Suppose an airplane is flying in air with a speed of u=100m/s . The wind is from the south west with a velocity of v= 70m/s
what direction should the airplane head in order to fly directly north ( with respect to the ground)? Give the angle in degrees of west of north
what is the plane's speed as measured by a bystander on the ground
A is heading angle west of north
Wind is taking us NE at 70
North speed over ground = 100 cos A + 70 cos 45
West speed over ground = 100 sin A - 70 cos 45
if we are making a course made good of north, west speed is 0
so
sin A = (70 /100)cos 45
A = 29.7 degrees west of North
Now for speed made good north calculate
100 cos 29.7 + 70 sin 45
by the way I unintentionally used the fact the sin 45 = cos 45 = sqrt2/2
where did the 45 come from ?
wind is from SW
That means it is pushing you NE
NE is 45 degrees east of north (halfway between north which is 0 and east which is 90)
DRAW THE PICTURE !!!!
To determine the direction the airplane should head in order to fly directly north with respect to the ground, we need to consider the vector addition of the airplane's velocity and the wind velocity.
First, let's break down the velocities into their x and y components. We can use trigonometry to do this.
The airplane's velocity, u=100 m/s, can be broken down into its components as follows:
u_x = u * cos(theta), where theta is the angle between the airplane's velocity and the positive x-axis.
u_y = u * sin(theta), where theta is the angle between the airplane's velocity and the positive y-axis.
Similarly, the wind velocity, v=70 m/s, can be broken down into its components as follows:
v_x = v * cos(phi), where phi is the angle between the wind velocity and the positive x-axis.
v_y = v * sin(phi), where phi is the angle between the wind velocity and the positive y-axis.
Since the problem states that the wind is coming from the south-west, we can determine the angles theta and phi:
theta = 90 degrees (since the airplane is flying north)
phi = 45 degrees (since the wind is coming from the south-west)
Now let's calculate the x and y components of the resultant velocity (sum of airplane's velocity and wind velocity):
resultant_x = u_x + v_x
resultant_y = u_y + v_y
To find the angle of the resultant velocity with respect to the positive x-axis, we can use the inverse tangent function:
angle = arctan(resultant_y / resultant_x)
Substituting the respective values, we have:
resultant_x = u * cos(theta) + v * cos(phi)
resultant_y = u * sin(theta) + v * sin(phi)
angle = arctan(resultant_y / resultant_x)
Calculating the values:
resultant_x = 100 * cos(90 degrees) + 70 * cos(45 degrees) = 0 + 70 * √2 / 2 = 35√2
resultant_y = 100 * sin(90 degrees) + 70 * sin(45 degrees) = 100 + 70 * √2 / 2 = 100 + 35√2
Now we can calculate the angle:
angle = arctan((100 + 35√2) / (35√2))
= arctan((100 + 35√2) / (35 * 1.414))
≈ 66.92 degrees
Therefore, the airplane should head at an angle of approximately 66.92 degrees west of north in order to fly directly north with respect to the ground.
To determine the plane's speed as measured by a bystander on the ground, we need to calculate the magnitude of the resultant velocity.
The magnitude of the resultant velocity, V_res, can be calculated using the Pythagorean theorem:
V_res = sqrt(resultant_x^2 + resultant_y^2)
Substituting the values, we have:
V_res = sqrt((35√2)^2 + (100 + 35√2)^2)
= sqrt(2 * 35^2 + (100 + 35√2)^2)
= sqrt(2 * 1225 + (100 + 35√2)^2)
≈ sqrt(2450 + (100 + 35√2)^2)
Calculating the value, we find:
V_res ≈ 200.21 m/s
Therefore, the plane's speed as measured by a bystander on the ground is approximately 200.21 m/s.