Which Equations constant I use here ?

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To find the original speed of a truck that covers 50 meters in 8 seconds while slowing down to a final speed of 2.5 m/s, the correct equation to use is Dx = (v + v0)t/2. Substituting the known values, the equation simplifies to 50 = (2.5 + v0) * 8 / 2. The calculations reveal that v0 equals 10 m/s, aligning with the answer provided in the book. The confusion arises from incorrect manipulation of the equation, emphasizing the importance of careful algebraic steps. Understanding the correct application of the equation leads to the accurate determination of the original speed.
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Hi ALL



which Equations constant I use here ?



a truck covers a distance of 50 m in 8 s while smoothly sowing down to a final speed of 2.5 m/s . find its original speed



Dx = 50 m

t = 8 s

v = 2.5 m/s



Now I take this equation bit I don't now why I didn't get the correct answer





Dx = (v + v0 ) t/2



=50 = 2.5 + v0 ) 8 /2

=2 X 50 + 92.5 + v0 ) 8 /2

=100 = 20 + 8v0/2

=128v

v = -128
 
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I think you've just added it up wrongly.
50m = (2.5m s^-1 + v0 ) 8s /2

is right.
Solve for v0.
 
but here in my book the answe is 10 m/s

??


they do v0 = 2DX/2-v = 2 X 50 /8 -2.5 = 10

??please I want the help beacuse I fell confused

I wait
 
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