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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?)
Quote:
Originally Posted by Infiniti
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
Let r be rate in still airQuote:
Originally Posted by Infiniti
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