Finding Displacement through integration

[GregA]

I'm having trouble with the following question:

at any time t, the acceleration of a particle P, traveling 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):

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

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

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

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

problem is...the books answer is s=\frac{32t^2}{15(t+1)} 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 ... what am I doing wrong?

 

[Astronuc]

If v is given by

\frac{-k}{2(t+1)^2}+C

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

C = \frac{k}{2(0+1)^2} = \frac{k}{2},

not that C = 0.

Then one needs the intial dispacement at t=0

 

[GregA]

agh...just kept overlooking that screw up in my working ...Thanks astronuc

 

[Astronuc]

I've done the same

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