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# Physics/Getting graphs from an acceleration vs time graph

Question

acceleration vs time
how would you proceed with plotting a velocity versus time graph and position versus time graph of the image provided.

v = v0+at
so velocity increases from 0 to 5, v = 2+2(5) = 12m/s
it stays constant from 5 to 15
velocity decreases from 15 to 25, v = 12 -2(10) = -8m/s
it stays constant from 25 to 35
velocity increases from 35 to 40, v = -8+2(5) = 2m/s
constant after 40

x(t) = x0+v0t+(1/2)at^2
x(5) = -1+2(5)+(1/2)(2)(5^2) = 34m
x(15) = 34+12(10) = 154m
x(25) = 154+12(10)-(1/2)2(10^2) = 174m
x(35) = 174 -8(10) = 94
x(40) = 94 +2(5) +(1/2)(2)(5^2) = 129m
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#### Richard J. Raridon

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