# College Physics 3.5 – Resultant of two vectors

#### 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 3.54, then this problem finds their sum R=A+B.) Figure 3.54. The two displacements A and B are added to give a total displacement of R with magnitude R and direction θ measured from the x-axis

## Solution:

To solve for the magnitude and direction of the resultant R, we can refer to the triangle formed by the tree vectors as shown in the figure below.

From the figure above, we are given the two sides A and B, and the included angle measured to be 70°. To solve for the magnitude of R, we can use the cosine law. $\text{R}^2=\text{A}^2+\text{B}^2-2\text{AB}\:\cos 70^{\circ}$ $\text{R}=\sqrt{\text{A}^2+\text{B}^2-2\text{AB}\:\text{cos}\:70^{\circ} }\:$ $\text{R}=\sqrt{\text{(12 m)}^2+\text{(20 m)}^2-2\text{(12 m)(20 m)}\:\text{cos}\:70^{\circ} }\:$ $\text{R}=19.4892\:\text{m}$

To solve for the direction of the resultant, we just need to solve for angle C, then subtract 70° to get the value of θ. We can solve for angle C using the sine law. $\displaystyle \frac{20\:\text{m}}{\sin \text{C}}=\frac{19.4892\:\text{m}}{\sin 70^{\circ} }$ $\displaystyle \sin \:\text{C}=\frac{20\:\text{m}\:\left(\sin \:\:70^{\circ \:}\right)}{19.4892\:\text{m}}$ $\displaystyle \text{C}=\text{sin}^{-1}\left(\frac{20\:\text{m}\:\left(\sin \:\:70^{\circ \:}\right)}{19.4892\:\text{m}}\right)$ $\text{C}=74.65^{\circ}$

Finally, to solve for θ, we subtract 70° from 74.65°. $\theta =74.65^{\circ} -70^{\circ} =4.65^{\circ}$

Therefore, the compass direction of the resultant displacement is 4.65° South of West.