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7nadin3 [17]
2 years ago
8

A light plane is headed due south with a speed of 200 km/h relative to still air. After 1.00 hour, the pilot notices that they h

ave covered only 137 km and their direction is not south but 15.0∘ east of south. What is the wind velocity?
Physics
1 answer:
Lilit [14]2 years ago
8 0

Answer: the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.

Explanation: The initial velocity of the plane is 200km/h south (supose that south is our positive x-axis here and east is the positive y-axis)

In one hour, the plane is located 137km away from the initial position, and the position in X is equal to 137km*cos(15°) = 132.33, this means that the velocity in the x axis is equal to 132.33 km/h, knowing that the initial velocity of the plane was 200km in the x-axis, this means that the velocity of the air must be:

132.33km/h - 200km/h = -67.69km/h

km and the position in "y" is equal to 137km*sin(15°) = 35.4km

This means that the velocity of the air in the y-axis is 35.4km/h

So the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.

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Given;

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The distance traveled by Charlotte in feet is calculated as;

Distance = Speed \ \times \ time \\\\D = ut\\\\D = (\frac{66.5 \ mi}{h} \times  \frac{5280 \ ft}{1 \ mi} \times \frac{1 \ h}{3600 \ s} )(3.47 \ s)\\\\D = 338.44 \ ft

Therefore, the distance traveled by Charlotte in feet is 338.44 ft

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