SOLUTION: A bicyclist accelerates uniformly from a speed of 2.50 m/s forward to 5.55 m/s forward at a rate of +o.525 m/s^2. How far did the bicyclist move during this acceleration?

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Question 1010399: A bicyclist accelerates uniformly from a speed of 2.50 m/s forward to 5.55 m/s forward at a rate of +o.525 m/s^2. How far did the bicyclist move during this acceleration?
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
We use the formula
(Vf)^2 - (Vo)^2 = 2ax
so that distance covered x is
x = [(Vf)^2 - (Vo)^2] / 2a
x = [5.55^2 - 2.50^2] / (2*0.525)
x = 23.4 m