Vector Addition and Subtraction|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 .)

3.5

Solution:

Solve for the x and y components of the displacement.

The x-component of the displacement is:

\displaystyle R_x=A_x+B_x

\displaystyle R_x=A\:sin\left(20^{\circ} \right)+B\:cos\left(40^{\circ} \right)

\displaystyle R_x=-\left(12.0\:m\right)sin\left(20^{\circ} \right)-\left(20.0\:m\right)cos\left(40^{\circ} \right)

\displaystyle R_x=-19.425\:m

The y-component of the displacement is:

\displaystyle R_y=A_y+B_y

\displaystyle y=Acos20^{\circ} -Bsin40^{\circ}

\displaystyle R_y=\left(12.0\:m\right)cos\left(20^{\circ} \right)-\left(20.0\:m\right)sin\left(40^{\circ} \right)

\displaystyle R_y=-1.579

The resultant displacement is therefore,

\displaystyle R=\sqrt{R_x+R_y}

\displaystyle R=\sqrt{\left(-19.425\:m\right)^2+\left(-1.579\:m\right)^2}

\displaystyle R=19.5\:m

The compass direction is:

\displaystyle \theta =tan^{-1}\left|\frac{R_y}{R_x}\right|

\displaystyle \theta =tan^{-1}\left|\frac{1.579\:m}{19.425\:m}\right|

\displaystyle \theta =4.65^{\circ} \:S\:of\:W