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Math Help - optimization car problem

  1. #1
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    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.
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  2. #2
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    What is the relationship between v(t),a(t), \text{ and }r(t)?
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  3. #3
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    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?
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  4. #4
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    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.
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  5. #5
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    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.
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  6. #6
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    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
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