Question

An airplane, flying horizontally at an altitude of 1km.,passes directly over an observer. If the constant speed of the plane is 240kph, how fast is its distance from the observer increasing 30 seconds later?

Answer 1

If we let t=0 when the plane passes over, the distance from the observer at time t will be Sqrt[1*1 + (240*t)^2] where t is in hours (via Pythagorean theorem). The change in distance is the derivative of this, or 57600t/Sqrt[1+57600*t^2]. 30 seconds means t=1/120 (since we're measuring t in hours), so the change in distance over time is 96*Sqrt[5] kph or about 214.66 kph.

Answer 2

wow thanks a lot

Answer 3

ooops i didn't get the same answer

Answer 4

isn't 2*57600t/Sqrt[1+57600*t^2] ???