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Chapter 5: Problem 27

An airplane goes into a turn \(3.6 \mathrm{km}\) in radius. If the banking angle required is \(28^{\circ}\) from the horizontal, what's the plane's speed?

Short Answer

Expert verified
The speed of the airplane in this banking turn is approximately 307.5 m/s.

Step by step solution

01

- Defining Variables

Let's denote the speed of the airplane as, \(v\), the radius of the circular flight path as, \(r = 3.6\) km (which is \(3600\) m after converting kilometers to meters), the banking angle as, \(\theta = 28^\circ\), and the acceleration due to gravity as, \(g = 9.8\) m/s².
02

- Apply Principles of Circular Motion

In a banking turn, two forces work on the airplane - gravity and the lift generated by the wings. The two forces combine to provide the centripetal force required for circular motion. This can be expressed as: \(\tan(\theta) = \frac{v^2}{g \times r}\)
03

- Solve for Speed

Rearranging the above equation, we can solve for \(v\): \[v = \sqrt{\tan(\theta) \times g \times r}\] Now, substitute the given values (\(g = 9.8\) m/s², \(r = 3600\) m and \(\theta = 28^\circ\)) into this formula and calculate.
04

- Speed Calculation

Remember to convert the angle from degrees to radians as follows: \(28^\circ = 0.49\) radian. Substituting the known values: \[v = \sqrt{\tan(0.49) \times 9.8 \times 3600}\] Calculate this expression to find the speed of the airplane.

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