Two airplanes are flying in the air at the same height: airplane a is flying east at 250 mi/h and airplane b is flying north at
300mi/h.300mi/h. if they are both heading to the same airport, located 30 miles east of airplane a and 40 miles north of airplane b, at what rate is the distance between the airplanes changing?
If the airport is located at the origin, for units of miles and hours, we can write the equations of position for airplanes "a" and "b" in rectangular coordinates as a = (-30 +250t, 0) b = (0, -40 +300t)
The distance between these (moving) points can be computed in the usual way using the Pythagorean theorem. d = √((-30 +250t - 0)² +(0 - (-40 +300t))²) d = √(2500 -39000t +152500t²)
Then the rate of change of d is the derivative of this. d'(t) = (-19500 +152500t)/√(2500 -39000t +152500t²)
At the present time (t=0), the rate of change of distance between the planes is d'(0) = -19500/√2500 = -390
The distance between the planes is decreasing at 390 mi/h.
So to get to 9:00, we need 3 minutes so now you have 27 minutes left of break. Then, you add 27 and you get 9:27. I hope this helped and please mark me brainliest!
