An aircraft maintains a speed of 500 miles per hour in a southwestern direction. The velocity of the jet stream is constant 50 miles per hour from the west. Find the resultant vector, the speed, and actual direction of the aircraft.
R = 500mi/h @ 225o + 50mi/h @ 0o
X = 500*cos225 + 50 = --303.6/303.6 mi/h.
Y = 500*sin225 = -353.6 mi/h.
tanAr = Y/X = -353.6/-303.6 = 49.35o =
Reference Angle.
A = 49.35 + 180 = 229.35o,CCW = 40.65o
West of South = Direction.
V = X/cosA = -303.6/cos229.35=466 mi/h.
To solve this problem, we need to understand vector addition. The resultant vector is the total displacement of an object when multiple vectors act upon it. In this case, we have two vectors: the velocity of the aircraft and the velocity of the jet stream.
First, let's represent the aircraft's velocity as a vector. Since it maintains a speed of 500 miles per hour in a southwestern direction, we can represent it as a vector with a magnitude of 500 miles per hour pointing in a southwestern direction. Let's call this vector A.
Next, let's represent the velocity of the jet stream as a vector. Since it is a constant 50 miles per hour from the west, we can represent it as a vector with a magnitude of 50 miles per hour pointing west. Let's call this vector B.
To find the resultant vector, we need to add vectors A and B. We can do this by breaking down each vector into its components, adding their corresponding components, and then combining them to find the resultant vector.
For vector A, since it is in a southwestern direction, we can break it down into two components: one in the southwest direction (x-component) and one in the south direction (y-component). The magnitudes of these components can be found using trigonometry.
Let's assume the x-component is A_x and the y-component is A_y. To find A_x, we use cosine since it's the adjacent side to the angle between vector A and the x-axis. To find A_y, we use sine since it's the opposite side to the angle.
A_x = A * cosθ
A_y = A * sinθ
Here, θ is the angle between vector A and the x-axis, which we can find using trigonometry. Since vector A is in a southwestern direction, we can consider it as 225 degrees with respect to the positive x-axis.
Let's calculate A_x and A_y:
A_x = 500 * cos(225°) ≈ -353.55 miles per hour
A_y = 500 * sin(225°) ≈ -353.55 miles per hour
Now, let's calculate the x and y components of vector B. Since vector B is constant at 50 miles per hour from the west, the x-component of B will be -50 miles per hour, and the y-component will be 0 since there is no north or south component.
Now, we can add the x-components and y-components separately:
Resultant x-component = A_x + B_x
Resultant y-component = A_y + B_y
Resultant x-component ≈ -353.55 miles per hour - 50 miles per hour ≈ -403.55 miles per hour
Resultant y-component ≈ -353.55 miles per hour + 0 miles per hour ≈ -353.55 miles per hour
Finally, use these resultant x and y components to find the magnitude and direction of the resultant vector:
Magnitude of the resultant vector = sqrt((Resultant x-component)² + (Resultant y-component)²)
Magnitude of the resultant vector ≈ sqrt((-403.55)² + (-353.55)²) ≈ 548.03 miles per hour
To find the direction of the resultant vector:
Direction = atan(Resultant y-component / Resultant x-component)
Direction ≈ atan(-353.55 / -403.55) ≈ atan(0.875) ≈ 41.17 degrees
Therefore, the resultant vector has a magnitude of approximately 548.03 miles per hour and a direction of approximately 41.17 degrees.