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Finding Displacement through integration

  • Thread starter GregA
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  • #1
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I'm having trouble with the following question:

at any time t, the acceleration of a particle P, travelling in a straight line, is inversely proportional to (t+1)3. Initially, when t = 0, P is at rest at a point O, and 3 seconds later it has a speed 2ms-1. Find in terms of t the displacement of P from O at any time.

My working (using u = t+1):

[tex]a=\frac{k}{(t+1)^3}[/tex]
[tex]v=\int\frac{k}{(t+1)^3}dt[/tex] = [tex]\frac{-k}{2(t+1)^2}+C[/tex]

Now as the particle starts from rest at t= 0, C = 0, and when t =3, v = 2 such that

[tex]2 = \frac{-k}{2(3+1)^2}[/tex] and so K= -64 giving [tex] v = \frac{32}{(t+1)^2}[/tex]

I should now integrate this again to find s...
[tex]s=\int\frac{32}{(t+1)^2}dt[/tex] = [tex]\frac{32}{t+1}+C[/tex]
again, I can lose the constant because when t = 0 P is at the origin so that
[tex]s=\frac{32}{t+1}[/tex]

problem is...the books answer is [tex]s=\frac{32t^2}{15(t+1)}[/tex] and even worse...by differentiating this I find that v does indeed = 2 when t = 3. I am stumped as to how I should derive this though :frown: ... what am I doing wrong? :confused:
 
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Answers and Replies

  • #2
Astronuc
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If v is given by

[tex]\frac{-k}{2(t+1)^2}+C[/tex]

and v = 0 at t = 0, then this would mean,

[tex]C = \frac{k}{2(0+1)^2} = \frac{k}{2}[/tex],

not that C = 0.

Then one needs the intial dispacement at t=0
 
  • #3
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agh...just kept overlooking that screw up in my working :redface: ...Thanks astronuc :smile:
 
  • #4
Astronuc
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Science Advisor
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I've done the same :biggrin:

Sometimes, one has to just walk away and not think about it, and that's usually when one has an "Aha!" moment. :cool:
 

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