the velocity of the wind would be:
(42km/h)cos(30)=36.37
After flying for 15 min in a wind blowing 42 km/h at an angle of 30° south of east, an airplane pilot is over a town that is 50 km due north of the starting point. What is the speed of the airplane relative to the air in km/h?
what formula do u start with?
To solve this problem, we can break down the velocity of the airplane into two components: one in the east direction and one in the north direction.
Let's start with the formula for vector addition. The magnitude of the resultant vector (in this case, the velocity of the airplane) can be found using the Pythagorean theorem:
R^2 = A^2 + B^2
Where:
R is the magnitude of the resultant vector (velocity of the airplane),
A is the east component of the velocity, and
B is the north component of the velocity.
Now, we need to find the east component of the velocity. We know that the wind speed is 42 km/h and the angle of the wind (relative to the east) is 30°. To find the east component, we can use trigonometry. The east component (A) of the velocity of the airplane can be calculated by:
A = Wind Speed * cos(Wind Angle)
Substituting the values:
A = 42 km/h * cos(30°)
Similarly, the north component of the velocity (B) can be found using trigonometry as well. The north component (B) of the velocity of the airplane can be calculated by:
B = Wind Speed * sin(Wind Angle)
Substituting the values:
B = 42 km/h * sin(30°)
Now, we can substitute the values of the east and north components (A and B) into the formula for the magnitude of the resultant vector (R) to find the speed of the airplane relative to the air (velocity of the airplane):
R = sqrt(A^2 + B^2)
Plugging in the values of A and B into the formula:
R = sqrt((42 km/h * cos(30°))^2 + (42 km/h * sin(30°))^2)
Simplifying the equation gives the answer in km/h.