SOLUTION: In a round trip, a plane flies 990 miles each way,one way with tailwind and one with headwind. The total trip time is 3 hours and 20 mins. The airspeed of the plane is 600 miles/ho

Algebra ->  Rational-functions -> SOLUTION: In a round trip, a plane flies 990 miles each way,one way with tailwind and one with headwind. The total trip time is 3 hours and 20 mins. The airspeed of the plane is 600 miles/ho      Log On


   



Question 881931: In a round trip, a plane flies 990 miles each way,one way with tailwind and one with headwind. The total trip time is 3 hours and 20 mins. The airspeed of the plane is 600 miles/hour. What is the speed of the wind?
To me, this seems extremely bizarre and I am wondering where the heck did the speed of wind come into the question
In no rush for the answer and thanks in advanced for your help!

Found 3 solutions by richwmiller, jim_thompson5910, MathTherapy:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
This is a common type travel problem
a+b=3 1/3
r*t=d
The speed one way will be 600 + plus the wind and the other way 660 minus the wind.
They assume the speed of the wind is constant.
We have three unknowns: the speed of the wind and the time in each direction.
(600+w)*a=990,
(600-w)*b=990,
a+b=3 1/3
a=3 1/3-b
(600+w)*(3 1/3-b)a=990,
(600-w)*b=990
a = 3/2, b = 11/6, w = 60

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let

p = speed of plane in still air (without any wind at all)
w = speed of wind

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The tailwind speeds the plane up (since it's coming from the tail and it's pushing the plane along where it wants to go). So the speed of the plane with a tailwind is p%2Bw mph.


D+=+r%2At


D+=+%28p%2Bw%29%2At1 Replace r with p%2Bw


D+=+%28600%2Bw%29%2At1 Plug in p+=+600 (the airspeed of the plane)


990+=+%28600%2Bw%29%2At1 Plug in the distance d = 990 and solve for t1


990%2F%28600%2Bw%29+=+t1


t1+=+990%2F%28600%2Bw%29


So the time it takes to travel with the tailwind is t1+=+990%2F%28600%2Bw%29

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The headwind slows the plane down (since it's coming from the head and it's going in the opposite direction of the plane). So the speed of the plane with a headwind is p-w mph.


D+=+r%2At


D+=+%28p-w%29%2At2 Replace r with p-w


D+=+%28600-w%29%2At2 Plug in p+=+600 (the airspeed of the plane)


990+=+%28600-w%29%2At2 Plug in the distance d = 990 and solve for t2


990%2F%28600-w%29+=+t2


t2+=+990%2F%28600-w%29


So the time it takes to travel with the headwind is t2+=+990%2F%28600-w%29

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The two times t1 and t2 add to "3 hours, 20 minutes" which is equivalent to "3 1/3 hours" (since 20/60 = 1/3)

That's equivalent to 10/3 hours (3 + 1/3 = 9/3 + 1/3 = 10/3)

So...

t1+%2B+t2+=+990%2F%28600%2Bw%29+%2B+990%2F%28600-w%29+=+10%2F3

which means

990%2F%28600%2Bw%29+%2B+990%2F%28600-w%29+=+10%2F3

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Now let's solve for w


990%2F%28600%2Bw%29+%2B+990%2F%28600-w%29+=+10%2F3


Multiply both sides by the LCD to clear out the fractions.











3%2A%28600-w%29990+%2B+3%2A%28600%2Bw%29990+=+10%28600%2Bw%29%28600-w%29


3%2A990%28600-w%29+%2B+3%2A990%28600%2Bw%29+=+10%28600%2Bw%29%28600-w%29


2970%28600-w%29+%2B+2970%28600%2Bw%29+=+10%28600%2Bw%29%28600-w%29


2970%28600-w%29+%2B+2970%28600%2Bw%29+=+10%28600%5E2-w%5E2%29


2970%28600-w%29+%2B+2970%28600%2Bw%29+=+10%28360000-w%5E2%29





1782000-990w%2B1782000%2B990w+=+3600000-10w%5E2


3564000=+3600000-10w%5E2


3564000%2B10w%5E2=+3600000


10w%5E2=+3600000-3564000


10w%5E2=+36000


w%5E2=+36000%2F10


w%5E2=+3600


w=+sqrt%283600%29


w=+60


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So the windspeed is 60 miles per hour

Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!
In a round trip, a plane flies 990 miles each way,one way with tailwind and one with headwind. The total trip time is 3 hours and 20 mins. The airspeed of the plane is 600 miles/hour. What is the speed of the wind?
To me, this seems extremely bizarre and I am wondering where the heck did the speed of wind come into the question
In no rush for the answer and thanks in advanced for your help!

Let speed of wind be S
With a tailwind, the plane's total speed is: 600 + S
With a headwind, the plane's total speed is: 600 - S
Since outbound TIME, plus return TIME, equals 3%2620%2F60, or 3%261%2F3, or 10%2F3 hours, then we can say that:
990%2F%28600+%2B+S%29+%2B+990%2F%28600+-+S%29+=+3%2620%2F60
990%2F%28600+%2B+S%29+%2B+990%2F%28600+-+S%29+=+10%2F3+
990(3)(600 - S) + 990(3)(600 + S) = 10(600 + S)(600 – S) ------- Multiplying by LCD, 3(600 + S)(600 – S)
1782000+-+2680S+%2B+1782000+%2B+2680S+=+10%28360000+-+S%5E2%29
3564000+=+3600000+-+10S%5E2
10S%5E2+%2B+3564000+-+3600000+=+0
10S%5E2+-+36000+=+0
10%28S%5E2+-+3600%29+=+10%280%29
S%5E2+-+3600+=+0
(S - 60)(S + 60) = 0
S, or speed of wind = highlight_green%28highlight_green%2860%29%29%29 mph OR S = - 60 (ignore)