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Math Help - Direct and Inverse Variation

  1. #1
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    Direct and Inverse Variation

    A car's stopping distance varies directly with the speed it travels, and inversely with the friction value of the road surface. If a car takes 60 feet to stop at 32 mph, on a road whose friction value is 4, what would be the stopping distance of a car traveling at 60 mph on a road with a friction value of 2?

    MY WORK:

    Let d = distance
    Let k = constant of variation
    Let s = speed
    Let f = friction

    I came up with the equation d = ks/f

    I then proceeded to do the following:

    60 = 32k/4

    240 = 32k

    240/32 = k

    8 = k

    Now that I found k, I proceeded to find the distance:

    d = 60(8)/2

    d = 240 feet

    The correct answer is 225 feet.

    Where did I go wrong?




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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by sologuitar View Post
    A car's stopping distance varies directly with the speed it travels, and inversely with the friction value of the road surface. If a car takes 60 feet to stop at 32 mph, on a road whose friction value is 4, what would be the stopping distance of a car traveling at 60 mph on a road with a friction value of 2?

    MY WORK:

    Let d = distance
    Let k = constant of variation
    Let s = speed
    Let f = friction

    I came up with the equation d = ks/f

    I then proceeded to do the following:

    60 = 32k/4

    240 = 32k

    240/32 = k

    8 = k

    Now that I found k, I proceeded to find the distance:

    d = 60(8)/2

    d = 240 feet

    The correct answer is 225 feet.

    Where did I go wrong?



    Hi sologuitar,

    \frac{240}{32}\ne 8
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  3. #3
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    I see...

    Quote Originally Posted by masters View Post
    Hi sologuitar,

    \frac{240}{32}\ne 8
    I see that it was a simple error on my part.
    We know that 240/32 = 7.5 not 8. I now see where the error occured.
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