Question 1201012: three airports, A, B, and C, are located in a north-south line. b is 645 mi north of A, and C is 540 mi north of B. a pilot flew from A to B, delayed 2 h, and continued to C,. the wind was blowing from the south at 15 mi/h during the first part of the trip, but during the delay it changed to the north with a velocity of 20 mi/h. if each flight required the same time, find the airspeed of the plane.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52803)   (Show Source):
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Let's draw a vertical number line to represent the locations of airports A, B, and C.
The order from top to bottom is: C, B, A.
We'll also mark the distances between each adjacent airport.
Each distance is in miles.
Add in the wind blowing from the south (i.e. the wind is blowing toward the north) at 15 mph.
This applies only when the plane is going from A to B.
Since the plane and the wind are both going north (when the plane goes from A to B), the plane's speed is increased.
x = speed of the plane with no wind
x+15 = speed after adding the 15 mph wind boost
distance = rate*time
time = distance/rate
time = (645 miles)/(x+15 mph)
time = 645/(x+15)
This is the time, in hours, it takes to go from A to B.
Side note:
A headwind slows the plane down (because it attacks the head or front of the plane).
A tailwind speeds the plane up (because it comes from the tail to push the plane forward).
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When the plane goes from B to C, the wind blows from the north.
The wind is now aiming south. It moves at 20 mph.
At this point the plane's speed x is now decreased to x-20 because it's fighting the wind (the plane goes north while the wind is going south).
Note: This assumes the plane's speed without wind (x) is the same as before.
Then,
distance = rate*time
time = distance/rate
time = (540 miles)/(x-20 mph)
time = 540/(x-20)
This is the time, in hours, it takes to go from B to C
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To summarize what we know so far:- time from A to B = 645/(x+15)
- time from B to C = 540/(x-20)
Each time value is in hours.
The instructions mention "each flight required the same time".
So we'll set each time expression equal to one another.
Then solve for x.
time from A to B = time from B to C
645/(x+15) = 540/(x-20)
645(x-20) = 540(x+15)
645x-12900 = 540x+8100
645x-540x = 8100+12900
105x = 21000
x = 21000/105
x = 200
The plane's speed without wind is 200 mph
When going from A to B, the time spent is:
time = 645/(x+15)
time = 645/(200+15)
time = 645/215
time = 3 hours
When going from B to C, the time spent is:
time = 540/(x-20)
time = 540/(200-20)
time = 540/180
time = 3 hours
Both portions of the trip are 3 hours each. This confirms we have the correct answer.
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Answer: The plane's speed without wind is 200 mph
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