SOLUTION: with a tail wind, a helicopter traveled 300mi in an hour and 40 min. the return trip aganst the same wind took 20 min longer. find the wind speed and also the air speed of the heli

Question 63918This question is from textbook algebra and trigonmetry struture method
: with a tail wind, a helicopter traveled 300mi in an hour and 40 min. the return trip aganst the same wind took 20 min longer. find the wind speed and also the air speed of the helicopter. This question is from textbook algebra and trigonmetry struture method

Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
with a tail wind, a helicopter traveled 300mi in an hour and 40 min. the return trip aganst the same wind took 20 min longer. find the wind speed and also the air speed of the helicopter.
The trick to this one is remembering to convert your time in minutes to time in hours.
The time with the tail wind is: 40min(1hr/60min)=40hr/60=(2/3)hr
Let the rate of the helicopter be: h
Let the rate of the wind be: w
The distance formula is: d=rt, where d=distance, r=rate, and t=time
so with the wind:
300=(2/3)(h+w)
:
The time against the wind is (40+20)min=60min=1 hr
Therefore:
300=1(h-w)
300=h-w
300+w=h Substitute this into the equation with the wind and solve for h:
300=(2/3)((300+w)+w)
300=(2/3)(300+2w)
(3/2)(300)=(3/2)(2/3)(300+2w)
450=300+2w
450-300=300-300+2w
150=2w
150/2=2w/2
75=w The speed of the wind is 75 m/h
Substitute that into the equation aganst the wind and solve for h.
300+75=h
375=h The speed of the helicopter is 375 m/h.
:
Sanity check:
If the helicopter went 375 m/h and the wind was 75 m/h, will the helicopter fly 300 miles in the allotted time with and against the wind?
With the wind: (2/3)(375+75)=300
(2/3)(450)=300
300=300 so far we're right.
Against the wind: (1)(375-75)=300
300=300 we're right if we understood the question and converted our time right.
Happy Calculating!!!

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