A plane travels 1000 km at an angle of 40 degrees East to North, then changes direction and travels 500 km at an angle of 20 degrees North to East. What is the magnitude and direction of the planes total displacement?
40o E. to N. = 50o N. t0 E.
D = 1000km[50o] + 500km[20o].
X = 1000*Cos50 + 500*Cos20o =
Y = 100*sin50 + 500*sin20 =
Tan A = Y/X, A = ? = Direction.
X/Cos A = Magnitude.
To find the magnitude and direction of the plane's total displacement, we can add the individual displacements vectorially.
First, we'll find the horizontal and vertical components of the two displacements:
The first displacement of 1000 km at an angle of 40 degrees East to North can be split into its horizontal (x) and vertical (y) components as follows:
x₁ = 1000 km * sin(40°)
y₁ = 1000 km * cos(40°)
Similarly, the second displacement of 500 km at an angle of 20 degrees North to East can be split into x and y components as follows:
x₂ = 500 km * cos(20°)
y₂ = 500 km * sin(20°)
Next, we'll add the x and y components to get the total displacement:
x_total = x₁ + x₂
y_total = y₁ + y₂
To find the magnitude of the total displacement, we'll use the Pythagorean theorem:
magnitude = √(x_total^2 + y_total^2)
Finally, we'll find the direction of the total displacement using trigonometry:
direction = atan(y_total / x_total), where atan calculates the inverse tangent.
By plugging the values into the equations, we can find the magnitude and direction of the plane's total displacement.