Vector applications help
I am not sure how to solve this problem, can someone please help me.
A plane is supposed to travel 820km on a bearing of 80 degrees. The plane's speed is 800km/h. The wind is blowing at 70km/h on a bearing of 100 degrees. What heading should the plane take and how long will the trip take?
What work have you done so far?
This is what I have done so far (the arrows are meant to be on top of the letter to indicate a vector):
Let ->V be the wind
Let ->U be the plane’s speed
Let ->W be the plane’s heading
Vx = 70cos10°
Vy = 70sin10°
Ux = 800cosθ
Uy = 800sinθ
but I don't know how to find the resultant vector without theta.
I'm assuming you meant
Vx = 70 cos(100°)
Vy = 70 sin(100°), right?
Don't you already know the desired resultant vector? Perhaps you could write out an equation that would help you find what you need...
That is what I meant.
I dont know the desired resultant vector, only the total distance not the speed. I need to find the resultant vector of the 800km/h speed and the 70km/h wind, but I don't know how to do this since i dont know the angle of the speed, I only know the angle of the resultant vector which is 10 degrees.
Is the plane's speed 800 km/hr in air, or relative to the ground? I suppose I'm also asking this: what are the assumptions regarding how the velocities combine? Do they just add vectorially?
How the speed is measured is irrelevant, they just add vectorially.
Ok. Suppose you do this:
800 cos(θ) + 70 cos(100°) = S cos(80°)
800 sin(θ) + 70 sin(100°) = S sin(80°),
where S is the unknown resultant speed. Now, it seems to me that you have here two equations in two unknowns. How could you go about solving this?
Oh!!! That makes sense, I don't know why I didn't think of that. Ok so now I have to use either substitution or elimination to solve.
Yeah, you could do that. Except that one of your variables does not appear in the equations in a linear fashion (which, I think, rules out elimination). I think the first thing I would do is divide one equation by the other to eliminate S.
Once I divide the equations how to I solve for theta since there are two of them.
Just take it one step at a time. What do you get for the division? And what do you suppose the next step after that would be?
800sin(θ) - 70sin(10) = sin(10)
800cos(θ) + 70cos(10) = cos(10)
I think the next step would be:
800tan(θ) - 0.18 = 0.176
θ = tan-1 ( (0.176 + 0.18) / 800)
Is that right?
Addition does not distribute over division. Your second step is therefore incorrect. You've got:
800 sin(θ) + 70 sin(10)
--------------------------- = tan(10),
800 cos(θ) + 70 cos(10)
Now you must multiply both sides by the denominator of the LHS:
800 sin(θ) + 70 sin(10) = tan(10) (800 cos(θ) + 70 cos(10)).
Get all the terms with θ over to the LHS, and all terms without θ over to the RHS.
One recommendation: don't substitute in numbers for anything until the very end. Why? Professors love to give you similar problems with slightly different initial conditions. If you've algebraically solved for the answer, you can just plug those new conditions in to your final answer. If you've plugged in numbers too early, you'll have to re-do everything. It's more work!
Once I have all the terms with theta to one side how do I solve for both?