Question on relative velocity -- Find time?

  • #1

Homework Statement


The distance between cities A and B it's L. An airplane flies back and forth between A and B, in a straight line, with velocity V in relation to the wind.
So, calculate the total time of the flight, if the wind blows with velocity v, in a direction which forms an angle theta with direction AB.

Homework Equations


x = x0 + v0*t
v12 = v2 - v1 (relative velocity of 2 in relation to 1 is the difference between velocities 2 and 1 in relation to the origin).

The Attempt at a Solution


Well, the time of the flight between A and B will be t = L/Ve (and I'm assuming that Ve is the velocity of the airplane in relation to the Earth). The back flight time will be different. I would calculate both times and add them both. But I don't know how to do this.
The answer in my book is:
https://www5b.wolframalpha.com/Calculate/MSP/MSP40222424iaf5566d1e00004f5598ab3d54c171?MSPStoreType=image/gif&s=12 [Broken]
Any hints?
 
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Answers and Replies

  • #2
haruspex
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First step is to figure out the heading of the plane relative to the air. Do you know how to add vectors?
 
  • #3
First step is to figure out the heading of the plane relative to the air. Do you know how to add vectors?
Yes, but I've got lost in this example. I have a straight line from A to B, and wind blows in some other direction with an angle theta. So I have to find where the plane is headed? Let me guess: it would be the sum of the AB position vector and the wind position vector? Well, I don't know how to handle the angle between the vector to add them up in this case...
 
  • #4
haruspex
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Yes, but I've got lost in this example. I have a straight line from A to B, and wind blows in some other direction with an angle theta. So I have to find where the plane is headed? Let me guess: it would be the sum of the AB position vector and the wind position vector? Well, I don't know how to handle the angle between the vector to add them up in this case...
The actual direction, AB, is the desired result. It is the result of two input vectors: the wind direction and the plane's heading.
Consider the conponents of the two speeds in the AB direction and orthogonal to it. What two equations can you write down?
 
  • #5
The actual direction, AB, is the desired result. It is the result of two input vectors: the wind direction and the plane's heading.
Consider the conponents of the two speeds in the AB direction and orthogonal to it. What two equations can you write down?
Well, as the wind blows an angle theta from AB direction, I could say that the two componentes for the wind would be: vx = v*cos(theta) and vy = v*sin(theta).
But I don't have an angle for the plane in relation to the the AB direction. That's what I'm struggling with.
 
  • #6
haruspex
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Well, as the wind blows an angle theta from AB direction, I could say that the two componentes for the wind would be: vx = v*cos(theta) and vy = v*sin(theta).
But I don't have an angle for the plane in relation to the the AB direction. That's what I'm struggling with.
So create a variable for it.
 
  • #7
So create a variable for it.
Yes, I did it, but I didn't managed to get anywhere with it. I tried to substitute it to put it in function of theta.
 
  • #8
haruspex
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Yes, I did it, but I didn't managed to get anywhere with it. I tried to substitute it to put it in function of theta.
I cannot help you if you do not post your attempts.
 
  • #9
I cannot help you if you do not post your attempts.
Sorry. As I said, I've found vx = v*cos(theta) and vy = v*sin(theta). For the angle between the airplane's direction and AB, I used some alpha angle, so I've got: Vx = V*cos(alpha) and Vy = V*sin(alpha). Then I added them up: Vab(the direction desired) would be = (v*cos(theta) + V*cos(alpha), v*sin(theta)+V*sin(alpha)).
I've put the AB vector in the x-axis, so I imagined that the y direction of the Vab vector would be 0, so v*sin(theta)+V*sin(alpha) = 0. Now, Vab = (v*cos(theta) + V*cos(alpha), 0). The magnitude of this vector would be: v*cos(theta) + V*cos(alpha).
I don't know what to do now.
 
