Question 620145: Hello, I need help in solving this problem:
A plane flew 150 kilometers with a 30 km/hr tailwind, then turned into the wind
and flew for another 65 kilometers. If the wind and the plane’s airspeed were
constant and the entire trip took 30 minutes, what was the plane’s airspeed?
(Ignore the time needed to turn the plane around).
Thank you.
Found 2 solutions by htmentor, MathTherapy:
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! Let s = the plane's airspeed
The total time of the trip is 30 mins = 1/2 hr; the speed of the wind = 30 km/hr
Since t = d/v, we can write the following for the total time of the trip:
150/(s+30) + 65/(s-30) = 0.5
Combine fractions using the common denominator, and cross-multiply:
150(s-30) + 65(s+30) = 0.5(s^2-900)
Simplify and collect terms:
s^2 - 430s + 4200 = 0
This can be factored as
(s-10)(s-420) = 0
The solutions are s = 10, s = 420
Since the speed of the plane cannot be less than the wind speed, we take the 2nd solution.
So s = 420 km/hr
Check:
150/450 + 65/390 = 1/3 + 1/6 = 3/6 = 1/2
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
Hello, I need help in solving this problem:
A plane flew 150 kilometers with a 30 km/hr tailwind, then turned into the wind
and flew for another 65 kilometers. If the wind and the plane’s airspeed were
constant and the entire trip took 30 minutes, what was the plane’s airspeed?
(Ignore the time needed to turn the plane around).
Thank you.
Let speed of plane in still air be S
Time taken to travel 150 km, plus time taken to fly 65 km, equals entire time taken ( , or ˝ hour)
300(S - 30) + 130(S + 30) = (S + 30)(S - 30) ------ Multiplying by LCD, 2(S + 30)(S – 30)
(S - 10)(S - 420) = 0
S, or speed in still air = 10 (ignore as this is IMPOSSIBLE, because the wind speed CANNOT EXCEED the plane’s speed), OR
S, or speed of plane in still air =  km/h
I’ll leave the check up to you.
Send comments and “thank-yous” to “D” at MathMadEzy@aol.com
|
|
|
