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Math Help - Grade 10 math Help

  1. #1
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    Grade 10 math Help

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    "An airliner traveling from Toronto to Vancouver took 5hrs to cover the 3900k trip against a headwind. The return trip, traveling with a tailwind that was twice the speed of the headwind took 4hours and 20 mins.

    Hint: You will need to use the formula, distance = speed x time.
    a) How fast were the headwind and the tailwind on the two trips
    b) How fast would the airliner have flown in still air?)
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  2. #2
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    if you meant 3900 kilometer by '3900k'....

    Quote Originally Posted by Infiniti View Post
    Long time reader, first time poster

    "An airliner traveling from Toronto to Vancouver took 5hrs to cover the 3900k trip against a headwind. The return trip, traveling with a tailwind that was twice the speed of the headwind took 4hours and 20 mins.

    Hint: You will need to use the formula, distance = speed x time.
    a) How fast were the headwind and the tailwind on the two trips
    b) How fast would the airliner have flown in still air?)

    Let the speed of the airplane as 'x km/hr'
    Let the speed of the headwind as 'y km/hr'
    Then the speed of the tailwind would be '2y km/hr'

    Then you can get two simultaneous equations since 'distance = speed x time'

    5(x-y)=3900
    5(x+2y)=4500

    If you solve the equation, it'll be clear that

    the speed of the airplane is 3600 km/hr
    the speed of the headwind is 60 km/hr
    and the speed of the tailwind is 120 km/hr.

    So the answer for the problem (a) would be 60km/hr and 120km/hr
    And the answer for the problem (b) would be 3600km/hr
    Last edited by FutureKSAStudent; January 13th 2010 at 07:40 PM.
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  3. #3
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    Quote Originally Posted by Infiniti View Post
    Long time reader, first time poster

    "An airliner traveling from Toronto to Vancouver took 5hrs to cover the 3900k trip against a headwind. The return trip, traveling with a tailwind that was twice the speed of the headwind took 4hours and 20 mins.

    Hint: You will need to use the formula, distance = speed x time.
    a) How fast were the headwind and the tailwind on the two trips
    b) How fast would the airliner have flown in still air?)
    Let r be rate in still air
    w be rate of headwind

    Traveling from Toronto -> Vancouver
    Distance = (rate)(time)
    3900 = (r-w)(5)

    Traveling from Vancouver -> Toronto
    3900 = (r+2w)(13/3)

    Thats two equations with two variables so you should be able to use methods for systems of equations to solve

    Headwind = 40
    Tailwind = 80
    Speed in still air = 820
    Last edited by S31J41; January 13th 2010 at 07:33 PM.
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