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Quadratic Word Problem

  1. Aug 27, 2006 #1
    I'm having trouble just figuring out how to set these problems up. Here is one of them.

    A jogger and a walker both cover a distance of 5 miles. The runner is traveling 1.5 times faster than the walker and finishes in 25 minutes less time. How fast is each going?

    I know it probably should include the r*t=d

    So I "tried" to make a table

    jogger = rate = 1.5x time = x-25 = 5
    walker I had tried x*x+25= 5 but
    I then tried to combine the two & got 2.5x^2-12.5x-10=0

    I know the answer is 4mph for the jogger & 6 for the walker. Any help on how to get the equation setup or where I'm going wrong would be appreciated.
  2. jcsd
  3. Aug 27, 2006 #2


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    "jogger = rate = 1.5x time = x-25 = 5 "
    What is x? in "rate= 1.5x" it appears to be the rate (speed) of the walker but in "time= x- 25" it appears to be the time the walker takes to walk the 5 miles. And, if that last "5" is the distance walked, "time= 5" makes no sense. Finally, whatever x is, in x*x+ 25= 5, you can't add 25 minutes to any quantity and get 5 miles as the result.

    Let r be the speed of the walker. since "r*t= d", t= d/r so the time required for the walker to go 5 miles is t= 5/r. The speed of the jogger is 1.5r so the time required for the jogger to go 5 miles is t= 5/(1.5r)= 10/(3r). The walker took 25 minutes longer and 25 minutes= 25/60= 5/12 hour so 5/r= 10/(3r)+ 5/12. Multiply the equation by 12r to get rid of the fractions. That does not give a quadratic equation.
  4. Aug 28, 2006 #3
    Thanks, that makes sense. What was throwing me off was the fact that all the other problems in the chapter were quadratic equations. So naturally I wanted to "try" & set the problem up as one... but as you can see I just got confused & "tried" to throw something together that would make it quadratic. This here helped a lot. Thanks again.
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