An airplane is flying on a bearing of 285 degrees at 480 mph. A wind is blowing with the bearing 265 degrees at 30?
a. find the component from of the velocity of the airplane
b. find the actual speed and direction of the airplane
485 sin 75 degrees
-485 cos 75 degrees (in 4 th quadrant)
30 sin -20 degrees
-30 cos -20 degrees
Then, added them for component form.
Here is what I got, but it is not correct....I'm really not sure how to start the problem and finish it with the actual speed and direction.
First, planes fly on headings, not bearings.
Second, winds are usually indicated by where they are blowing from, but I'll assume your wind is indeed blowing roughly westward.
285° is measured clockwise from due north, which is 90°. So, it is in QII. Due west is a heading of 270°, so it is 15° north of west.
480@285° = <480cos285°,480sin285°> = <-480cos15°,480sin15°>
30@265° = <30cos265°,30sin265°> = <-30cos5°,-30sin5°>
To find the actual speed and direction of the plane, convert the values to decimal values, add the components, and then use the Pythagorean theorem to find the speed, and then figure the angle (tanθ=y/x). That θ is measured counterclockwise from the x-axis, so you need to convert that to actual course heading.
Well, solving this problem is definitely "plane" sailing! Let's tackle it step by step.
a. To find the component form of the velocity of the airplane, we can break it down into two components: the north-south component and the east-west component. We can use trigonometry to solve this.
For the north-south component, we can use the sine function: (485 mph) * sin(75 degrees).
For the east-west component, we can use the cosine function: (485 mph) * cos(75 degrees).
b. To find the actual speed and direction of the airplane, we need to consider the effect of the wind. The wind affects the east-west component of the airplane's velocity.
To find the new east-west velocity, we need to take into account the wind's east-west component: (30 mph) * cos(-20 degrees).
To find the new north-south velocity, we also need to take into account the wind's north-south component: (30 mph) * sin(-20 degrees).
Finally, we can calculate the actual speed using the Pythagorean theorem: sqrt((north-south component + wind north-south component)^2 + (east-west component + wind east-west component)^2)
To find the direction, we can use the inverse tangent function (arctan) of the east-west component divided by the north-south component.
Give these calculations a try, and remember, laughter is the best airplane "wing"man!
To solve this problem, we need to break down the velocities into their x- and y-components.
Let's start by calculating the x- and y-components of the airplane's velocity.
a. Component form of the velocity of the airplane:
Given that the airplane is flying on a bearing of 285 degrees at 480 mph, we can find the x and y-components using trigonometry.
The x-component, Vx_A, of the airplane's velocity can be calculated by:
Vx_A = 480 * cos(285°)
The y-component, Vy_A, of the airplane's velocity can be calculated by:
Vy_A = 480 * sin(285°)
b. Actual speed and direction of the airplane:
To find the actual speed and direction of the airplane, we need to add the wind vector to the airplane's velocity vector.
The x-component of the wind velocity, Vx_W, can be calculated by:
Vx_W = 30 * cos(265°)
The y-component of the wind velocity, Vy_W, can be calculated by:
Vy_W = 30 * sin(265°)
Now, the x-component of the total velocity, Vx_T, can be found by adding the x-components of the airplane's velocity and the wind velocity:
Vx_T = Vx_A + Vx_W
Similarly, the y-component of the total velocity, Vy_T, can be found by adding the y-components of the airplane's velocity and the wind velocity:
Vy_T = Vy_A + Vy_W
The actual speed, V_T, of the airplane can be found by taking the square root of the sum of the squares of the x- and y-components of the total velocity:
V_T = sqrt(Vx_T^2 + Vy_T^2)
The actual direction, θ_T, of the airplane can be calculated using the inverse tangent function:
θ_T = atan(Vy_T / Vx_T) (considering any necessary quadrant adjustments)
Finally,
Vx_A = 480 * cos(285°) = -378.5366
Vy_A = 480 * sin(285°) = -246.509
Vx_W = 30 * cos(265°) = 27.396
Vy_W = 30 * sin(265°) = -11.598
Vx_T = Vx_A + Vx_W = -378.5366 + 27.396 = -351.1406
Vy_T = Vy_A + Vy_W = -246.509 + (-11.598) = -258.107
V_T = sqrt(Vx_T^2 + Vy_T^2) = sqrt((-351.1406)^2 + (-258.107)^2) = 439.506 mph (approximately)
θ_T = atan(Vy_T / Vx_T) = atan(-258.107 / -351.1406) ≈ 35.11°
Therefore, the actual speed of the airplane is approximately 439.506 mph, and the direction of the airplane is approximately 35.11 degrees.
To solve this problem, we can break down the velocities into their components and then add them up.
a. To find the component form of the velocity of the airplane, we can use trigonometry. The given bearing of the airplane is 285 degrees, which means the angle with the horizontal axis is 90 degrees - 285 degrees = -195 degrees. The magnitude of the velocity is given as 480 mph, so the component form of the airplane's velocity is (-480 cos 195 degrees, -480 sin 195 degrees).
b. To find the actual speed and direction of the airplane, we need to consider the effect of the wind. Since the wind is blowing at a bearing of 265 degrees, we can subtract 265 degrees from the bearing of the airplane (285 degrees) to find the relative bearing between the airplane and the wind. The relative bearing is 285 degrees - 265 degrees = 20 degrees.
Now, we can use trigonometry to find the resulting speed and direction of the airplane. The magnitude of the resulting velocity can be found using the Pythagorean theorem:
Resulting speed = sqrt[(component of airplane velocity)^2 + (component of wind velocity)^2]
component of airplane velocity = -480 cos 195 degrees
component of wind velocity = 30 sin 20 degrees (since the wind is blowing in the opposite direction of the airplane)
Plugging in the values, we have:
Resulting speed = sqrt[(-480 cos 195 degrees)^2 + (30 sin 20 degrees)^2]
To find the direction, we can use the tangent function:
Direction = arctan[(component of wind velocity)/(component of airplane velocity)]
Direction = arctan[(30 sin 20 degrees)/(-480 cos 195 degrees)]
Remember to convert the angle from radians back to degrees.
By calculating the above expressions, you will be able to find the actual speed and direction of the airplane.