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 63919This 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

Found 2 solutions by funmath, TurnerFam:
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!!!

Answer by TurnerFam(1) (Show Source): You can put this solution on YOUR website!
The formula to use is: Distance = rate x time or d = rt
The distance traveled was 300 mi. The rate on the first trip {WITH tail wind} is h {helicopter} + w {wind}. The time is 1 hour and 40 minutes, or 1.67 hours
So, you have 300 = (h + w)1.67
The rate on the second trip {AGAINST wind} is {h - w}, and an additional 20 minutes, so 2 hours. So you have 300 ={h - w)2
Solve for h in the second trip:
300 = 2(h-w)
300 = 2h - 2w
Add the 2w to both sides to isolate the 2h
300 + 2w = 2h
Divide both sides by 2 to isolate the h
150 + w = h OR h = 150 + w
Now, plug that info into Trip 1:
300 = 1.67(h+w) OR 300 = 1.67 ([150 + h) + w)
300 = 250.5 + 1.67w + 1.67w
300 = 250.5 + 3.34w
Subtract the 250.5 from both sides to isolate the 3.34w
49.5 = 3.34w
Divide both sides by 3.34 to isolate the w
14.8 = w.... round up to w = 15 Wind speed is 15
Plug that into the formula:
300 = (h + 15)1.67
300 = 1.67h + 25.05
Subtract 25.05 from 300 to isolate the 1.67h
274.75 = 1.67h
164.6 = h
Round up to 165 is Air Speed.
So... Wind speed is 15; Air speed is 165.

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