I'm having trouble with the following question:(adsbygoogle = window.adsbygoogle || []).push({});

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 ... what am I doing wrong?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Finding Displacement through integration

**Physics Forums | Science Articles, Homework Help, Discussion**