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