Non uniform circular motion

1. Oct 7, 2013

Panphobia

1. The problem statement, all variables and given/known data

3. The attempt at a solution
So I just want to ask any of you who know physics pretty well if I am on the right track on this question, so it is asking for the magnitude of the acceleration of the car, and the direction. The magnitude of acceleration is (v^2/R) v = velocity and R = radius, so it is actually (at)^2/R where a = tangential acceleration, then plug in the numbers (0.75*20)^2/60 = 3.75 that is the centripetal acceleration. Now the magnitude of the acceleration of the car is just adding the radial acceleration and the tangential so sqrt(3.75^2 + 0.75^2) = 3.82 m/s^2. Now to get the direction, would I just use trigonometry for this?

Last edited: Oct 7, 2013
2. Oct 7, 2013

nasu

Yes, trigonometry will do.

3. Oct 7, 2013

Panphobia

But when I use trigonometry I just get the angle between the centripetal acceleration and resultant, what does the angle need to be with respect to?

4. Oct 7, 2013

Staff: Mentor

I suppose that since they want the "angle with respect to the track at this time", they'd want the angle with respect to the local tangent to the track, since the tangent represents the instantaneous direction of the track (and the direction of motion of the car).

5. Oct 7, 2013

Panphobia

Ahhhhh so basically it would be theta = 90 - cos^-1((Atot^2 - Ar^2 - Ac^2)/(-2(Ar)(Ac)))

6. Oct 7, 2013

Staff: Mentor

Should be a bit simpler than that, no? You're dealing with two vectors which are at right angles to each other, and one of them is pointing in the direction of travel.

7. Oct 7, 2013

Panphobia

ahhhhh!!!! I was thinking in relative motion still, both of those things are on the assignment, yea I know I figured it out right after I posted that, just soh cah toa :P

8. Oct 7, 2013