The cruising speed of a Boeing 767 in still air is 530 mph. Suppose that a 767 is cruising directly east when it encounters an 80 mph wind blowing 40 degrees south of west. Sketch the vectors for the velocities of the airplane and the wind. Express both vectors in ordered pair notation. Find the sum of the vectors. Find the magnitude and direction of the resultant vector.
edit to reiny's answer, they had an error in the final step.
first vector = (530cos0, 530sin0
= (530,0)
2nd vector = (80cos220 , 80sin 220
= (-61.28355, -51.423
resultant = ((468.71644 , -51.423)
maginitude = √((468.71644^2 + (-51.423)^2 )
= 471.529
direction:
tan ฿ = -51.423/468.71644 = -0.11
฿ = 6° S of E
To sketch the vectors for the velocities of the airplane and the wind, we can use a coordinate system where the positive x-axis points east and the positive y-axis points north.
The velocity of the airplane in still air can be represented by the vector (530, 0) mph, since it is cruising directly east with no northward or southward component.
The velocity of the wind can be represented by the vector (-80cos(40°), -80sin(40°)) mph, where -80cos(40°) represents the westward component and -80sin(40°) represents the southward component.
To find the sum of the vectors, simply add the corresponding components:
Sum vector = (530 + -80cos(40°), 0 + -80sin(40°))
Now we can calculate the magnitude and direction of the resultant vector.
Magnitude of the resultant vector = sqrt((530 + -80cos(40°))^2 + (-80sin(40°))^2) mph
Direction (angle) of the resultant vector = arctan((-80sin(40°)) / (530 + -80cos(40°)))
Simplifying these equations, we have:
Magnitude of the resultant vector ≈ 521.10 mph
Direction of the resultant vector ≈ 143.97° (measured clockwise from the positive x-axis)
To find the sum of two vectors, we can add their corresponding components. Let's break down the problem step by step.
1. Sketch the vectors:
a) Velocity of the airplane:
The cruising speed of the Boeing 767 in still air is 530 mph directly east. Therefore, the velocity vector for the airplane can be represented as (530, 0).
b) Velocity of the wind:
The wind is blowing 80 mph at an angle of 40 degrees south of west. To represent this vector, we need to find its horizontal and vertical components.
The horizontal component can be calculated as 80*cos(40°), which is approximately 61.16 mph. Since the wind is blowing west, the horizontal component will be negative (-61.16).
The vertical component can be calculated as 80*sin(40°), which is approximately 51.27 mph. Since the wind is blowing south, the vertical component will also be negative (-51.27).
Therefore, the velocity vector for the wind can be represented as (-61.16, -51.27).
2. Find the sum of the vectors:
To find the sum of the airplane velocity and wind velocity vectors, we add their corresponding components:
(530, 0) + (-61.16, -51.27) = (530-61.16, 0-51.27) = (468.84, -51.27).
3. Find the magnitude and direction of the resultant vector:
The magnitude of a vector can be found using the formula: magnitude = sqrt(x^2 + y^2).
In our case, the magnitude of the resultant vector is:
magnitude = sqrt((468.84)^2 + (-51.27)^2) = sqrt(219885.5852 + 2628.6729) = sqrt(222514.2581) = 471.34 mph (approximately).
The direction of the resultant vector can be found using the inverse tangent function:
direction = tan^(-1)(y/x).
In our case, the direction of the resultant vector is:
direction = tan^(-1)(-51.27/468.84) = tan^(-1)(-0.1094) = -6.230 degrees (approximately).
Therefore, the magnitude of the resultant vector is approximately 471.34 mph, and its direction is approximately 6.230 degrees south of west.
first vector = (530cos0, 530sin0
= (530,0)
2nd vector = (80cos220 , 80sin 220
= (-61.28355, -51.423
resultant = ((468.71644 , -51.423)
maginitude = √((468.71644^2 + (-51.423)^2 )
= 471.529
direction:
tan ฿ = -51.423/468.71644 = -.839..
฿ = E 40° S