Kinematic Equations Used to Find A.) Average Speed & B.)Acceleration

AI Thread Summary
The discussion revolves around solving a physics problem involving a truck's motion, specifically finding its original speed and acceleration while it slows down. The user applied the kinematic equations but arrived at incorrect answers for both parts. For part A, the original speed (Vo) was calculated as 12.5 m/s, but there was an algebraic mistake in the setup. For part B, the acceleration was computed as -1.26 m/s², which is also contingent on correcting the original speed. The community suggests reviewing the algebra to ensure accurate results.
lethalfresa
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Homework Statement



A truck covers 40.0m in 7.15 s while uniformly slowing down to a final velocity of 3.50 m/s. a.) Find the truck's original speed b.) Find it's acceleration



Homework Equations



Δx = Vav(Δt) = (V+Vo/2)Δt

a=(V-Vo)/(Δt)



The Attempt at a Solution



Part A.)


1.) I used the formula Δx = Vav(Δt) = (V+Vo/2)Δt

2.) Plugged in what is given & solved for Vo

40.0m=((3.50m/s=Vo)/(2))(7.15s)

Vo=12.5 m/s

Part B.)

1.) I used the formula a=(V-Vo)/(Δt)

2.)Plugged in what I have

(3.50 m/s-12.5 m/s)/(7.15s) = -1.26 m/s2

Apparently both my answers are wrong and I am not sure what I am doing wrong. I tried this multiple times, as well as other different versions of this problem which I get correctly.

Your input would be appreciated. Thanks :)
 
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lethalfresa said:
40.0m=((3.50m/s+Vo)/(2))(7.15s)

Vo=12.5 m/s
Check your algebra here.

Part B.)

1.) I used the formula a=(V-Vo)/(Δt)

2.)Plugged in what I have

(3.50 m/s-12.5 m/s)/(7.15s) = -1.26 m/s2
It will be correct when you fix your Vo. :smile:
 
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