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

a) This never happens. b) This happens at $t = \frac{2c}{3d}$.
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: $4c - 6dt = 0 \\ t = \frac{2c}{3d}$