An airplane flies with a velocity of 45.0 m/s [ W 35o N] with respect to the air (this is known as air speed). If the velocity of the airplane according to a stationary observer on the ground is 30.0 m/s [W 60o N], what was the wind velocity?
Would you find the answer by adding or subtracting the vectors?
A marble rolls with a velocity of 14 mm/s [W] on a game board that is being pulled [E 60o N] at 20.0 mm/s. What is the velocity of the marble relative to the floor?
what do i do for the first. I am stuck at this point. let me know if i did something wrong
VAa= 45.00 m/s [W35 degrees N]
VAG= 30.0 m/s (W 60 degrees N]
try to do law of cosine
angle c= 60-35=25
convert to radians = (25pi/180)
c squared= 45 squared + 30 sqaured - 2 (45) (30) cost (25pi/180)
=225.1
c=15.0 m/s (wind speed)
vx= 45 x cos 35 = 36.9 m/s
ux= 30 x cos 60 =15.0 m/s
vx-ux= 36.9-15=21.9 m/s
theta=cos-1(21.9/15)
but I get an error ( what did I do wrong)
@MITer94 for the first question how come this way would not work?:
VAa= 45.00 m/s [W35 degrees N]
VAG= 30.0 m/s (W 60 degrees N]
try to do law of cosine
angle c= 60-35=25
convert to radians = (25pi/180)
c squared= 45 squared + 30 sqaured - 2 (45) (30) cost (25pi/180)
=225.1
c=15.0 m/s (wind speed)
vx= 45 x cos 35 = 36.9 m/s
ux= 30 x cos 60 =15.0 m/s
vx-ux= 36.9-15=21.9 m/s
theta=cos-1(21.9/15)
but I get an error ( what did I do wrong)
I thought using law of cosines would work to solve the problem
@zxcvbnm1216 as I said before, just solve it component by component (e.g. find the x-component, then the y-component). You should get a vector for V_W.
To find its direction, take the inverse tangent of (VWx / VWy) where VWx, VWy are the x- and y-components of the vector.
@zxcvbnm1216 you have to be very careful with +/- signs and your choice of coordinate systems. I don’t even know what coordinate system you are using.
I would rather use headings (simply because it’s less confusing to me, but also because it’s more standardized). The “air speed” direction is 305° and the velocity of the plane (relative to observer) has heading 330°. A slightly better choice might be to use 0° = east, 90° = north, etc., but headings are pretty standard when flying.
Define the +x and +y in the usual way, where +x corresponds to “east” and +y corresponds to “north” (as if you were looking at a compass).
Then:
VAG[x] = -30 cos 60° = -15 m/s (I put the - sign because VAG points westward, or in the -x direction).
V_AG[y] = 30 sin 60° = 25.98 m/s
VAW[x] = -45 cos 35° = -36.86 m/s
VAW[y] = 45 sin 35° = 25.81 m/s
VW = VAG - V_AW
Therefore:
VW[x] = -15 m/s - (-36.86) m/s = 21.86 m/s (this is positive, which means the wind is headed eastward)
VW[y] = 25.98 m/s - 25.81 m/s = 0.18 m/s (also positive, meaning the wind is headed slightly northward)
|V_W| = sqrt((21.86)^2 + (0.18)^2) ≈ 21.9 m/s
So this part is good.
For the direction, if you draw out the VW vector, you will see that tan^(-1) (0.18/21.86) gives the angle x relative to east (E x N using your notation). The 0.5° means that the actual wind velocity has direction E 0.5° N, or an 89.5° heading relative to north. If you draw out the VAW and V_AG vectors to scale, you should see that the direction makes sense.