At a certain instant an aircraft flying due east at 240 miles per hour passes directly over a car traveling due southeast at 60 miles per hour on a straight, level road. If the aircraft is flying at an altitude of .5mile, how fast is the distance between the aircraft and the car increasing 36 seconds after the aircraft passes directly over the car?

Question

At a certain instant an aircraft flying due east at 240 miles per hour passes directly over a car traveling due southeast at 60 miles per hour on a straight, level road. If the aircraft is flying at an altitude of .5mile, how fast is the distance between the aircraft and the car increasing 36 seconds after the aircraft passes directly over the car?

Answer 1

at time t=0,

let the plane be at (0,0,.5)
let the car be at (0,0,0)
let the +x axis be due east, so +y is due north

at time t hours,

plane is at (240t,0,.5)
car is at (60/√2 t,-60/√2 t,0)

the distance d between car and plane is

d^2 = (240t - 60/√2 t)^2 + (60/√2 t)^2 + .5^2

36 seconds = 0.01 hours, so

d^2 = (2.4-.6/√2)^2 + (.6/√2)^2 + .5^2
d = 2.0817

2d dd/dt = 7200(17-4√2)t
2(2.0817) dd/dt = 7200(17-4√2)(.01)
dd/dt = 196.16 mph

Answer 2

To find the rate at which the distance between the aircraft and the car is changing, we can first create a diagram to visualize the situation.

Let's call the moment the aircraft passed over the car as time t=0. After 36 seconds, the aircraft would have traveled 36 * 240 = 8640 miles from its initial position (since it is flying at 240 miles per hour). Similarly, the car would have traveled 36 * 60 = 2160 miles from its initial position (since it is traveling at 60 miles per hour).

Now, let's consider a right triangle formed by the aircraft, the car, and the distance between them. The distance of the aircraft from its initial position forms the base of the triangle, which is 8640 miles. The distance of the car from its initial position forms the perpendicular side of the triangle, which is 2160 miles. The distance between the aircraft and the car forms the hypotenuse of the triangle, which we can label as D(t) where t represents time.

To find D(t), we can use the Pythagorean theorem: D(t)^2 = (8640)^2 + (2160)^2

Differentiating both sides of the equation with respect to time (t) using implicit differentiation, we get:

2D(t) * (dD/dt) = 2(8640)(dD/dt) + 2(2160)(0)
(dD/dt) = (8640)(0) + (2160)(0) / (2D(t))

Since the altitude of the aircraft is 0.5 mile, we can convert the distances from miles to miles + 0.5 mile. Hence, D(t) = sqrt((8640)^2 + (2160)^2 + (0.5)^2)

We can now plug in the values to find dD/dt at t=36 seconds:

D(36) = sqrt((8640)^2 + (2160)^2 + (0.5)^2)

(dD/dt) = (8640)(0) + (2160)(0) / [2 * sqrt((8640)^2 + (2160)^2 + (0.5)^2)]

Simplifying the equation, we find the rate at which the distance between the aircraft and the car is increasing 36 seconds after the aircraft passes directly over the car.