Lesson Airspeed, Time, and Distance

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This Lesson (Airspeed, Time, and Distance) was created by by oberobic(2304) About Me : View Source, Show
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Problem: A light aircraft flies from A to B, 450km away, and returns from B and A in a total time of 5 hours and 30 minutes. Suppose that during the whole journey there is a constant wind blowing from A to B. The speed of the aircraft in still air is 165 km/h. Find the speed of the wind.
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Solution: The basic approach to solving problems such as these is to begin with the fundamental distance equation and plug in the known values. Then solve for the unknowns.
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d = rt is the basic distance equation.
d = 450 km
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We are told the wind is blowing from A to B, so on the trip from A to B the airplane's speed is 165 km/h + w, where 'w' is the wind speed in km/h. That means the speed across the ground (the ground speed) is faster than the indicated airspeed. A wind from behind is called a 'tailwind.'
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Similarly, the ground speed going from B to A will be reduced by the wind. The indicated airspeed will still be 165, but the speed across the ground will be 165 - w. Such a wind is called a 'headwind.'
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We must assume the distance from A to B equals the distance from B to A. That is a reasonable mathematical assumption, although there are many factors in aviation that could force the path to be quite different going and coming.
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We assume the plane is flying at an indicated airspeed of 165 km/h in both directions.
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We are told the roundtrip time is 5 hr 30 min, which = 5.5 hr.
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A to B trip:
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+450+=+%28165%2Bw%29%2At+
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+t+=+450%2F%28165%2Bw%29+
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B to A trip:
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+450+=+%28165-w%29%2A%285.5-t%29+
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+5.5-t+=+450%2F%28165-w%29+
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+-t+=+450%2F%28165-w%29+-+5.5+
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+t+=+-450%2F%28165-w%29+%2B+5.5+
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+t+=+t+
so
+450%2F%28165%2Bw%29+=+-450%2F%28165-w%29+%2B+5.5+
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+450%2F%28165%2Bw%29+%2B+450%2F%28165-w%29+=+5.5+
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+450%2A%28165-w%2B165%2Bw%29+=+%28165%5E2+-165w+%2B+165w+-w%5E2%29+%2A+5.5+
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+450%2A330+=+%28165%5E2+-165w+%2B+165w+-w%5E2%29+%2A+5.5+
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+450%2A330%2F5.5+=+%28165%5E2+-165w+%2B+165w+-w%5E2%29+
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+27000+=+-w%5E2+%2B+165%5E2+
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+w%5E2+=+165%5E2+-27000+
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+w%5E2+=+27225+-27000+=225+
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+w+=+15+
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Note that sqrt(225) = + or - 15, but a negative wind speed is not applicable.
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Substitute w = 15
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+450+=+%28165%2Bw%29%2At+
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+450+=+%28165%2B15%29%2At+
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+450+=+180t+
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+180t+=+450+
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+t+=+450%2F180+=+45%2F18+=+5%2F2+=+2.5+
The flying time from A to B with a tailwind is 2.5 hr.
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+5.5-t+=+5.5-2.5+=+3+
The flying time from B to A with a headwind is 3 hr.
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Check the distances traveled to be sure this is the answer.
+%28165%2B15%29%2A2.5+=+450+
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+%28165-15%29%2A3+=+450+
Correct.
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Answer: The wind speed is 15 km/h.
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