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Homework Help: I have no idea how to solve this problem.

  1. Aug 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Driving along a crowded freeway, you notice that it takes a time t to go from one mile marker to the next. When you increase your speed by 7.0 mi/h, the time to go one mile decreases by 12 s. What was your original speed?


    2. Relevant equations
    D = VT

    3. The attempt at a solution

    D = VT
    1 = (Vi+7)(T-0.2)
    Vi = 1/(T-0.2) - 7

    I have 2 unknowns and I don't know how to solve????
     
  2. jcsd
  3. Aug 24, 2010 #2
    You are almost done! You know the value of D, and by eliminating T from the first two equations you shold get the right answer.
     
  4. Aug 24, 2010 #3
    I can't solve it. I ended up with

    V = 1/(1/V-0.2) - 7
     
  5. Aug 25, 2010 #4

    ehild

    User Avatar
    Homework Helper

    Originally, it took T hours to travel 1 mile with Vi speed.

    ehild
     
  6. Aug 25, 2010 #5
    Can someone please tell me how to solve this instead of beating around the bush?
     
  7. Aug 25, 2010 #6
    The distance [itex]D = 1 \, \mathrm{mi}[/itex]. Also, [itex]12 \, \mathrm{s} = 1/5 \, \mathrm{min} = 1/300 \, \mathrm{h}[/itex].
     
  8. Aug 25, 2010 #7
    Hint: There's another relation between [itex]v_{i}[/itex] and [itex]t[/itex] that you have not taken into account.
     
  9. Aug 26, 2010 #8
    I have tried that, I just cannot find the answer.
     
  10. Aug 26, 2010 #9

    presbyope

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    Gold Member

    You tried solving both equations for V_i and you couldn't find the answer? Did you get a quadratic equation?
     
  11. Aug 26, 2010 #10
    This is an equation with one unknown that can be further simplified. However, it is wrong because you had not converted 12 s into hours correctly.
     
  12. Aug 26, 2010 #11
    Sorry, it would be

    V = 1/(1/V-0.00333) - 7

    I still don't know how to solve for V
     
  13. Aug 26, 2010 #12
    Get rid of the double fractions first and then multiply everything with the common denominator to get rid of fractions.
     
  14. Aug 26, 2010 #13
    V = 1/(1/V-0.00333) - 7
    V = -V/0.003333 - 7
    V + 7 = -V/0.0033333
    0.003333V + 0.0233333 = -V
    0.003333V + V = 0.023333
    V(0.00333 + 1) = 0.023333
    V = 0.023333 / (0.00333 + 1)
    V = 0.022 mph?
     
  15. Aug 26, 2010 #14
    This is incorrect.
     
  16. Aug 26, 2010 #15
    How is it incorrect? If you take

    1 / 2 / 3 = 3 / 2
     
  17. Aug 27, 2010 #16
    But, this is not what you have. There is a "-" sign in there. Also, if you take:

    as it is written, it might mean:

    (1/2)/3 = 1/6 or 1/(2/3) = 3/2

    so you have to be careful with your notation. I suggest learning LaTeX:

    [tex]
    \frac{1}{\frac{2}{3}} = \frac{3}{2}
    [/tex]
     
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