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Flight 501 is at an altitude of 6 miles

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Flight 501 is at an altitude of 6 miles

by abhi332 » Thu Feb 25, 2010 12:59 pm

Timer

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Your Answer

A

B

C

D

E

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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 Manhattan Airport, which is located exactly 8 miles due north of the plane's current position. Flight 501 is scheduled to land at Manhattan 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 Manhattan 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?

100 miles/hr
5sqrt {13}-10 miles/hr
100 miles/hr
100 sqrt {13}-200 miles/hr
100 sqrt {13} miles/hr

[spoiler]
OA:D[/spoiler]
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by harsh.champ » Thu Feb 25, 2010 2:20 pm
abhi332 wrote: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 Manhattan Airport, which is located exactly 8 miles due north of the plane's current position. Flight 501 is scheduled to land at Manhattan 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 Manhattan 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?

100 miles/hr
5sqrt {13}-10 miles/hr
100 miles/hr
100 sqrt {13}-200 miles/hr
100 sqrt {13} miles/hr

[spoiler]
OA:D[/spoiler]
New destination = sqrt(8^2 + 15^2)
=17 miles
Previous speed = 8/3 miles /min
New speed = 17/3 miles/min.
Increase in speed = 9/3 = 3 miles/min. = 180 miles/hr

I guess A and C are same.
So,180 should be either A or C.
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by kstv » Fri Feb 26, 2010 12:18 am
The distance to be covered when flying due North (N) . Is
√(8^2 + 6^2 ) = 10

But now it has to fly at a destination East (E) of (N) in a north easterly (NE) direction

(NE) is at a distance √(8^2 + 15^2 ) = 17 miles but the plane has to descend 6 miles

So it has to travel √(17^2 + 6^2) = ↓5√13

It covers 10 miles in 3 minutes but now has to cover 5√13 miles in the same time

So increase in speed in 60 (20 x 3) mins is = 20( 5√13 - 10)
= 100√13-200
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by alanforde800Maximus » Mon Jul 30, 2018 11:10 pm
Is there any other approach that can be deployed on this problem?
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