Question

An air traffic controller spots two airplanes at the same altitude converging to a point as they fly at right angles to each other. One airplane is 70 miles from the point and has a speed of 420 miles per hour. The other is 240 miles from the point and has a speed of 1440 miles per hour.
(a) At what rate is the distance between the planes changing? in mph?
(b) How much time does the controller have to get one of the airplanes on a different flight path?

Answer 1

Answer:

(a) ðz/ðt = 1500miles/hr

(b) 10minutes

Step-by-step explanation:

Let x rep plane one

Let y rep plane two

ðx/ðt = 1440miles/hr = speed of plane one

ðy/ðt = 420miles/hr = speed of plane 2

ðz/ðt = ? = rate of change of distance btw two plane moving at right angle to meet at point o...

See attachment

Since x² + y² = z²... equation 1

So therefore (240² + 70²) ^½= z = 250 miles = distance between two planes

Differential rate of equation 1 gives

(2x* ðx/ðt ) + (2y* ðy/ðt) = 2z* ðz/ðt

Substituting to get ðz/ðt

(2*240*1440) + (2*70*420) = 2*250*ðz/ðt

750000 = 500 ðz/ðt

ðz/ðt = 1500miles/hr

B.

Since v = d/t where d = distance = 250miles

t = time = ? = time needed for controller to divert one of e planes , v = 1500miles/hr

t = d/v = 250/1500 = 10minutes

Answer 2

Answer:

The rate at which the distance between the planes changing is 1500mph

The time the controller have to get one of the airplanes on a different flight path is 0.207hours(12.4minutes)

Step-by-step explanation:

The two planes are approaching the same point.

Plane1 Velocity = 420miles per hour

Plane2 Velocity = 1440miles per hour.

Since the two planes are flying perpendicular to each other and Velocity is a vector quantity, their resultant Velocity can be calculated with vector addition.

The rate at which the distance between the planes is changing represent the resultant velocity of the two planes.

Therefore,

Vr = sqrt(420^2 + 1440^2)

Vr = 1500mph

Time the controller has to get one of the airplanes on a different flight path is calculated thus:

Time = distance/velocity

Distance that will be travelled by the two airplanes to the converging point is 70 + 240 = 310miles

Therefore, Time = 310/1500 = 0.2066hours = 12.4minutes