Nonuniform circular motion problem.

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feathermoon
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Homework Statement



A car moves on a circular track of constant radius b. If speed of car varies with time t according to v = ct, where c is a positive constant, show that the angle between the velocity vector and the acceleration vector is 45 degrees at time t=√b/c. (Hint: At this time the tangential and normal components of the acceleration are equal in magnitude.

Homework Equations



r= ib sin(ωt) + jb cost(ωt)
r=r[itex]_{e}r[/itex]
a=v^2/b

The Attempt at a Solution



Few ideas:
r=1/2ct^2, v=ct, a=c
At time t=√b/c, [itex]_{a}r[/itex]=[itex]_{a}t[/itex], so v^2/b=c?

I just really need a hint to get started in the right direction is all.. :[
 
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Seems to me that if you can show that the tangential and normal components are equal, you are done. And you've just about done that.

Since they give you the answer, all you have to do is plug in the value and determine the angle with respect to the tangent.
 
I'm just worried because I found the tangential component, v^2/b as a hint online, and I don't know how it was arrived at.

Thanks for your reply too, once I read it I saw what you meant immediately. I guess I just needed a nudge.
 
At first I thought I had to find equations of motion and position for it to finish this problem, and I couldn't think of a way to go from either r in polar or Cartesian to an absolute value v=ct. So I was worrying a lot.
 
feathermoon said:
I'm just worried because I found the tangential component, v^2/b as a hint online, and I don't know how it was arrived at.

Thanks for your reply too, once I read it I saw what you meant immediately. I guess I just needed a nudge.

Compare the standard motion equation v = u + at to the expression v = ct to see what c means.

Centripetal acceleration is given by a = v2/R and you are given the size of the radius.

It is just a case of combining those formulae/expressions really.