a plane is flying southwest at 155 mi/h and encounters a wind from the west at 45.0 mi/h. what is the plane velocity in respect to the ground in standard position?
The west component of velocity is reduced becasue of the west wind to 155cos45 - 45 = 64.6 mph (relative to the ground).
The south component of speed remains 155 sin45 = 109.6 mph
The new ground speed is the hypotenuse:
sqrt[(109.6)^2 + (64.6)^2] = 127.2 mph
To find the plane's velocity in respect to the ground, we need to combine the effect of its own velocity and the wind velocity.
Let's break it down into components:
1. Let the plane's velocity be represented by vector A.
2. The magnitude of vector A is 155 mi/h (speed of the plane).
3. The angle of vector A with respect to the positive x-axis is 225 degrees (counter-clockwise from the positive x-axis to the direction of the plane's velocity).
Next, let's analyze the wind velocity:
1. Let the wind velocity be represented by vector B.
2. The magnitude of vector B is 45.0 mi/h (speed of the wind).
3. The angle of vector B with respect to the positive x-axis is 270 degrees (counter-clockwise from the positive x-axis to the direction of the wind's velocity).
Now, add the two vectors together to find the resultant velocity (plane velocity in respect to the ground):
1. Add the x-components of the two vectors: A_x = A * cos(angle_A) and B_x = B * cos(angle_B).
A_x = 155 mi/h * cos(225 degrees) = -109.85 mi/h (negative because it's in the opposite direction of the positive x-axis)
B_x = 45.0 mi/h * cos(270 degrees) = 0 mi/h (wind is directed along the y-axis, no x-component)
2. Add the y-components of the two vectors: A_y = A * sin(angle_A) and B_y = B * sin(angle_B).
A_y = 155 mi/h * sin(225 degrees) = -109.85 mi/h (negative because it's in the opposite direction of the positive y-axis)
B_y = 45.0 mi/h * sin(270 degrees) = -45.00 mi/h (negative because it's in the opposite direction of the positive y-axis)
3. Add the x-components and y-components separately to get the resultant vector:
Resultant_x = A_x + B_x = -109.85 mi/h + 0 mi/h = -109.85 mi/h
Resultant_y = A_y + B_y = -109.85 mi/h + (-45.00 mi/h) = -154.85 mi/h
Therefore, the plane's velocity in respect to the ground, in standard position, is -109.85 mi/h in the x-direction and -154.85 mi/h in the y-direction.
To find the velocity of the plane with respect to the ground, we need to consider the vector sum of the plane's velocity and the wind's velocity.
The velocity of the plane can be represented as a vector pointing southwest, with a magnitude of 155 mi/h.
The velocity of the wind can be represented as a vector pointing west, with a magnitude of 45.0 mi/h.
To find the plane's velocity with respect to the ground, we need to add these two vectors together.
Step 1: Convert the given information into vector form.
The velocity of the plane is 155 mi/h southwest, which can be represented as a vector (-155, -155) mi/h.
The velocity of the wind is 45.0 mi/h west, which can be represented as a vector (0, -45.0) mi/h.
Step 2: Add the two vectors.
To add these vectors, we simply add their corresponding components. The x-components are added together, and the y-components are added together.
(-155, -155) + (0, -45) = (-155 + 0, -155 + (-45)) = (-155, -200) mi/h
Therefore, the plane's velocity with respect to the ground, in standard position, is (-155, -200) mi/h.