Magnitudes and Directions In Uniform Circular Motion

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SUMMARY

The discussion focuses on calculating the magnitude and direction of displacement, velocity, and acceleration for a biker moving in uniform circular motion. The biker starts 40.0m east of a flag, traveling south at 10m/s, and after 30 seconds, is 40.0m north of the flag, heading east at the same speed. Key equations include a = v²/r for acceleration, emphasizing that both velocity and acceleration are not constant, leading to the interpretation of average values for these quantities.

PREREQUISITES
  • Understanding of uniform circular motion principles
  • Familiarity with vector displacement and velocity
  • Knowledge of average acceleration calculations
  • Proficiency in trigonometry for angle calculations
NEXT STEPS
  • Study the concept of average velocity in non-uniform motion
  • Learn about vector addition in displacement calculations
  • Explore the relationship between linear and angular acceleration
  • Investigate the effects of changing direction on velocity in circular motion
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Physics students, educators, and anyone interested in understanding the dynamics of motion in circular paths, particularly in solving problems related to displacement, velocity, and acceleration.

Rodriesk
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Homework Statement


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?

Homework Equations


a=v^2/r

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.

Thank you in advance for your replies.
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Rodriesk said:
multiply time and velocity to solve for distance (magnitude)
Distance travelled, ∫|v|.dt, is not the same as the magnitude of the displacement, |∫v.dt|.
Rodriesk said:
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.
Yes.
Rodriesk said:
the magnitude and direction of the velocity?
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?
Rodriesk said:
the magnitude and direction of the acceleration?
Again, acceleration is not constant, and might not be constant in magnitude, so I would interpret this as the average acceleration.
 

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