How do you find the final velocity when acceleration is changing between two values over some distance? How do you calculate a final velocity of an object when given its initial velocity and the object is accelerating between an initial and final acceleration over some given distance?
 A: Your problem here is that your equation has the form:
$$ \frac{dv}{dt} = f(x) $$
i.e. on the left side you have a derivative wrt time, but on the right side you have a function of distance. Solving this requires one of the (many) tricks that physicists only learn from experience. You need to use the chain rule to rewrite:
$$ \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = \frac{dv}{dx}v $$
So now you can rewrite your equation as:
$$ v \frac{dv}{dx} = f(x) $$
and then integrate:
$$ \int v dv = \int f(x) dx $$
A: Here is an example to show that the problem as stated in the original question is underdetermined.
Suppose the distance $s$ travelled by the object at time $t$ is given by
$s(t) = 91t^3 -49t^4 +21t^5$
Then the object's velocity and acceleration are
$v(t) = 273t^2 - 196t^3 + 105t^4 \\ a(t) = 546t - 588t^2 + 420t^3$
So we have $s(0)=v(0)=a(0)=0$ and $s(1)=63$, $v(1)=182$, $a(1)=378$
But now suppose the object's distance, velocity and acceleration are:
$s(t) = 51t^3 +21t^4 -9t^5 \\ v(t)=153^2+84t^3-45t^4 \\ a(t)=306t+252t^2-180t^3$
So now we have $s(0)=v(0)=a(0)=0$ and $s(1)=63$, $v(1)=144$, $a(1)=378$
So we have two different scenarios where the object travels a distance of $63$, its initial velocity and initial acceleration are both $0$, its final acceleration is $378$, but its final velocity is $182$ in one case and $144$ in the other.
A: As others have said, if you have an acceleration function given explicitly as a function of distance you can use mathematical tricks to do it, or use the
relation between work and kinetic energy
$1/2mv^2 = \int F(x) dx$
$1/2v^2 = \int a(x) dx$
However given that you actually have acceleration as a function of time, we can  do this to solve:
$V=\int a(t) dt$
Plug in $(t=0, v=v_{0})$
Then use this expression for velocity as a function of time to calculate
S(t)= $\int v dt$
Because we want to find out the velocity over a specific distance, we can use this equation to find the time at which the particle reaches a certain distance by setting s(t) to be a certain value, and solving the equation for t
Defining the inverse of s(t) that returns a time for a given distance $S^{-1}(s_{0})$
We can then plug this value for time back into our original equation for velocity
