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Driving speed physics problem

  1. Jun 14, 2009 #1
    Here is my problem as stated:

    Driving along a crowded freeway, you notice that it takes a time (t) to go from 1 mile marker to the next. When you increase your speed by 7.8 mi/h, the time to go one mile decreases by 14 seconds. What was your original speed?

    I have tried to think of it as if I were going 60 mi/hr it would take me 1 minute to travel 1 mile and then I tried using the conversions I have just learned, but I keep getting stuck. Any help would be much appreciated.
     
  2. jcsd
  3. Jun 14, 2009 #2

    LowlyPion

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    Welcome to PF.

    Just write out they conditions as they are.

    d = v*t

    So ...

    1 mile = v * t = (v + 7.8) * (t - 14)

    Now the units are all wrong there, but I'm sure you can figure out how to make it all good right?
     
  4. Jun 14, 2009 #3

    Cyosis

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    Could you write down an equation for both situations, using x=vt?
     
  5. Jun 14, 2009 #4
    Thanks for the help. I'll give it a try and let you know what I come up with.
     
  6. Jun 14, 2009 #5
    So 1 mile = 5280 ft. So am I trying to find (v+7.8) * (t-14) an equation that will equal 5280? I am confused.
     
  7. Jun 14, 2009 #6
    What I am confused about is what numbers do I need to convert. For example, I know my time has decreased by 14 seconds for me to travel 1 mile. So instead of miles per hour should I convert this to miles per second?
     
  8. Jun 14, 2009 #7

    LowlyPion

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    Maybe try to convert mph to m/s? Then convert back when you solve?

    V* t = (V+7.8*(.447))*(t - 14)

    14*V = 3.487*(t - 14)

    V * t = 1

    So substituting gives

    14/t = 3.487*(t - 14)

    Solve a quadratic looks like to me.

    Edit: Oops: need to convert the v*t = 1609 m as well.

    14*1609/t = 3.487*(t - 14)
     
    Last edited: Jun 14, 2009
  9. Jun 14, 2009 #8
    This is an answer I got from another website. Does this make sense?

    v= 1/ (t - 6.9 x 10-4 h) - 7.8 mph

    I was told that 14 s was converted into 6.9 x 10-4 by converting seconds into hours, but I am not getting that number in my conversions. I was also told the original speed did not have to posted, but that did not seem right either.
     
  10. Jun 14, 2009 #9
    I'm afraid you lost me, for I am not sure what to do with the equation you came up with. I am a physics beginner so I apologize for being a little slow to catch on.
     
  11. Jun 14, 2009 #10

    LowlyPion

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    As far as converting I suggest that you convert mph to meters/sec. That ratio is 1 mph = .447 m/s.

    Then work the equations in meters and seconds. That should yield ...

    V* t = (V+7.8*(.447))*(t - 14)

    Rearranging that equation (canceling out the V*t term and moving 14V to the other side) you should get

    14V = 7.8*(.447)*(t - 14) = 3.487*(t - 14)

    Since you also know that V*t = 1 mile = 1690 m then you can substitute with V = 1690/t

    14*1690/t = 3.487*(t-14)

    3.487*t2 - 3.487*14*t - 14*1690 = 0

    t2 -14*t - 6785.2 = 0

    This is a quadratic in t. The positive root of the equation is the time t.
    With that time and knowing that v*t = 1690, that should give you v in m/s.

    Dividing that by the mph to m/s ratio (.447) should give you the initial speed in mph.
     
  12. Jun 15, 2009 #11
    Thanks. That helped a lot. I got like 41mph and my teacher said today that is the correct answer.
     
  13. Sep 27, 2009 #12


    I'm doing the same problem and have 5.5 instead of 7.8 as well as 10 instead of 14. Either way, whichever numbers you use to solve this problem, my calculations are always 10% off. I have online physics problems and I can enter my answer and it tells me if it's correct or what percentage wrong it is. I've tried it multiple times and it always tells me my answer is 10% off. Somewhere in this formula there is a small error. I'm just letting all who are using this formula know.
    Thanks
     
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