Displacement from an Acceleration Time Graph

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SUMMARY

The discussion focuses on calculating displacement from an acceleration time graph using kinematic equations. The user has determined the time at which the velocity returns to 54 km/h (15 m/s) but struggles with finding the correct displacement value. They attempted to use the equations \(x_0 = v_0 t + \frac{1}{2} a t^2\) and \(x = x_0 + vt\), yielding an incorrect result of 270 m instead of the correct 178.8 m. The suggested solution involves setting up acceleration functions for the two regions and integrating them twice with respect to time.

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Homework Statement


upload_2016-2-5_1-42-37.png

Homework Equations


Kinematic equations
t = 18s when the velocity is back to 54km/h or 15m/s
at t = 2, V = 13.3 m/s
at t = 4.5, velocity is 1.5 m/s
at t = 1, V = 15m/s

The Attempt at a Solution


I have solved the first part and I have the time at which it is 54km/h again, but now I'm trying to figure out
the position. I think I could make a v-t graph and find the area under that curve, but it seems like that would be too hard to accurately represent the graph. Is there any other way to find displacement? I've tried
xo = vot + 1/2at2 but that doesn't seem to help
I also tried
x = xo + vt, but I get 270m as an answer. The correct answer is 178.8, but I just want to understand how to do it. Any help would be much appreciated!
 

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What you can try is set up a(t) functions for the two regions and integrate the functions twice wrt time.
 
You'll have to show your attempt(s) in detail.
 

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