Answer
$(a)\space\frac{4c}{3d}$
$(b)\space \frac{c}{3d}$
Work Step by Step
To find the velocity vector $\vec V$, we have to differentiate equation (1) by $t$
$\frac{d}{dt}\vec r=\frac{d}{dt}(ct^{2}-2dt^{3})\space i+\frac{d}{dt}(2ct^{2}-dt^{3})\space j$
$\vec V=(2ct-6dt^{2})\space i+(4ct-3dt^{2})\space j$
(a) When the particle is moving in the x-direction, the y-component of the velocity vector must be zero. So we can write,
$4ct-3dt^{2}=0$
$t=\frac{4c}{3d}$
(b) When the particle is moving in the y-direction, the x-component of the velocity vector must be zero. So we can write,
$2ct-6dt^{2}=0$
$t=\frac{c}{3d}$
