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I'm having trouble with the following question:

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?

**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?

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