Average acceleration from average velocity?

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SUMMARY

The discussion centers on calculating the average acceleration of a particle traveling along a semicircular path with a radius of 5 m at a constant speed of 1 m/s. The average acceleration is derived as (2 / (5π)) m/s², while the average velocity is determined to be (2/π) m/s. The participant clarifies that average acceleration is not equal to average velocity divided by time, as this would introduce an incorrect factor of π² in the denominator. The relationship between average acceleration and average velocity is further explored through the context of constant acceleration in one-dimensional motion.

PREREQUISITES
  • Understanding of kinematic equations, particularly average acceleration and average velocity.
  • Familiarity with circular motion concepts and the relationship between distance, speed, and time.
  • Basic knowledge of calculus, particularly in relation to motion equations.
  • Ability to manipulate algebraic expressions involving π and other constants.
NEXT STEPS
  • Study the derivation of average acceleration in circular motion scenarios.
  • Learn about the differences between average and instantaneous acceleration.
  • Explore the implications of constant speed on average velocity and acceleration in various motion types.
  • Investigate one-dimensional motion equations, specifically how they relate to average velocity and acceleration.
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and motion analysis, as well as educators seeking to clarify concepts of average acceleration and velocity in circular motion.

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Homework Statement



A particle travels from point A to point B following a semicircular path (a half-circle) of radius 5 m, and it travels at a constant speed of 1 m/s. Find the average acceleration of the particle.


Homework Equations



Average acceleration = change in velocity/ change in time

Total time = distance / speed

The Attempt at a Solution



I already know the answer, and I understand why it's true:

v0 = 1 m/s
vf = -1 m/s

total time = (5 pi m) / (1 m/s) = 5pi seconds

Avg acceleration = (2 / (5pi) ) m/s^2

My question is: Is the average acceleration ever equal to the (average velocity/ time) ? In this case, I know it's not because:

Avg velocity = displacement / time = 10 m / 5pi sec = (2/ pi) m/s

And, if I divided by the time, I would get a factor of pi^2 in the denominator, which is not in my original answer.

I would really appreciate any help sorting this out! :smile:
 
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Consider a 1-dimensional motion, ##v(t)=v_0 + at## with constant a, from t=0 to T. The average acceleration is a, the average velocity is v0+aT/2. As you are free to choose v0, you can get your equality.

Particles at rest have zero average velocity and acceleration, they are a special case of my example for a=0.
 

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