Question

1. An airplane is flying with the velocity of 459km/h [E40S], however its ground velocity is 498km/h [E24S]. Determine the velocity of the blowing wind. (Use the vector subtraction diagram and geometry to solve this question).

Answer 1

where's your diagram?

Answer 2

ill attach it
sorry

Answer 3

what is E40S? and E24S? Do you mean 130 degrees for the first and 114 degress for the second?

Answer 4

so like East40(degrees)South. so on a compass plane it would be 40 degrees below east towards south

Answer 5

So what i did so far..:
Vwg = 498km/h [E24S]-459km/h[E40S]
Vwg = 498km/h [E24S]+ 459km/h[W40N]
Vwg = 498km/h*cos24 +298km/h*sin24 + 459*cos40 + 459*sin40
Vwg = 455km/h[E]+203km/h[S]+351km/h[W] + 295km/h[N]
Vwg = 104km/h[E]+ 92km/h[N]
Vwg = 138.86km/h[E41.5N]

so i got 138.86km/h

Vwg is wind velocity relative to ground
that's what im looking for..but i dont know if that's right. also i tried vector subtraction using geometry but i got a wholeeeee different answer

Answer 6

you have to break up into components or use the law of cosines or sines to figure out the needed vector. Also, the angles should be negative (4th quadrant).

Answer 7

and if \(\vec{u}=459\frac{\text{km}}{\text{hr}}\text{ @ }-40^{\circ}\) and \(\vec{v}=498\frac{\text{km}}{\text{hr}}\text{ @ }-24^{\circ}\) then you want \(\vec{v}-\vec{u}\).

Answer 8

ok i broke it into components like the teacher taught and my final answer was 138.61..is that right??

Answer 9

ok. i get that...but i dont get what the teacher meant by using geometry. like in class he did something like. v-u is the same as v+(-u) so he flipped the u on the diagram then he found the answer. that's what im really confused with

Answer 10

\[\vec{u}_x=459 \cdot \cos(-40), \text{ }\vec{u}_x=459 \cdot \sin(-40)\]
\[\vec{v}_x=498 \cdot \cos(-24), \text{ }\vec{u}_x=498 \cdot \sin(-24)\]

Answer 11

ooh ok i think i get it. ill try it again

Answer 12

i get wind = 138.675 km/h @ E 41.830 degrees N

Answer 13

ok so i get the answer...sorry one more question..but i dont understand how to get the orientation?..like how did you get the E41.830N??