A pilot flies her route in two straight-line segments. The displacement vector A for the first segment has a magnitude of 246 km and a direction 30.0o north of east. The displacement vector B for the second segment has a magnitude of 178 km and a direction due west. The resultant displacement vector is R = A + B and makes an angle è with the direction due east. Using the component method, find (a) the magnitude of R and (b) the directional angle è.
a. R = 240km @ 30o+178km @ 180o
X = 240*cos30 + (-)178 = 29.8 km.
Y = 240*sin30 = 120 km.
R^2 = X^2 + Y^2 = (29.8)^2+(120)^2=15288
R = 123.6 km.
tanA = Y/X = 120/29.8 = 4.03.
A = 76.1o North of East = Direction
Correction: Change 240 km to 246 km and redo all calculations.
To find the magnitude of the resultant displacement vector R, we can use the Pythagorean theorem. According to the given information, the displacement vector A has a magnitude of 246 km, and the displacement vector B has a magnitude of 178 km.
(a) Finding the magnitude of R:
Let's find the x-component and y-component of the resultant displacement vector R:
For vector A:
Magnitude of A = 246 km
Component of A in the x-direction = 246 km * cos(30°) ≈ 213.09 km
Component of A in the y-direction = 246 km * sin(30°) ≈ 123 km
For vector B:
Magnitude of B = 178 km
Component of B in the x-direction = -178 km (since it is due west)
Component of B in the y-direction = 0 km (since it is due west)
Now, let's add the x-components and y-components separately:
x-component of R = (213.09 km) + (-178 km) ≈ 35.09 km
y-component of R = (123 km) + (0 km) = 123 km
Using the Pythagorean theorem, the magnitude of R is:
|R| = sqrt((35.09 km)^2 + (123 km)^2)
|R| ≈ sqrt(1225.6681 + 15129)
|R| ≈ sqrt(16354.6681)
|R| ≈ 127.97 km
Therefore, the magnitude of R is approximately 127.97 km.
(b) Finding the directional angle è:
To find the directional angle è, we can use the inverse tangent function (tan⁻¹). The tangent of the angle è can be calculated by dividing the y-component of R by the x-component of R:
tan(è) = (123 km) / (35.09 km)
è = tan⁻¹(123 km / 35.09 km)
è ≈ 72.33°
Therefore, the directional angle è is approximately 72.33°.
To summarize:
(a) The magnitude of R is approximately 127.97 km.
(b) The directional angle è is approximately 72.33°.
To find the magnitude of the resultant displacement vector R, we can use the component method.
First, we need to resolve each displacement vector into its x and y components.
For vector A:
Magnitude = 246 km
Direction = 30.0° north of east
To find the x-component of A, we can use the cosine function:
Ax = Magnitude of A * cos(Direction of A)
Ax = 246 km * cos(30.0°)
Ax ≈ 213.13 km
To find the y-component of A, we can use the sine function:
Ay = Magnitude of A * sin(Direction of A)
Ay = 246 km * sin(30.0°)
Ay ≈ 123.00 km
For vector B:
Magnitude = 178 km
Direction = due west
The x-component of B is in the negative x-direction, so Bx = -178 km.
The y-component of B is zero since it is in the direction perpendicular to the y-axis.
Now, we can add the x and y components of A and B to find the components of the resultant vector R.
Rx = Ax + Bx
Rx ≈ 213.13 km + (-178 km)
Rx ≈ 35.13 km
Ry = Ay + By
Ry ≈ 123.00 km + 0 km
Ry ≈ 123.00 km
The magnitude of vector R can be found using the Pythagorean theorem:
|R| = sqrt(Rx^2 + Ry^2)
|R| = sqrt((35.13 km)^2 + (123.00 km)^2)
|R| ≈ sqrt(1235.8 km^2 + 15129.0 km^2)
|R| ≈ sqrt(16364.8 km^2)
|R| ≈ 127.97 km
Therefore, the magnitude of the resultant displacement vector R is approximately 127.97 km.
To find the directional angle θ, we can use the tangent function:
θ = atan(Ry / Rx)
θ = atan(123.00 km / 35.13 km)
θ ≈ atan(3.50)
θ ≈ 74.74°
Therefore, the directional angle θ is approximately 74.74°.