# Finding Displacement through integration

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):

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

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? Last edited:

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Astronuc
Staff Emeritus
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

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