Question

A plane with a speed of 800.0 km/hr due east passes directly above a train traveling due southeast at 100.0 km/hr on a straight level road. The altitude of the aircraft is 10.0 km. How fast is the distance between the plane and the train increasing 45 s after the aircraft passes directly above the train?

Answer 1

distance between the train and plane is given as

d = sqrt(x² + y² + z²)

here x = distance between the plane and train along x-direction

y = distance between plane and train along Y-direction

z = height of the plane

x = distance between train and plane along X-direction at any time "t" = relative velocity x time

x = (800 - 100 Cos45) (5/18) t in m

y = distance between train and plane along y-direction at any time "t" = relative velocity x time

y = (100 Sin45) (5/18) t in m

distance is given as

D = sqrt(x² + y² + z²)

taking derivative both side relative to "t"

dD/dt = (0.5) (1/sqrt(x² + y² + z²)) (2x (dx/dt) + 2y (dy/dt) )

V = (0.5) (1/sqrt(x² + y² + z²)) (2x (dx/dt) + 2y (dy/dt) )

at t = 45

x = (800 - 100 Cos45) (5/18) (45) = 9116.12 m

dx/dt = (800 - 100 Cos45) (5/18) = 202.6

y = (100 Sin45) (5/18) (45) = 883.88 m

dy/dt = (100 Sin45) (5/18) = 19.64

V = (0.5) (1/sqrt((9116.12)² + (883.88)² + (10000)²)) (2(9116.12) (202.6) + 2 (883.88) (19.64) )

V = 137.5 m/s