Find time at which position vector points towards origin

[ehild]

I understand what you're saying but isn't this a little too complicated for a 1 mark A'Level question?

You are looking for one t that satisfies both the i and j components at the same time.
Let v = cR, where c is any non-zero constant. If c is negative then the velocity is pointing toward the origin. When c is positive, the velocity is pointing away from the origin.

$\vec v = c \vec R$ means that v_x=c R_x and v_y =c R_y. Dividing the first equation with the second one, v_x/v_y=R_x/Ry . Substitute the expressions in terms of t,
\frac{1-t^2}{3t-5t^2}=\frac{-2t}{3-10t}
and solve for t. You get two solutions. One of them means a positive c , the other solution means a negative one.

 

[Taniaz]

Thank you RUber, Haruspex and ehild for your help! Much appreciated! :)

 

[haruspex]

Thank you RUber, Haruspex and ehild for your help! Much appreciated! :)

If you are familiar with cross products, another route is to note that the collinearity of the position and velocity vectors means their cross product is zero. This avoids bringing in c and gives you an equation in t only (though it might be nasty). Having extracted the possible values of t, you could then evaluate the two vectors at those times to find out which has velocity towards the origin and which away.

 

[Taniaz]

So I got that t= 1/3 and t = 3
At t=3, c = 3/4 and at t=1/3, c = -3/4
So at t=1/3 it's pointing towards the origin.

 

[haruspex]

So I got that t= 1/3 and t = 3
At t=3, c = 3/4 and at t=1/3, c = -3/4
So at t=1/3 it's pointing towards the origin.

Looks right.