What Is the Correct Calculation for the Running Dog Vector Problem?

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The discussion revolves around the correct calculation for the Running Dog Vector Problem, where a dog runs in specific directions and distances, and the goal is to determine how far and in what direction it must run to end up 10.0m south of its starting position. The initial calculations provided by the user yield a resultant vector magnitude of 30.9m and an angle of 28.2° east of south, which differ from the book's answers of 32.6m and 26.5° east of south. Participants suggest that the user may have misapplied the coordinate system in their calculations and emphasize the importance of drawing a diagram for clarity. Ultimately, it is revealed that the book's answer may be incorrect, as it appears to reflect a different final position than specified in the problem. The conversation highlights the complexities of vector calculations and the necessity of careful interpretation of problem statements.
andylatham82
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I've been racking my brain for a while now over what I feel should be a simple problem to solve, but my answer is wrong. I'm not wildly wrong, but wrong enough to think I've made a proper error and it's not just a rounding error or something. The question goes like this:

A dog in an open field runs 10.0m east and then 30.0m in a direction 55.0° west of north. In what direction and how far must the dog run to end up 10.0m south of her original position?
I decided to work this out by making the starting position of the dog coincide with the position the dog is aiming for. She'll follow vector A up to the original staring position, then travel east on vector B, followed by the north-westerly direction along vector C. Then I'm looking for the resultant vector D. I worked out the vector components as follows:

Ax = 0m
Ay = 10.0m
Bx = 10.0m
By = 0m
Cx = -30 sin 55.0° = -24.57m
Cy = -30 cos 55.0° = 17.21m​

I then calculated the resultant components of vector D:

Dx = Ax + Bx + Cx = 14.57m
Dy = Ay + By + Cy = -27.21m​

Then the magnitude of vector D is:

D = √(Dx2 + Dy2) = 30.9m​

And the direction is:

θ = tan-1(Dx / Dy) = 28.2° east of south.​

However the answers given to the problem are 32.6 m, 26.5° east of south. So you can see I'm not a million miles out, but I can't for the life of me spot my error! If anyone could put me out of my misery I'd be very grateful!

Thanks!
Andy
 
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[edit] matter of reading more carefully what you did. o:)
I don't understand how your Bx and By come about. You should use the same coordinate system as you used for A.

Goes to show making a drawing of the situation is a life saver. Even when you intend to do things in a different way, a check is always nice ! :smile: [edit] still good advice

[edit]me and my arrogant posts! The book answer is off and -- at least as stated -- you're not.
 
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BvU said:
[edit] matter of reading more carefully what you did. o:)
I don't understand how your Bx and By come about. You should use the same coordinate system as you used for A.

Goes to show making a drawing of the situation is a life saver. Even when you intend to do things in a different way, a check is always nice ! :smile: [edit] still good advice

[edit]me and my arrogant posts! The book answer is off and -- at least as stated -- you're not.

Ah! I'm correct? Thank heavens for that! I thought my brain had stopped working! Thanks for putting my mind at ease!

As for a diagram, I always draw them, I just didn't include my diagram here. I'd be lost without a diagram haha.
 
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I did it a slightly different way, but I get the same answer you do.

I assume the dog starts at (0,0). She goes to P1 = (10,0). (Units of course in meters.)

Then she goes to P2 by walking a distance of 30 m at an angle of 55 degrees west of North. So P2x = 10 - 30 (sin 55 deg) = -14.575 and P2y = 30(cos 55 deg) = 17.207.

So now the dog is at P2 = (-14.575, 17.207). The problem is how far does she walk and at what angle to get to P3 = (0, -10). Think of a right triangle formed by drawing a vertical line down from P2 to y = -10, and from there draw a horizontal line to P3. The vertical side has length 27.207 and the horizontal side has length 14.575. The hypotenuse is the final leg of the journey, from P2 to P3.

The distance traveled on the last leg is the hypotenuse = sqrt (27.207^2 + 14.575^2) = 30.865 m.

The angle between the vertical side and the hypotenuse = sin^-1(14.575/30.865) = 28.177 degrees.

Rounding gives your answer.

Does the book give a worked out solution, or just the answer?
 
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andylatham82 said:
However the answers given to the problem are 32.6 m, 26.5° east of south
The book's answer is consistent with the dog finishing 12 m south of starting position, rather than 10 m.
Can you check the goal description again?
 
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Is the book old. At one time, books assumed the student did the calculation with slide rules. I taught physics in the old days and I often saw different answers depending on the tools that did the calculation.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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