
optimization car problem
A car brakes to avoid a truck stopped in the road ahead. The car brakes with a constant deceleration of $\displaystyle 16 \frac{ft}{sec^2}$, producing skid marks that measure 200 feet before it is able to stop. How fast was the car moving (in ft/sec) when the brakes were first applied?
I gathered that a=16 and 200 will be some part of s(t). So I need a t value to plug into v(t), but how should I do it? answer is 80 ft/sec.

What is the relationship between $\displaystyle v(t),a(t), \text{ and }r(t)$?

You differentiate/integrate to get them. But if I integrate to the position function to use 200, I'll have two variables:$\displaystyle 200=8t^2 +ct$. So, I'm guessing one of the variables should be 0 since the car is stopping?

Okay, let's say the initial speed is $\displaystyle v(0) = v_0$ then $\displaystyle v(t) = \int \! 16 \,dt 16t + c$. From here we solve for the integration constant.
$\displaystyle v(0) = 16\cdot 0 + c = v_0$, so $\displaystyle c=v_0$. Now we integrate again
$\displaystyle r(t) = \int \! 16 t + v_0 \,dt = 8t^2 + v_0 t + d$. From here we will assume the initial position $\displaystyle r(0) = 0$ which leads to $\displaystyle d=0$ and $\displaystyle r(t) = 8t^2 + v_0 t$. From here I am sure you can finish.

Okay one more hint. We still need to determine the time at which the car stops. If the car has stopped this means that $\displaystyle v(t) = 0$. Using this we can determine the stopping time.
$\displaystyle 16t + v_0 = 0 \Rightarrow t = \frac{v_0}{16}$.
Now we solve the equation $\displaystyle r(\frac{v_0}{16}) = 200$.

Aha!
$\displaystyle \frac{8v^2}{256} + \frac{v^2}{16}=200$
$\displaystyle \frac{2v^2}{64} + \frac{4v^2}{64}=200$
$\displaystyle \frac{2v^2}{64}=200$
$\displaystyle 2v^2=12800$
$\displaystyle v=80$
(Happy)