# Find the magnitude and direction of the net displacement of the player

1. Sep 5, 2011

### matt@USA

1. The problem statement, all variables and given/known data

A basketball player runs down the court, following the path indicated by the vectors A,B, and C in the figure. The magnitudes of these three vectors are 10, 20 and 7 m . Assume the positive x axis is directed to the right.

There is a 45degree angle outside of where A and B meet, and there is a 30degree angle outside of where B and C meet.

I have found what Dx and Dy are. Dx and Dy are 20m, and 0m. In order to find the degree of the equation you would have to take the tan^-1(Dy/Dx). My question is, how do you solve this when Dy = 0? The answer is not 0 degrees, I know this for a fact. Thanks for all the help!

2. Sep 5, 2011

### Rayquesto

if dy is zero, then the resultant angle should also be zero.

You must have added the components incorrectly. It's hard for me to imagine your details. Basically, I think youre trying to say that the angle vector A makes with the horizontal is 45 degrees, the angle vector B makes with the horizontal is 30 degrees, but what about the angle that vector c makes with the horizontal? whats that?

3. Sep 5, 2011

### matt@USA

No, I know my answers are correct because I had to solve for the absolute value of vectors A, B, and C, and they were correct. Imagine vector A going south. Then vector B heads off to the North East. Then vector C heads to the south east.

4. Sep 5, 2011

### Rayquesto

so dy= -10 + .707(20) + -.5(7)=.64
and dx=.707(20) + (sqrt3)/2 times 7=20.2

resultant theta=tan-1(.64/20.2)= 1.81degrees.

the only thing I'm scepticle about is the who idea of the 30 degrees south of east you gave. It could actually mean 60 degrees above horizontal, but I can't see the problem in your book or the source you got it from.