After flying for 15 min in a wind blowing 42 km/h at an angle of 20° south of east, an airplane pilot is over a town that is 42 km due north of the starting point. What is the speed of the airplane relative to the air?
There are three vectors involved:
A. the flight vector as yet unknown.
B. The wind at 42 km/h at -20° with respect to the x-axis.
C. the resultant: (0,42) km. from the starting point.
Since the vectorial sum of A and B equals C, you can obtain A by adding C and -B, both of which are known.
i need more help
You will need to know how to add and subtract vectors. Have you done that at school?
There are different methods, depending on what you have learned. It could be graphical by drawing directed arrows and joining them together in the right order. It could be by resolving the vectors into the x- and y-components.
We need to know more about what you are working on to provide more detailed help.
To find the speed of the airplane relative to the air, we need to break down the velocity vectors into the horizontal and vertical components.
First, let's calculate the horizontal component of the wind velocity. The angle of 20° south of east means the angle between the wind direction and the positive x-axis is (180° - 20°) = 160°. We can find the horizontal component using trigonometry:
Horizontal component of wind velocity = wind speed * cos(angle)
Horizontal component of wind velocity = 42 km/h * cos(160°)
Next, let's calculate the vertical component of the wind velocity using trigonometry:
Vertical component of wind velocity = wind speed * sin(angle)
Vertical component of wind velocity = 42 km/h * sin(160°)
Now, let's calculate the net velocity of the airplane by subtracting the wind velocity from the velocity of the plane relative to the ground:
Net velocity of airplane = √((horizontal component of wind velocity)^2 + (vertical component of wind velocity)^2)
Net velocity of airplane = √((velocity of the plane relative to the ground)^2 - (wind speed)^2)
Since the airplane is flying due north, the net horizontal velocity is zero. Therefore:
0 = (velocity of the plane relative to the ground)^2 - (horizontal component of wind velocity)^2
Let P be the velocity of the plane relative to the ground. We can solve the equation for P:
P^2 = (horizontal component of wind velocity)^2
P^2 = (42 km/h * cos(160°))^2
P = 42 km/h * cos(160°)
So, the speed of the airplane relative to the air is approximately 19.8 km/h.