# Homework Help: Magnitudes and Directions In Uniform Circular Motion

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1. Feb 21, 2017

### Rodriesk

1. The problem statement, all variables and given/known data
A biker is 40.0m to the east of a flag in a park, heading south at 10m/s. 30.0 seconds later, the biker is 40.0m north of the flag and heading east at 10.0m/s. For the biker in this 30.0s interval, what are:
a) the magnitude and direction of the displacement?
b) the magnitude and direction of the velocity?
c) the magnitude and direction of the acceleration?

2. Relevant equations
a=v^2/r

3. The attempt at a solution
The image attached shows how I've set up the problem. For a), my first thought was to multiply time and velocity to solve for distance (magnitude). For the direction, I thought about the triangle P1 and P2 form, and solved for the angle that the line joining those two points forms with the x-axis. My attempt at solving for magnitude makes sense to me, but I'm not sure about the answer for the direction.

For b), I already have the velocity, which is 10m/s, and I believe that would be the magnitude, however I'm unsure about the direction. The direction changes every second, and so does the velocity, so I think there would be multiple answers for the direction depending on the time we're taking into account.

For c), I think it's the easiest part, since I have the velocity and radius, I can use a = v^2 / r to solve for the acceleration magnitude, and if I understand the concept of circular motion, the direction of the acceleration is always headed toward the center. I'm confident about this one.

2. Feb 21, 2017

### haruspex

Distance travelled, ∫|v|.dt, is not the same as the magnitude of the displacement, |∫v.dt|.
Yes.
The velocity is not constant. We are not even told whether the magnitude of the velocity is constant. So I suggest the question is asking about the average velocity. How is that defined?
Again, acceleration is not constant, and might not be constant in magnitude, so I would interpret this as the average acceleration.