Using acceleration to find distance traveled

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SUMMARY

The discussion focuses on calculating the distance traveled by a car accelerating from 18 m/s to 45 m/s over 12 seconds with constant acceleration. The correct acceleration is determined to be 2.25 m/s². The participant initially misapplied the formula for distance, using the wrong variables. The accurate formula for distance, incorporating both initial velocity and acceleration, is s = ut + 1/2 at², which leads to a distance of 70.875 meters when applied correctly.

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Homework Statement


A car accelerates from 18 m/s to 45 m/s in 12 s. Assume constant acceleration.
What was its acceleration? How far did it travel?

I know that the acceleration is 2.25 m/s/s but I'm not sure how to go from acceleration to distance traveled.


Homework Equations


I'm guessing that its this one
vf^2 = vi^2 + 2a * ChangePosition

The Attempt at a Solution



(vf^2 - vi^2)/2a = change in position

So I get 45^2 - 18^2 / 24 which equals 70.875 meters. When I submitted the problem I got the acceleration right but I screwed up the change in position.
 
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d = 1/2 at^2
 
You accidentally substituted the time instead of the acceleration into your rearranged equation:

\Delta x=\frac{v_{f}^{2}-v_{i}^{2}}{2a}
 
ussfletcher said:
d = 1/2 at^2

To use this formula, the entire equation will need to be put to use. i.e.

s=ut+\frac{1}{2}at^{2}
s = displacement
u = initial velocity
 
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