1. ## variable acceleration

Bit of help with this one if poss..

A vehicle has acceleration

f(t) = a/(1+t)^n

where a > 0, n>1(n not equal 2) are const.

Find x dot of (t) and x(t)

x(0)=x dot x(0)=0

what is the max speed poss??

2. so we have ...

$\displaystyle \frac{dv}{dt}=\frac{a}{(1+t)^n}$

This is a seperable ODE

$\displaystyle \int dv=\int \frac{a}{(1+t)^n}dt$

solving gives

$\displaystyle v=\frac{a(1+t)^{-n+1}}{-n+1}+C$

Using the intial condition v(0)=0

we get

$\displaystyle 0=\frac{a(1+0)^{-n+1}}{-n+1}+C$

so solving for C we get

$\displaystyle C=\frac{-a}{-n+1}=\frac{a}{1-n}$

So

$\displaystyle v=\frac{a(1+t)^{-n+1}}{-n+1}+\frac{a}{1-n}$

so now we get

$\displaystyle v=\frac{dx}{dt}=\frac{a(1+t)^{-n+1}}{-n+1}+\frac{a}{1-n}$

This is also a seperable ODE. So just repeate the process from above.

You should be able to finish from here

3. I don't understand how you know you get $\displaystyle -a$ on the top when you're solving for C... in fact, $\displaystyle n>1$ so the power of the (1+0) is bound to be positive surely? Surely it should just be $\displaystyle a$?

Also, I don't get where $\displaystyle -n+1$ has come from either