optimization car problem

**hazecraze** · December 5th 2009, 12:02 PM

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.

**lvleph** · December 5th 2009, 12:05 PM

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

**hazecraze** · December 5th 2009, 12:29 PM

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?

**lvleph** · December 5th 2009, 12:36 PM

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.

**lvleph** · December 5th 2009, 01:07 PM

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 $r(\frac{v_0}{16}) = 200$ .