Where Did I Go Wrong in Calculating Average Acceleration?

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SUMMARY

The average acceleration of a man standing still for the first 5 minutes and then walking at a constant speed of 2.5 m/s from 5 to 10 minutes is calculated incorrectly in the forum discussion. The correct average acceleration (a_avg) during the interval from 2.00 minutes to 8.00 minutes is 0.00694 m/s². The user mistakenly attempted to calculate the final velocity instead of recognizing that the final velocity is already provided. The confusion arose from misapplying the average velocity formula instead of focusing on the average acceleration formula, which is defined as ∆V/∆t.

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praecox
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Homework Statement



From t = 0 to t = 5.00 min, a man stands still, and from t = 5.00 min to t = 10.0 min, he walks briskly in a straight line at a constant speed of 2.5 m/s.

What is his average acceleration aavg in the interval 2.00 min to 8.00 min?

Homework Equations



I know that in the formula the A_avg is ∆V/∆t... but I can't make it work for me. It says the answer is .00694 m/s^2, but I keep getting a different answer.

The Attempt at a Solution



I tried to calculate the final velocity (V_f): ∆d/∆t: [2.5 m/s x 180 s]/480s. (or 2.5 m/s x 3 minutes)/8 minutes). This give me 0.975 m/s^2.
The initial velocity (V_i) seems to me that it would be 0 at t=2 minutes.
So A = [.975-0]/∆t = [.975 m/s]/360s = .0027 m/s^2.

but this is wrong.

Help please. where did I mess up?
 
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praecox said:
I tried to calculate the final velocity (V_f): ∆d/∆t: [2.5 m/s x 180 s]/480s. (or 2.5 m/s x 3 minutes)/8 minutes). This give me 0.975 m/s^2.
The final velocity is given--no need to calculate it! (You're calculating the average velocity, which is not needed.)
 
!
Thank you so much. I knew it was something silly I was messing up.
:blush:
 

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