quantskillsgmat wrote:Q at 7;57 am, flight 501 is at an altitude of 6 miles above the ground and is on direct approach (flying in direct line of runway) towards the airport,which is located exactly 8 miles due north of planes current position.flight 501 is scheduled to land at airport at 8 am but at 7;57 am, control tower radios the plane and changes the landing location to an airport 15 miles due east of airport.assuming a direct approach by how many miles per hour the pilot have to increase speed in order to arrive new location on time.
a)100root3-200 b)100root13-200 c)100 d)100root13
First, let's calculate the plane's initial speed (in mph)...
Direct-line distance of the plane to the north airport is 10 miles (3,4,5 right triangle with 6 height and 8 base). Therefore, since the plane will reach this airport in 3 minutes, it is traveling at 200 mph.
Next, we need to calculate the direct-line distance to the secondary airport, which is due east from the first airport (creating a right triangle on the x-y plane). This time, the "height" of the triangle is the north-south difference (8 miles), and the base of the triangle is the east-west difference (15 miles). From the (8,15,17) triplet for right triangles, we now have the 2d straight-line distance from the place to the secondary airport. Note that 17 can also be calculated using the Pythagorean Theorem.
To get the 3-d straight-line distance from the plane to the secondary airport, we use the Pythagorean Theorem: dist = sqrt(6^2+17^2) = sqrt(325) = sqrt(25*13) = 5*sqrt(13). Again, we multiply the speed by 20 to get the mph needed to reach this new airport in 3 minutes (the intended landing time): 100*sqrt(13) mph.
Finally, we take the difference in speeds to answer the question: 100*sqrt(13)-200 mph increase needed.
Answer is B
