The distance between the airplanes changing is -390.
<h3>Calculation:</h3>
The miles from the airport to Airplane A can be calculated as follows: a = 30 -250t, where t is in hours.
<h3>Similar to airplane A, airplane B's
distance from the airport can be expressed as...</h3>
b = 40 -300t
Using the Pythagorean theorem, one can calculate the separation (d) between the aircraft:
d^2 = a^2 + b^2
<h3>Depending on the passage of time, we have...</h3>
2dd' = 2aa' + 2bb', and d' = (aa' + bb')/d.
<h3>The values of its variables at t=0 must be determined in order to determine the numerical value of this</h3>
a = 30 -2500 = 30 a' = -250 b = 40 -3000 = 40 b' = -300
d = √(a²+b²) = √(900+1600) = 50
Then, d' = (30(-250) +40(-300))/50 = -19500/50 = -390
At a speed of 390 miles per hour, the space between the aircraft is closing.
<h3>How far apart are the equation's two planes?</h3>
The distance between the two planes is going to be the square root of six, therefore if we solve for d, multiply both sides of this equation by the square root of six, we obtain six is equal to negative d, or d is equal to negative six.
<h3>How widely apart are commercial airplanes?</h3>
For a commercial airliner, separation will typically be at least 3 miles laterally or 1,000 feet vertically (as stated in the question). The 5 mile rule is applied laterally in the enroute environment, at greater operating speeds over 10,000 feet, depending on the kind of radar and distance from the antennae.
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