- #1
andylatham82
- 11
- 4
New user has been reminded to post schoolwork in the Homework Help forums and fill out the HH Templatge showing their work
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:
I then calculated the resultant components of vector D:
Then the magnitude of vector D is:
And the direction is:
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
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
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
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