Average Acceleration Confusion

[praecox]

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?

 

[Doc Al]

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.)

 

[praecox]

!
Thank you so much. I knew it was something silly I was messing up.
:blush: