An airplane flies on a compass heading of 90° at 310 mph. The wind affecting the plane is blowing from 332° at 40 mph. What is the true course and ground speed of the airplane? '
I know to use the cosine and sine formula but I can get a right answer..
If you dont want to solve the above problem, please just show me HOW to solve this.
thank you!
I know to use the cosine and sine formula but I can get a right answer..
If you dont want to solve the above problem, please just show me HOW to solve this.
thank you!
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Always start a vector problem with a sketch. This helps, even if you think you know where the problem is going. It's crucial if you don't. You'll have an x-y coordinate system with the airplane at the origin.
Navigators measure headings clockwise from due North. Mathematicians measure angles counter-clockwise from the +x axis. That difference can cause problems if you're not aware of it. In this case, you'll get the right answer if you're consistent. I like to keep the sketch close to the problem being modeled, A heading of 90 is due East, and that's along the +x axis to me.
So, you have the airplane's velocity relative to still air as 310 mph due East. This is called the airspeed, even though it's the speed of the plane, not the air, and is a velocity when the direction is known. The wind velocity is 40 mph at (360-332) = 28 degrees counter-clockwise from due North. That's at an angle 90+28=118 counter-clockwise from +x, in the 2nd quadrant.
No trig needed for the first velocity, V1 = = <310, 0>.
The second is V2 = = <40*cos 118, 40*sin 118> ~~ <-18.8, 35.3>
The ground speed (velocity of the airplane relative to the ground) is the sum V1+V2:
V = = <310 - 18.8, 0 + 35.3> = <291.2, 35.3>
That's the answer in Cartesian form. You want polar form to report speed and direction, so:
||V|| = sqrt(vx^2 + vy^2> ~~ sqrt(291.2^2 + 35.3^2) ~~ 293.3
V is in the first quadrant, so the angle theta is arctan(vy/vx):
theta ~~ arctan(35.3 / 292.2) ~~ 6.9 degrees
That corresponds to a heading of 90-6.9 = 83.1 for the heading and the speed is 293.3 mph.
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Confession: That diagram keeps everything in map orientation, which is nice, but under pressure I'd simply label +x as N and+y as E (giving a rotated mirror image of the usual map view) and then
= <310 cos 90, 310 sin 90> + <40 cos 332, 40 sin 332>
~~ <0 + 35.3, 310 - 18.8> = <35.3, 291.2>
||V|| ~~ 293.3 mph, as above
theta = arctan(291.2/35.3) ~~ 83.1 degrees
That saves the steps of translating from and back to compass orientation, and gets the same answer.
Navigators measure headings clockwise from due North. Mathematicians measure angles counter-clockwise from the +x axis. That difference can cause problems if you're not aware of it. In this case, you'll get the right answer if you're consistent. I like to keep the sketch close to the problem being modeled, A heading of 90 is due East, and that's along the +x axis to me.
So, you have the airplane's velocity relative to still air as 310 mph due East. This is called the airspeed, even though it's the speed of the plane, not the air, and is a velocity when the direction is known. The wind velocity is 40 mph at (360-332) = 28 degrees counter-clockwise from due North. That's at an angle 90+28=118 counter-clockwise from +x, in the 2nd quadrant.
No trig needed for the first velocity, V1 =
The second is V2 =
The ground speed (velocity of the airplane relative to the ground) is the sum V1+V2:
V =
That's the answer in Cartesian form. You want polar form to report speed and direction, so:
||V|| = sqrt(vx^2 + vy^2> ~~ sqrt(291.2^2 + 35.3^2) ~~ 293.3
V is in the first quadrant, so the angle theta is arctan(vy/vx):
theta ~~ arctan(35.3 / 292.2) ~~ 6.9 degrees
That corresponds to a heading of 90-6.9 = 83.1 for the heading and the speed is 293.3 mph.
- - - - -
Confession: That diagram keeps everything in map orientation, which is nice, but under pressure I'd simply label +x as N and+y as E (giving a rotated mirror image of the usual map view) and then
||V|| ~~ 293.3 mph, as above
theta = arctan(291.2/35.3) ~~ 83.1 degrees
That saves the steps of translating from and back to compass orientation, and gets the same answer.