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Math problem involving reciprocal of linear functions

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img97.imageshack.us/img97/1521/lastscano.jpg [Broken]

    2. Relevant equations

    avg speed = distance / time


    3. The attempt at a solution

    not sure how to go about this one, looking for a few hints on how to get the equation.

    answer is :
    t = 3850 /v

    But im not sure why.

    thank you
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 6, 2009 #2

    Mark44

    Staff: Mentor

    You show the equation above as being relevant, but I'm not sure that you are really convinced. Can you solve this equation for time in terms of the other two variables.
    I'm not sure why either. There's either a typo in your answer sheet or you wrote it incorrectly.
     
    Last edited by a moderator: May 4, 2017
  4. Oct 6, 2009 #3
    just confirmed the answer it is as i typed it.
     
  5. Oct 6, 2009 #4
    t = distance / v

    distance = t * v
    distance = 11 * 350
    distance = 3850

    We know distance is 3850km since it takes 11 hours to reach from Quebec City to Montreal at a speed of 350km/h. Time is also inversely proportional to speed so that means as time goes up, speed goes down and vice-versa.

    t = 3850/v

    To conform this is correct, we will plug the original speed into the equation.
    t = 3850/350
    t = 11h

    If time was lower than 11 (we'll use 5h), speed should go up.
    t = 3850/v
    5 = 3850/v
    v = 3850/5
    v = 770 km/h

    If time was higher, speed should go down. So let's use 20 hours.
    t = 3850/v
    20 = 3850/v
    v = 192.5 km/h
     
  6. Oct 6, 2009 #5
    thank you very much, that helped alot. Pretty easy i jsut didnt see it. :)
     
  7. Oct 6, 2009 #6
    Anytime! Good luck. :)
     
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