Running dog vector problem

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1. Apr 4, 2017

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

2. Apr 4, 2017

BvU

 matter of reading more carefully what you did.
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 !  still good advice

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

Last edited: Apr 4, 2017
3. Apr 4, 2017

andylatham82

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.

4. Apr 4, 2017

Aufbauwerk 2045

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.

Does the book give a worked out solution, or just the answer?

Last edited: Apr 4, 2017
5. Apr 4, 2017

SlowThinker

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?

6. Apr 5, 2017

mpresic

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.