An airplane is flying with a ground speed of 212 meters per second North encounters a wind at 10
meters per second East. What is the resultant speed of the airplane? What direction?
I assume the wind is 10 m/s FROM the east (an east wind takes you west). Therefore it is moving us west.
we are then moving at a speed s
10^2 + 212^2 = s^2
s^2 = 45044
s = 212.2
tan A = 10/212 where A is angle west of north
A = 2.7 degrees west of north
The question is not phrased very well (mathematicians seldom do navigation) so it may mean the wind is going TO the east in which case the vector is 2.7 degrees east of north
To find the resultant speed and direction of the airplane, we need to use vector addition.
1. Convert the given speeds to vector form:
- The ground speed of the airplane is 212 m/s North, which can be represented as (+212) North.
- The wind speed is 10 m/s East, which can be represented as (+10) East.
2. Add the vectors:
- Since the wind is blowing to the East, which is perpendicular to the North direction, we can treat the two speeds as orthogonal vectors. To find the resultant, we can use the Pythagorean theorem.
- The magnitude of the resultant speed is given by the square root of the sum of the squares of the two speeds:
resultant speed = sqrt((212^2) + (10^2)) = sqrt(44944 + 100) = sqrt(45044) ≈ 212.3 m/s
- The direction of the resultant can be found by taking the inverse tangent of the ratio of the North and East components:
direction = arctan(North component / East component) = arctan(212/10) ≈ 87.2° North of East
Therefore, the resultant speed of the airplane is approximately 212.3 m/s, and the direction is approximately 87.2° North of East.