# optimization car problem

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• Dec 5th 2009, 12:02 PM
hazecraze
optimization car problem
A car brakes to avoid a truck stopped in the road ahead. The car brakes with a constant deceleration of $16 \frac{ft}{sec^2}$, 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.
• Dec 5th 2009, 12:05 PM
lvleph
What is the relationship between $v(t),a(t), \text{ and }r(t)$?
• Dec 5th 2009, 12:29 PM
hazecraze
You differentiate/integrate to get them. But if I integrate to the position function to use 200, I'll have two variables: $200=-8t^2 +ct$. So, I'm guessing one of the variables should be 0 since the car is stopping?
• Dec 5th 2009, 12:36 PM
lvleph
Okay, let's say the initial speed is $v(0) = v_0$ then $v(t) = \int \! -16 \,dt -16t + c$. From here we solve for the integration constant.
$v(0) = -16\cdot 0 + c = v_0$, so $c=v_0$. Now we integrate again
$r(t) = \int \! -16 t + v_0 \,dt = -8t^2 + v_0 t + d$. From here we will assume the initial position $r(0) = 0$ which leads to $d=0$ and $r(t) = -8t^2 + v_0 t$. From here I am sure you can finish.
• Dec 5th 2009, 01:07 PM
lvleph
Okay one more hint. We still need to determine the time at which the car stops. If the car has stopped this means that $v(t) = 0$. Using this we can determine the stopping time.
$-16t + v_0 = 0 \Rightarrow t = \frac{v_0}{16}$.
Now we solve the equation $r(\frac{v_0}{16}) = 200$.
• Dec 5th 2009, 03:13 PM
hazecraze
Aha!

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