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At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a

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At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink]

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New post 11 Feb 2014, 06:15
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At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground and is on a direct approach (i.e., flying in a direct line to the runway) towards The Airport, which is located exactly 8 miles due north of the plane’s current position. Flight 501 is scheduled to land at The Airport at 8:00 am, but, at 7:57 am, the control tower radios the plane and changes the landing location to an airport 15 miles directly due east of The Airport. Assuming a direct approach (and negligible time to shift direction), by how many miles per hour does the pilot have to increase her speed in order to arrive at the new location on time?


A. \(5\sqrt{13} - 10\) miles/hr

B. 100 miles/hr

C. \(100\sqrt{13} - 200\) miles/hr

D. 200 miles/hr

E. \(100\sqrt{13}\) miles/hr
[Reveal] Spoiler: OA

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Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink]

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New post 11 Feb 2014, 06:40
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Assume Euclidean geometry. If the original airport is 8 miles due north, and the new airport is 15 miles due east of the original airport, then the new airport is 17 miles away. (17² = 8² + 15²)

Now the plane is 6 miles above ground, and the original airport is 8 miles away. Then the Euclidean distance from the plane to the original airport is 10 miles (10² = 6² + 8²). Because it would have taken 3 minutes to travel the 10 miles, We can say it would have taken 10/3 miles/min.

The new airport is 17 miles away and the plane is 6 miles above ground. So the distance the plane must now travel is 5√13 miles( [5√13]² = 17² + 6²). To do this in 3 minutes, the plane must travel at 5√13/3 miles/min.

The difference in speed is thus 5√13/3 - 10/3 miles/min. But we need to find the answer in miles/hour, so we need to multiply by 60 to get 60(5√13/3 - 10/3) miles/hour or 20(5√13 - 10) miles/hour.
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Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink]

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New post 14 Feb 2014, 01:08
what if like this:
R*T=D
8/3m/min*3min=8miles
5m/min*3min=15miles
5-8/3=7/3m/min or 140m/hour
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Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink]

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New post 14 Feb 2014, 03:51
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Temurkhon wrote:
what if like this:
R*T=D
8/3m/min*3min=8miles
5m/min*3min=15miles
5-8/3=7/3m/min or 140m/hour


Given the flight has an altitude of 6 miles and the airport is exactly 8 miles due north. The flight has to descend to the airport along the hypotenuse of a right angled triangle with legs 6 & 8 miles. hypotenuse = 10 miles. after that we can apply your approach => 3 mins per 10 miles => 60 mins is 60*10/3 = 200 mph. ---original speed.

Now since the flight got a new destination exactly 15 miles east of the current airport. we have 2nd right angled triangle with 8 & 15 as legs. hypotenuse is 17. Further, now the flight has to descend to the new airport along the hypotenuse of a third triangle with legs 6 miles(altitude of the plane) & 17 miles which is \(\sqrt{6^2+17^2} = 5\sqrt{13}\)

The plane has to descend \(5\sqrt{13}\) miles in 3 mins. hence in 60 mins it has to have a new speed of \(60*\frac{5}{3}\sqrt{13}\)

\(= 20*5\sqrt{13}\) --- new speed

Increase in speed is \(20*5\sqrt{13}-20*10\)

\(20(5\sqrt{13}-10)\)
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Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a [#permalink]

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New post 21 Nov 2017, 04:02
Temurkhon wrote:
what if like this:
R*T=D
8/3m/min*3min=8miles
5m/min*3min=15miles
5-8/3=7/3m/min or 140m/hour



If you get 140m/hr, thats good enough since 100 √ 13−200 miles/hr is approximately 140 miles an hour.

√ 13 = between 3 (3*3=9) and 4(4*4=16). take 3.5 for example.

you'd get 100*3.5 - 200

you get a difference of approximately 150 m/hr which is the closest to the answer you got.
Re: At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a   [#permalink] 21 Nov 2017, 04:02
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At 7:57 am, Flight 501 is at an altitude of 6 miles above the ground a

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