# Finding Displacement through integration

1. May 31, 2006

### GregA

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: May 31, 2006
2. May 31, 2006

### Staff: Mentor

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

3. May 31, 2006

### GregA

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

4. May 31, 2006

### Staff: Mentor

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.

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