Math Geniuses; lil help

[The Truth]

I need a little help with this. Cal 3 vectors problem. Lets see what you guys got.


"A plane is capable of flying at a speed of 180km/h in still air. The pilot takes off from an airfield and heads due north according to the plane's compass. After 30 minutes of flight time, the pilot notices that, due to the wind, the plane has actually traveled 80km at an angle of 5 degrees east of north.

A) What is the wind velocity?

B) In what direction should the pilot have headed to reach the intended destination?"


thanks
truth

[Darryl_P]

Unless I misunderstood something it seems pretty straightforward...

A) Draw a triangle with one side of length 90 in a vertical direction. Label the lower end A and the upper point B. Draw another side originating from A with length 80 making a 5 degree angle with AB. Label the end point of this side C. Connect B and C and now you have a triangle which can be solved using the cosine law because you have 2 sides and a contained angle. Side BC will be the distance the wind traveled in half an hour (double it to get the wind speed per hour) and angle B will give the direction of the wind.

B)

Extend line segment BC to a point D on the opposite side of B from C so that BD=BC. Connect D to A completing triangle ADB. That triangle can also be solved using the cosine law since you know side AB=90, side BD which equals BC, and the contained angle ABD which is just 180-angle ABC since DBC is a straight line. Angle DAB will be the required direction.

[ QUOTE ]
Unless I misunderstood something it seems pretty straightforward...

A) Draw a triangle with one side of length 90 in a vertical direction. Label the lower end A and the upper point B. Draw another side originating from A with length 80 making a 5 degree angle with AB. Label the end point of this side C. Connect B and C and now you have a triangle which can be solved using the cosine law because you have 2 sides and a contained angle. Side BC will be the distance the wind traveled in half an hour (double it to get the wind speed per hour) and angle B will give the direction of the wind.

B)

Extend line segment BC to a point D on the opposite side of B from C so that BD=BC. Connect D to A completing triangle ADB. That triangle can also be solved using the cosine law since you know side AB=90, side BD which equals BC, and the contained angle ABD which is just 180-angle ABC since DBC is a straight line. Angle DAB will be the required direction.

[/ QUOTE ]

Only read (A), but that won't work. The plane actually traveled 80 km. Airspeed was 90 km/half-hr, so there is a velocity component acting as headwind, too. Gotta run.

[Darryl_P]

Naturally angle B, since it's on the top of the triangle, is to be interpreted as that many degrees east from south, which has a headwind component since it's an acute angle and you're going north.

[Bradyams]

My guess is the wind speed is 82.4 km/hr, however I just used trig. I'm only in Calc 2, so I' probably missing something here.

[purnell]

Simply calculate the vector difference between the airplane's velocity and the sum of the vectors, and you have the wind velocity. I find it simpler to let my calculator transform from polar coordinates to rectangular coordinates.

Thus you have:

Vector sum = (160 km/hr, 5 degrees east) [polar] = (159.4 km/hr N, 13.9 km/hr E) [rectangular]

Subtacting the plane's vector (180 km/hr N) gives (-20.6 km/hr N, 13.9 km/hr E), or (20.6 S, 13.9 E) for the wind velocity, which is 24.9 km/hr, 145 degrees (clockwise from north). (again using the calculator to transform back into polar coordinates)

For the second question, if we want to go due north, and the wind's easterly component is 13.9 km/hr, we need to give the plane's velocity an equal westerly component. The sin of this angle is (13.94km/hr)/(180km/hr) = .077444..., thus our angle is 4.44 degrees west. Thus, if the pilot drives at 4.44 degees west according to his compass, the plane will fly due north at 158.9 km/hr. There will of course be some rounding error because I was not careful with my significant digits.

I don't think this problem requires a genius to solve it.

edit: I'm 38, college was a long, long time ago for me, and I haven't used this stuff since then. Somebody please show me my errors if I have made any here.

[WillyTrailer]

you don't need any fancy calculus to solve this problem, just the good ol' law of cosines!

you have the following triangle:


|\_
|--\_
|---/
|--/
|-/
|/

just pretend that top line is straight the left side has length 90 and represents HALF the velocity of the plane.

the jagged top line is the vector representing HALF the wind velocity (should be pointing lower right)

the right side is the actual path and has length 80.

now just use law of cosines to solve for the length and direction (angle) of the wind vector.

I made a brief calculation and the TOTAL magnitude of the wind vector should be about 25 miles per hour. The direction should be about 5 degrees south of east, but these aren't exact and I don't have a calculator.

good luck,
WT

[WillyTrailer]

or even easier is to take

[(0,90) - (80*sin(5),80*cos(5))]*2

this will give you the vector you want.

just another way of looking at the problem of solving for one side and angle of a triangle when you know two sides and the opposing anle of the side in question.

-WT