Well, we can simply use this one formula:

\(\displaystyle a = c\)

Meaning the acceleration is constant, in this case, it is \(\displaystyle -22 \frac{ft}{s^2}\)

So, we simply take the antiderivative to find the velocity function:

\(\displaystyle \int -22dx\)

\(\displaystyle = -22x + C\)

Well, we know that this equals the velocity, but it's in \(\displaystyle \frac{mi}{h}\) and not \(\displaystyle \frac{ft}{s}\), so we have to convert it.

\(\displaystyle 50 \frac{mi}{h} = \frac{220}{3} \frac{ft}{s}\)

Now we set the constant equal to the initial velocity, seeing as it fits the equation to find the final equation:

\(\displaystyle v_0 = \frac{220}{3} = C\)

And we know it slows down to 0, so:

\(\displaystyle v_f = 0\)

\(\displaystyle 0 = -22x + \frac{220}{3}\)

We can generalize a formula from this:

\(\displaystyle v_f = v_0 + at\)

\(\displaystyle -22t = at\)

\(\displaystyle C = v_0\)

\(\displaystyle 0 = v_f\)

Now we take the antiderivative of the right side of that function:

\(\displaystyle \int [v_0 + at]dt\)

\(\displaystyle = v_0t + \frac{1}{2}at^2 + D\)

This is the position function, and we can assume that since the initial velocity was the C value, that the initial position value would be the D value:

\(\displaystyle D = h_0\)

So, this function is a function of position, which is commonly called s(t):

\(\displaystyle s(t) = h_0 + v_0t + \frac{1}{2}at^2\)

Now we plug in what we know (Assuming we started at position = 0):

\(\displaystyle s(t) = \frac{220}{3}t - 11t^2\)

There is your final equation.