An air traffic controller spots two planes at the same altitude flying toward each other. their flight paths form a right angle at point p. one plane is 150 miles from point p and is moving at 450 miles per hour. the other plane is 200 miles from point p and is moving at 450 miles per hour. what is the distance s between the planes as a function of time t?

The problem states that the paths of the two planes form a right triangle, therefore this means that the distance between the two is the hypotenuse. Given that information, we can use the hypotenuse formula for finding the distance formula:

c^2 = a^2 + b^2             ---> 1

Where c is the hypotenuse, in this case the distance between the two planes.

First let us find the value of a. We know that one plane is 150 miles from point P and this distance is decreasing by a rate of 450 miles per hour, therefore a is:

a = 150 – 450 t              ---> 2

We also know that the other plane is 200 miles away and this distance decreases by a rate of 450 miles per hour also, therefore b is:

b = 200 – 450 t              ---> 3

Substituting equations 2 and 3 to 1:

c^2 = (150 - 450 t)^2 + (200 – 450 t)^2

c^2 = 22500 – 135000t + 202500t^2 + 40000 – 180000t +  + 202500t^2

c^2 = 405,000 t^2 – 315,000 t + 62,500               (ANSWER)

on your paper it doesn't say the answer but the the answer is altitude

step-by-step explanation:

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