Distance and velocity question I know that speed is the derivative of distance. So integrating speed should give you distance. Let's suppose we have a speed which obeys this function:
$$ v(x) = 2^{2^x} $$ 
So at time 0 the speed is 2 m/s, at time 1 4 m/s, at time 2 16 m/s, and so on.
I would like to calculate the amount of time required to travel a distance $Y$ (let's say 1 km for the sake of the example).
Would I be correct to do this?
$ \int 2^{2^x} dx $ from 0 to t = 1000 m.
 A: No.
First of all, the expression for the velocity that you have there is dimensionally wrong: if $x$ is position, the units of $v$ will turn out to be something$^{metres}$ instead of metres/seconds.
So you should stick a constant $C$ in front of your expression so that it takes care of the units, like:
$$ v = C2^{2^x} $$
Velocity =  derivative of distance with respect to time, so (in 1D) $ v = \frac{dx}{dt}$;
This is a differential equation, use the method of separation of variables, basically you just consider $dx$ and $dt$ as factors instead of as an operator:
$ v\cdot dt = dx $
$ C2^{2^x} \cdot dt = dx $
$ dt = \frac{dx}{C2^{2^x}} $
and now you integrate:
$\int_0 ^T dt = \int_0 ^Y \frac{dx}{C2^{2^x}} $
so $$ T  = \int_0 ^Y \frac{dx}{C2^{2^x}} $$
But that integral does not look nice...
IF YOU MEANT $$ v(t) = 2^{2^t} $$ then I would again suggest sticking a constant in front, so $ v(t) = C2^{2^t} $, and then do the same thing:
$v\cdot dt = dx $
$ C2^{2^t} \cdot dt = dx $
$\int_0 ^Y dx = \int_0 ^T C2^{2^t} \cdot dt$
and you should get $T$ from that, but the integral again does not look nice...
