# Non uniform circular motion

## 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?

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Yes, trigonometry will do.

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?

gneill
Mentor
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?
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).

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

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

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

gneill
Mentor
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 