How far down the runway does the plane take off? An airplane goes from $0$ to $220$ mph on a runway to take off. Assuming constant acceleration, how far down the runway does the plane travel before take-off? So I know I have to integrate until I get the distance formula which is $s(t)= \frac{1}{2}at^2 + C$. Since I am not given a time, how do I solve for this? Thank you!

With the provided data you can't find a single solution, you'll have to deal with a parametric one. This is what you can do:

$s(t) = \frac{1}{2}at^2$

$\dot{s}(t) = at$

From these 2, we can get the time:

$220 = at \Rightarrow t = \frac{220}{a}$

so, the solution is

$s(\frac{220}{a}) = \frac{220^2}{2a} = \frac{24200}{a}$