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??
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
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