  • #10
haruspex
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Sorry. As I said, I've found vx = v*cos(theta) and vy = v*sin(theta). For the angle between the airplane's direction and AB, I used some alpha angle, so I've got: Vx = V*cos(alpha) and Vy = V*sin(alpha). Then I added them up: Vab(the direction desired) would be = (v*cos(theta) + V*cos(alpha), v*sin(theta)+V*sin(alpha)).
I've put the AB vector in the x-axis, so I imagined that the y direction of the Vab vector would be 0, so v*sin(theta)+V*sin(alpha) = 0. Now, Vab = (v*cos(theta) + V*cos(alpha), 0). The magnitude of this vector would be: v*cos(theta) + V*cos(alpha).
I don't know what to do now.
Good.
You have an equation that you can use to find an expression for the unknown angle, alpha. Use that to eliminate alpha from the Vab expression.
 
  • #11
Good.
You have an equation that you can use to find an expression for the unknown angle, alpha. Use that to eliminate alpha from the Vab expression.
Okay. I used v*sin(theta)+V*sin(alpha) = 0. So sin(alpha) = -(v/V)*cos(theta), and alpha = arcsin((-v/V)*sin(theta))).
I've put that into the expression for Vab: v*cos(theta) + V*cos(arcsin((-v/V)*sin(theta)))).
 
  • #12
haruspex
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Okay. I used v*sin(theta)+V*sin(alpha) = 0. So sin(alpha) = -(v/V)*cos(theta), and alpha = arcsin((-v/V)*sin(theta))).
I've put that into the expression for Vab: v*cos(theta) + V*cos(arcsin((-v/V)*sin(theta)))).
No, don't resort to inverse trig functions - that's nearly always a blind alley.
What is the algebraic relationship between an angle's sine and its cosine?
 
  • #13
No, don't resort to inverse trig functions - that's nearly always a blind alley.
What is the algebraic relationship between an angle's sine and its cosine?
Of course! V*sin(alpha) = -v*sin(theta). So, sin(alpha) = -(v*sin(theta))/V. Using sin^2(alpha) + cos^2(alpha) = 1, I get the cosine of alpha = sqrt(1-(v^2*sin^2*theta)/V^2). Then answer is starting to show up!
Let me guess: now I have to do the same analysis for the back trip?
 
  • #14
haruspex
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Of course! V*sin(alpha) = -v*sin(theta). So, sin(alpha) = -(v*sin(theta))/V. Using sin^2(alpha) + cos^2(alpha) = 1, I get the cosine of alpha = sqrt(1-(v^2*sin^2*theta)/V^2). Then answer is starting to show up!
Let me guess: now I have to do the same analysis for the back trip?
Yes.
 
  • #15
Yes.
Well, the wind direction doesn't change in the back trip, and the AB direction will be inverse - BA, let's say - so the airplane velocity vector will be the different. How do I handle this?
 
  • #16
haruspex
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Well, the wind direction doesn't change in the back trip, and the AB direction will be inverse - BA, let's say - so the airplane velocity vector will be the different. How do I handle this?
It's the same problem, but with a different wind angle, θ', say. What is the relationship between θ and θ'?
 
  • #17
It's the same problem, but with a different wind angle, θ', say. What is the relationship between θ and θ'?
θ+θ' = 180 degrees?
The question says that V is the velocity of the airplane in relation to the wind. So the plane must retain this velocity back and forth, what means that Vba(velocity from B to A) will be different. I'm trying to sketch the new problem.
 
  • #18
It's the same problem, but with a different wind angle, θ', say. What is the relationship between θ and θ'?
I'm not sure about how I would use the information: θ+θ' = 180 to calculate the components of the vectors...

EDIT: Okay, think I almost got it. Used some beta angle between the new direction of vector V (velocity of airplane in relation to the wind), and I've got that Vba = v*cos(theta) - V*sqrt(1-(v^2*sin^2(theta)/V^2). But I think that, somehow, I switched signs, because I can reach the answer if Vba = V*sqrt(1-(v^2*sin^2(theta)/V^2) -v*cos(theta). This negative sign appeared because I've assigned a negative direction for the vector V.
 
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  • #19
haruspex
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θ+θ' = 180 degrees?
Not quite. As you say, all that has happened is that we have rotated everything around 180 degrees. If you start facing one direction, rotate through an angle theta, then through 180 degrees (either way), what is your rotation compared to where you started?

Take the equation you had at post #13 and use it to write an expression for the time taken for the outward trip.
Then use the 180 rotation to get the time for the return trip.
 

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