Let vector A = (6.0 m, 300 south of east), vector B = (13.9 m, north), vector C = (5.0 m, 450 south of west). Find the magnitude of vector D = A+B+C?
300 south of east?
that is 60 degrees short of all the way around to east again
maybe 30.0 ?
same with 450 south of west
My apologies, the degrees copied as zeroes. It's 30 degrees, 45 degrees.
ok
x is east
y is north
Ax = 6 cos 30
Ay = -6 sin 30
Bx = 0
By = 13.9
Cx = -5 cos 45
Cy = -5 sin 45
so
Dx = +6 cos 30 - 5 cos 45
Dy = -6 sin 30 + 13.9 -5 sin 45
|D| = sqrt (Dx^2 + Dy^2)
Thank you
To find the magnitude of vector D, you need to calculate the sum of vectors A, B, and C and then find the magnitude of the resulting vector.
First, let's break down the given vectors into their horizontal (x) and vertical (y) components:
Vector A:
Magnitude of A = 6.0 m
Theta (angle) of A = 300 degrees (south of east)
To find the x and y components of vector A:
Ax = A * cos(theta) = 6.0 m * cos(300 degrees)
Ay = A * sin(theta) = 6.0 m * sin(300 degrees)
Vector B:
Magnitude of B = 13.9 m
Theta (angle) of B = 0 degrees (north)
To find the x and y components of vector B:
Bx = B * cos(theta) = 13.9 m * cos(0 degrees)
By = B * sin(theta) = 13.9 m * sin(0 degrees)
Vector C:
Magnitude of C = 5.0 m
Theta (angle) of C = 450 degrees (south of west) or equivalently, -90 degrees (west)
To find the x and y components of vector C:
Cx = C * cos(theta) = 5.0 m * cos(-90 degrees)
Cy = C * sin(theta) = 5.0 m * sin(-90 degrees)
After calculating the x and y components of each vector, add the corresponding components together:
Dx = Ax + Bx + Cx
Dy = Ay + By + Cy
Once you have the x and y components of vector D, you can find its magnitude using the Pythagorean theorem:
Magnitude of D = sqrt(Dx^2 + Dy^2)
Plug in the values into these equations and solve to find the magnitude of vector D.