Question

A particle’s position is ~r = (ct2 − 2dt3 )iˆ+ (2ct2 − dt2 )jˆ, where c and d are positive constants. Find the expressions for times t > 0 when the particle is moving in (a) x-direction (b) y-direction

Answer 1

The velocity of the particle is given by the derivative of the position vector:

$\vec v = \dfrac{\mathrm d\vec r}{\mathrm dt} = (2ct-6dt^2)\,\vec\imath + (4ct-2dt)\,\vec\jmath$

(a) The particle is moving in the x-direction when the y-component of velocity is zero:

$4ct-2dt = 2t (2c - d) = 0 \implies t=0$

But we want t > 0, so this never happens, unless 2c = d is given, in which case the y-component is always zero.

(b) Similarly, the particle moves in the y-direction when the x-component vanishes:

$2ct-6dt^2 = 2t (c - 3dt) = 0 \implies t=0 \text{ or }c-3dt = 0$

We drop the zero solution, and we're left with

$c-3dt = 0 \implies c=3dt \implies \boxed{t = \dfrac c{3d}}$

In the case of 2c = d, this times reduces to t = c/(6c) = 1/6.