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Distance/Rate/Time word problem

  1. Sep 8, 2012 #1
    1. The problem statement, all variables and given/known data
    A Cessna 150 averages 150 mph in still air. With a tailwind it is able to make a trip in 2 1/3 hours. Because of the headwind, it is only able to make the return trip in 3 1/2 hours. What is the average wind speed?



    2. Relevant equations
    (1) If x = wind speed and y = still air speed
    ∴ resultant speed (tailwind) = x + y
    ∴ resultant speed (headwind) = x - y

    (2) d = rt


    3. The attempt at a solution
    let x = wind speed
    still air speed = 150 mph
    tailwind time = 2 1/3 hours
    tailwind speed = 150 + x
    headwind time = 3/12 hours
    headwind speed = 150 - x

    I used these as given for the problem, expressed distance in terms of tailwind/headwind time and rates, set distance as constant (equates), solves for x then gets 57.69 as answer.
    but in the book, the answer is 30 mph. What is the problem with my solution?

    Thanks in advance.
     
  2. jcsd
  3. Sep 8, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Well, your solution is wrong! And we can tell you what you did wrong because you do not say what you did. Please write out the equations exactly as you used them and how you solved.
     
  4. Sep 8, 2012 #3
    let x = wind speed
    still air speed = 150 mph
    tailwind time = 2 1/3 hours
    tailwind speed = 150 + x
    headwind time = 3 1/2 hours
    headwind speed = 150 - x

    d = rt

    tailwind:
    d = (150 + x)(2 1/3)
    d = 100 + 2 1/3 x

    headwind:
    d = (150 - x)(3 1/2)
    d = 225 - 3 1/2 x

    d = d

    100 + 2x/3 = 225 - 3x/2
    2x/3 + 3x/2 = 125
    4x + 9x = 125(6)
    13x = 750
    x = 57.69

    @HallsOfIvy, sorry..
     
  5. Sep 8, 2012 #4

    ehild

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    Homework Helper
    Gold Member

    Note that 3 1/2=3.5=7/2 and 2 1/3 =7/3.

    ehild
     
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