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Homework Statement
A plot of acceleration versus time for a particle is shown. Whats the difference between its position at t=4s and t=0s if v(0)=-4m/s.
http://img441.imageshack.us/img441/532/dynmc8.th.png
Homework Equations
The Attempt at a Solution
Ive been working on this problem for hours until I thought I solved it...I got the right answer but I realized after that I made a mistake..but still got the right answer. After I tried to do it the right way i keep getting the wrong answer.
v(0)=-4m/s
v(1)=16m/s (since constant acceleration from 1s to 2s)
[tex]a_{12} = -20t+40 = dv/dt[/tex] - y=mx+b, for the non-uniform section (1s-2s)
now I rearrange the above and integrated to find v(2)
[tex]\int_{v(1)}^{v(2)}dv = \int_1^2 -20t+40dt[/tex]
This is where I made my mistake, instead of using v(1) I used v(0)=-4m/s
continuing...[tex]v(2)-(-4m/s)=-10t^2+40t |_1^2[/tex]
solving gives v(2)=26m/s
Now I integrated again...[tex]-10t^2+40t = dx/dt[/tex] to find the displacement for the non-uniform section...giving me an answer of 68/3.
Anyways..add all the displacements up and I end up with 80.66 which is the right answer. How come when I try and use v(1)=16m/s for my lower limit in the first integration I keep getting wrong numbers?
Any help would be appreciated
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