two small planes take off from the same airport at the same time. One travels north at 200 km/h, and the other, west at 150 km/h. If the planes fly at the same altitude, how fast are they separating after 2 hours?
the separation distance z after t hours is found using
z^2 = (200t)^2 + (150t)^2
so, find dz/dt when t=2
To determine how fast the planes are separating after 2 hours, we can use the concept of relative velocity. Let's break down the information we have:
Plane A (travelling north) speed = 200 km/h
Plane B (travelling west) speed = 150 km/h
After 2 hours, the distance travelled by Plane A can be calculated by multiplying its speed by the time:
Distance travelled by Plane A = 200 km/h * 2 hours = 400 km
Similarly, the distance travelled by Plane B can be calculated as:
Distance travelled by Plane B = 150 km/h * 2 hours = 300 km
Now, we can form a right-angled triangle where the distances travelled by Plane A and Plane B represent the triangle's two sides. The separation distance between the two planes represents the hypotenuse of the triangle.
Using the Pythagorean theorem, we can calculate the separation distance (D) as follows:
D^2 = (Distance travelled by Plane A)^2 + (Distance travelled by Plane B)^2
D^2 = 400 km^2 + 300 km^2
D^2 = 160,000 km^2 + 90,000 km^2
D^2 = 250,000 km^2
Taking the square root of both sides, we find that D ≈ 500 km.
Therefore, the planes are separating at a rate of approximately 500 km after 2 hours.