Average acceleration from average velocity?

AI Thread Summary
A particle moving along a semicircular path of radius 5 m at a constant speed of 1 m/s experiences an average acceleration of (2 / (5π)) m/s². The total time taken for the journey is calculated as 5π seconds. The average velocity is determined to be (2/π) m/s, leading to a discrepancy when compared to the average acceleration formula. The discussion highlights that average acceleration is not equal to average velocity divided by time in this scenario. The relationship between average acceleration and average velocity can vary based on the chosen initial conditions and motion type.
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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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