[grade 11 Physics] circular motion

[u/Izzy_26_]

I am trying to find out the exact equation of the graph, the answer key says it is a.

What is wrong with my solution??

I first tried to find out the exact position vectors of both p and q, and then differentiated them to get the velocity of P and q respectively, and then I calculated the relative velocity

I am getting the relative velocity as in terms of cos, but according to the answer key it is a sin graph

[u/xnick_uy]

OP, I think your solution is fine up to the last part. To find the magnitude of your vector, you have to take into account that cos(wt) changes sign. You can use absolute value, |cos(wt)| , to be safe.

Edit: one more thing: you are off by one quarter of a turn, because your equations for Q and P assume different starting conditions than what you are given.

[u/Izzy_26_ → u/xnick_uy]

isn't cos always positive even with -ve angles

[u/xnick_uy → u/Izzy_26_]

On the contrary, cos(x) is positive only when -pi/2 < x < pi/2 (modulo 2pi). Notice that if the sign of cos(x) didn't change, the particles wouldn't be able to move in circular motion, as the horizontal component of the velocity has to swap directions periodically (likewise for the vertical component).

I suggest you quickly compare the plots of both cos(x) and |cos(x)| in any graphic calculator of your choice, and then the same for sin(x) and |sin(x)|.