# Vector Displacement| Vector Addition and Subtraction: Graphical Method| Two-Dimensional Kinematics| College Physics Problem 3.5

### Suppose you first walk 12.0 m in a direction 20º west of north and then 20.0 m in a direction 40.0º south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B , as in Figure 3.56, then this problem finds their sum R=A+B.)




SOLUTION:

First, we solve for the x and y components of vectors A and B.

For vector A, the components are:

$A_x=-\left(12.0\:m\right)sin\:\left(20^{\circ} \right)=-4.1042\:m$

$A_y=\left(12.0\:m\right)cos\left(20^{\circ} \right)=11.2763\:m$

For vector B, the components are:

$B_x=-\left(20\:m\right)cos\left(40^{\circ} \right)=-15.3209\:m$

$B_y=-\left(20\:m\right)sin\:\left(40^{\circ} \right)=-12.8557\:m$

Since we know the components, we can now solve for the resultant vector R by adding the components together.

The resultant of the x-components is:

$R_x=A_x+B_x=-4.1042+\left(-15.3209\right)=-19.4251\:m$

The resultant of the y-components is:

$R_y=A_y+B_y=11.2763+\left(-12.8557\right)=-1.5794\:m$

The magnitude of the resultant can be solved using the Pythagorean Theorem. That is:

$R=\sqrt{\left(R_x\right)^2+\left(R_y\right)^2}=\sqrt{\left(-19.4251\right)^2+\left(-1.5794\right)^2}=19.4892\:m$

The compass direction is

$\theta =tan^{-1}\left|\frac{R_y}{R_x}\right|=tan^{-1}\left(\frac{1.5794}{19.4251}\right)=4.6483^{\circ}$

The compass direction is $4.6483^{\circ} \:South\:of\:West$.