SOLUTION: Two aircraft leave simultaneously from an airport, one flying due north and the other due east. The northbound aircraft averages a speed of 100 miles per hour fater than the eatbou

Algebra ->  Trigonometry-basics -> SOLUTION: Two aircraft leave simultaneously from an airport, one flying due north and the other due east. The northbound aircraft averages a speed of 100 miles per hour fater than the eatbou      Log On


   

Question 486694: Two aircraft leave simultaneously from an airport, one flying due north and the other due east. The northbound aircraft averages a speed of 100 miles per hour fater than the eatbound aircraft. After 3 hours, the aircraft are 1500 miles apart. Find the average speed of each aircraft.
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
they are travelling at right angles to each other.
At any given instant they form a right triangle with their starting point
speed of DISABLED_event_one= x
other = x+100
using Pythagoras theorem
x^2+(x+100)^2 = 1500^2
x^2+x^2+200x+10000=2250000
2x^2+200x-2240000
Find the roots of the equation by quadratic formula

a= 2 ,b= 200 ,c= -2240000

b^2-4ac= 40000 - 17920000
b^2-4ac= 17960000
sqrt%28%0917960000%09%29= 4237.92
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=( -200 + 4237.92 )/ 4
x1= 1009.48
x2=( -200 -4237.92 ) / 4
x2= -1109.48
Ignore negative value
1009.48 mph= speed of one plane
add 100 to this speed to get speed of other
m.ananth@hotmail.ca