Calculate velocity after certain displacement Given is a function $a(t)$ for the acceleration. Starting from an initial velocity $v_0$ I want to calculate the velocity $v_1$ after a certain displacement $s$.
Is this calculation possible since there is no time involved although $a$ is a function of time?
What I actually know:
$$\frac{\mathrm{d}s}{\mathrm{d}t} = \int a(t) \:\mathrm{d}t$$
But does this help me since it's the derivation/integration over time.
 A: Yes, you can, if the velocity is always positive.
If that is the case, then the function $s(t)$ is invertible, and you are at the position $s$ only at the time $t(s)$. Moreover, the derivative of $t(s)$ is:
$$
\frac{dt}{ds} = \dfrac{1}{\dfrac{ds}{dt}}.
$$
Therefore:
$$
a(t) = \frac{dv}{dt} = \frac{dv}{dx}\,\frac{dx}{dt} = \frac{dv(x)}{dx}\, v(t),
$$
for some function $v(x)$ (that is exactly what you want to find), such that $v(x(t))= v(t)$.
In Leibniz notation:
$$
a\cdot dx = v\cdot dv,
$$
which means:
$$
\int_{x_0}^x a(x)\cdot dx =  \int_{v_0}^v v\cdot dv = \frac{1}{2}v^2 - \frac{1}{2}v_0^2.
$$
So:
$$
v^2 = v_0^2 + 2\int_{x_0}^x a\cdot dx,
$$
and to get $v$ just take the square root.
For example, if $a$ is constant, you get the well known relation:
$$
v^2 = v_0^2 + 2a\cdot \Delta x.
$$
A: As suggested in the comments for a given function of displacement with time $t$ can be find as a inverse function of displacement.
$$s=f(t) \implies t=f^{-1}(s)$$  .
Now You can use the expression
$$v_2=v_1+\int_{t_1}^{{t_2}} adt$$$$v_2=v_1+\int_{f^{-1}(s_1)}^{f^{-1}{(s_2)}} adt$$
