An airplane flying at an altitude of 6 miles passes directly over a radar antenna. When the airplane is 12 miles away, the radar detects that the distance s is changing at a rate of 240mi/hr. What is the speed of the airplane?
Please show work
Thanks
Please show work
Thanks
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The statement is a little confusing. I'll have to assume that the "12 miles away" is along the horizontal direction (parallel to the Earth's surface) as opposed to the straight-line distance from the antenna. It's not really clear if this interpretation is correct.
You can construct a right triangle with one leg of length 6 mi straight up from the radar antenna, one leg is the path of the plane parallel to the earth (6 mi. up), and hypotenuse a straight line from antenna to plane.
Call the length of the plane's path y, and straight line distance from antenna to plane s. Then
(6 mi)² + (y mi)² = (s mi)².
Given ds/dt = 240 mi/hr when y = 12 mi, the question is to find dy/dt when y = 12 mi.
2y dy/dt = 2s ds/dt ==> dy/dt = (s/y) ds/dt.
When y = 12, s = √(12² + 6²) = 6√(5). So when the plane is 12 mi. away its speed is
dy/dt = 6√(5)/12 (240) mi/hr = 120√(5) mi/hr ≈ 268 mi/hr.
You can construct a right triangle with one leg of length 6 mi straight up from the radar antenna, one leg is the path of the plane parallel to the earth (6 mi. up), and hypotenuse a straight line from antenna to plane.
Call the length of the plane's path y, and straight line distance from antenna to plane s. Then
(6 mi)² + (y mi)² = (s mi)².
Given ds/dt = 240 mi/hr when y = 12 mi, the question is to find dy/dt when y = 12 mi.
2y dy/dt = 2s ds/dt ==> dy/dt = (s/y) ds/dt.
When y = 12, s = √(12² + 6²) = 6√(5). So when the plane is 12 mi. away its speed is
dy/dt = 6√(5)/12 (240) mi/hr = 120√(5) mi/hr ≈ 268 mi/hr.