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Math Help - Quadratic formula word problem

  1. #1
    Junior Member
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    Quadratic formula word problem

    I'm stuck on this problem...what did i do wrong?
    d(t)=(velocity) (t)+1/2 (acceleration) (t)^2

    cycles 4 meters per second
    rate of .4 meters per second squared
    vertical distance from top to bottom of hill is 25 meters

    How long until goes down the hill?

    I have...

    (1/5)t^2+4t-25

    From the quad formula I got 1/4 and -5/2, but the answer is 5 seconds....why?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by RaphaelB30 View Post
    I'm stuck on this problem...what did i do wrong?
    d(t)=(velocity) (t)+1/2 (acceleration) (t)^2

    cycles 4 meters per second
    rate of .4 meters per second squared
    vertical distance from top to bottom of hill is 25 meters

    How long until goes down the hill?

    I have...

    (1/5)t^2+4t-25

    From the quad formula I got 1/4 and -5/2, but the answer is 5 seconds....why?
    s=ut+\frac{1}{2}at^2

    25=4t+\frac{1}{2}(0.4)t^2

     <br />
0.2t^2+4t-25=0<br />

    then t=\frac{-4\pm\sqrt{4^2-4(0.2)(-25)}}{2(0.2)}

    i got t=5 , -25

    so its 5 coz time cant be negative .
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Chicago, IL
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    Quote Originally Posted by RaphaelB30 View Post
    I'm stuck on this problem...what did i do wrong?
    d(t)=(velocity) (t)+1/2 (acceleration) (t)^2

    cycles 4 meters per second
    rate of .4 meters per second squared
    vertical distance from top to bottom of hill is 25 meters

    How long until goes down the hill?

    I have...

    (1/5)t^2+4t-25

    From the quad formula I got 1/4 and -5/2, but the answer is 5 seconds....why?
    Note that \tfrac{1}{5}t^2+4t-25=0\implies t^2+20t-125=0

    The expression factors into (t-5)(t+25)=0\implies t=5 or t=-25. We disregard the second answer, since t>0.
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