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Physics/Getting graphs from an acceleration vs time graph

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

Answer
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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I can answer most questions in undergraduate physics courses, including electricity and magnetism, atomic and nuclear, mechanics and optics.

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I have taught undergraduate physics courses

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Sigma Xi, AAAS, SE section of APS

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BA in math, MA in physics, PhD in physical chemistry

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