1. An airplane flies 200 km on course 110 degrees from point A to point B. At point B the pilot changes course to 37 degrees. The plane then flies 100 km to point C. Find the magnitude of the net displacement from A to C.
2. A plane is flying on a bearing of 25 degrees west of north at 530 mph. Express the velocity of the plane as a vector.
1. AB+BC = 200km[110o] + 100km[37o]
X = 200*cos110 + 100*cos37 = 11.46 km.
Y = 200*sin110 + 100*sin37 = 248.1 km.
D^2 = X^2 + Y^2 = (11.46)^2+(248.1)^2 =
61,685
D = 248.4 km
2. V = 530mph[N25oW] = 530mph[115o] CCW.
1. To find the magnitude of the net displacement from A to C, we need to find the x and y components of the displacement vector and then use the Pythagorean theorem.
First, let's find the x and y components of the displacement from A to B. We can use trigonometry to find these components.
The x-component can be found using the cosine function:
x_component_AB = 200 km * cos(110 degrees)
The y-component can be found using the sine function:
y_component_AB = 200 km * sin(110 degrees)
Next, let's find the x and y components of the displacement from B to C. Again, we can use trigonometry to find these components.
The x-component can be found using the cosine function:
x_component_BC = 100 km * cos(37 degrees)
The y-component can be found using the sine function:
y_component_BC = 100 km * sin(37 degrees)
Now, let's calculate the net displacement by adding the x and y components of both displacements:
x_component_net_displacement = x_component_AB + x_component_BC
y_component_net_displacement = y_component_AB + y_component_BC
Finally, we can calculate the magnitude of the net displacement using the Pythagorean theorem:
magnitude_net_displacement = sqrt((x_component_net_displacement)^2 + (y_component_net_displacement)^2)
2. To express the velocity of the plane as a vector, we need to consider its magnitude (speed) and direction.
The magnitude or speed of the plane is given as 530 mph.
The direction is given as 25 degrees west of north. To express this as a vector, we can consider north as the positive y-direction and west as the negative x-direction.
The x-component of the velocity vector can be found using the cosine function:
x_component_velocity = 530 mph * cos(25 degrees)
The y-component of the velocity vector can be found using the sine function:
y_component_velocity = 530 mph * sin(25 degrees)
Therefore, the velocity vector of the plane can be expressed as (x_component_velocity, y_component_velocity).
1. To find the magnitude of the net displacement from A to C, we can break down the displacement into its horizontal and vertical components.
First, let's find the horizontal and vertical components of the displacement from A to B. We can use trigonometry to do this.
The horizontal component (H₁) is given by H₁ = 200 km * cos(110 degrees).
The vertical component (V₁) is given by V₁ = 200 km * sin(110 degrees).
Next, let's find the horizontal and vertical components of the displacement from B to C.
The horizontal component (H₂) is given by H₂ = 100 km * cos(37 degrees).
The vertical component (V₂) is given by V₂ = 100 km * sin(37 degrees).
Now, we can find the net horizontal component (Hₙ) and net vertical component (Vₙ) by adding the corresponding components together.
Hₙ = H₁ + H₂
Vₙ = V₁ + V₂
Finally, we can find the magnitude of the net displacement (D) using the Pythagorean theorem.
D = sqrt(Hₙ^2 + Vₙ^2)
Let's substitute the values and calculate.
2. To express the velocity of the plane as a vector, we can break it down into its horizontal and vertical components.
The horizontal component (Vx) is given by Vx = velocity * cos(bearing).
The vertical component (Vy) is given by Vy = velocity * sin(bearing).
Now, we can express the velocity of the plane as a vector by combining both components.
Velocity vector = Vx î + Vy ĵ
Let's substitute the values and calculate.