleration is 0. Thus, we take the second derivative of the y-position to find:
$4c - 6dt = 0 \\ t = \frac{2c}{3d}$