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 find out how fast the planes are separating after 2 hours, we can use the concept of vector addition.
First, let's calculate the distances each plane has traveled in 2 hours.
The plane traveling north is moving at a speed of 200 km/h, so in 2 hours, it would have traveled a distance of 200 km/h * 2 h = 400 km directly north.
The plane traveling west is moving at a speed of 150 km/h, so in 2 hours, it would have traveled a distance of 150 km/h * 2 h = 300 km directly west.
Now, let's represent the distances traveled by each plane as vectors:
- The northward distance traveled by the first plane can be represented as a vector: 400 km north = 400 km directly upwards.
- The westward distance traveled by the second plane can be represented as a vector: 300 km west = 300 km directly to the left.
Since the planes are traveling at the same altitude, the vectors representing their distances are perpendicular to each other. To find how fast they are separating, we need to find the magnitude of the resultant vector formed by adding these two vectors.
Using the Pythagorean theorem, we can calculate the magnitude of the resultant vector:
Resultant magnitude = √((400 km)^2 + (300 km)^2)
Resultant magnitude = √(160,000 km^2 + 90,000 km^2)
Resultant magnitude = √250,000 km^2
Resultant magnitude ≈ 500 km
Therefore, the planes are separating at a speed of approximately 500 km after 2 hours.