question 9 of 14, step 1 of 1\na passenger airplane, flying at an altitude of 6.8 miles at a ground speed of…

question 9 of 14, step 1 of 1\na passenger airplane, flying at an altitude of 6.8 miles at a ground speed of 565 miles per hour, passes directly over an observer who is on the ground. how fast is the\ndistance between the observer and the plane increasing 3 minutes later? round any intermediate calculations, if needed, to no less than six decimal places, and round\nyour final answer to two decimal places.\nanswer\nkeypad\nkeyboard shortcuts

question 9 of 14, step 1 of 1\na passenger airplane, flying at an altitude of 6.8 miles at a ground speed of 565 miles per hour, passes directly over an observer who is on the ground. how fast is the\ndistance between the observer and the plane increasing 3 minutes later? round any intermediate calculations, if needed, to no less than six decimal places, and round\nyour final answer to two decimal places.\nanswer\nkeypad\nkeyboard shortcuts

Answer

Explanation:

Step1: Calculate the horizontal distance

The ground - speed of the plane is (v = 565) miles per hour. In (t=\frac{3}{60}=\frac{1}{20}) hours (since (3) minutes (=\frac{3}{60}) hours), the horizontal distance (x) that the plane travels is given by the formula (x = vt). So, (x=565\times\frac{1}{20}=28.25) miles.

Step2: Use the Pythagorean theorem

Let (y = 6.8) miles (altitude, constant) and (z) be the distance between the observer and the plane. By the Pythagorean theorem (z=\sqrt{x^{2}+y^{2}}). Substituting (x = 28.25) and (y = 6.8), we have (z=\sqrt{(28.25)^{2}+(6.8)^{2}}=\sqrt{798.0625 + 46.24}=\sqrt{844.3025}).

Step3: Differentiate the Pythagorean theorem with respect to time

We have (z^{2}=x^{2}+y^{2}). Differentiating both sides with respect to time (t): (2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}). Since (y) (altitude) is constant, (\frac{dy}{dt} = 0). So, (\frac{dz}{dt}=\frac{x}{z}\cdot\frac{dx}{dt}). We know (x = 28.25), (y = 6.8), so (z=\sqrt{28.25^{2}+6.8^{2}}\approx29.04) (from step 2) and (\frac{dx}{dt}=565) (ground - speed of the plane). Substituting the values: (\frac{dz}{dt}=\frac{28.25}{29.04}\times565) (\frac{dz}{dt}=\frac{28.25\times565}{29.04}=\frac{15961.25}{29.04}\approx549.63)

Answer:

(549.63)