3.3 Vector Addition and Subtraction: Analytical Methods

Repeat the exercise using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, $\textbf{B }+\textbf{ A }=\textbf{ A }+\textbf{ B}$.) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking your other path.

Do the exercise again using analytical techniques and change the second leg of the walk to $\text{25}.\text{0 m}$ straight south. (This is equivalent to subtracting $\mathbf{B}$ from $\mathbf{A}$ —that is, finding $\textbf{R} ' = \textbf{A }–\textbf{ B}$) (b) Repeat again, but now you first walk $\text{25} . \text{0 m}$ north and then $\text{18} . \text{0 m}$ east. (This is equivalent to subtract $\mathbf{A}$ from $\mathbf{B}$ —that is, to find $\mathbf{A} = \mathbf{B} + \mathbf{C}$. Is that consistent with your result?)

You fly $\text{32} . \text{0 km}$ in a straight line in still air in the direction $35.0º$ south of west. (a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.) (b) Find the distances you would have to fly first in a direction $45.0º$ south of west and then i…

In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: $2 . \text{50 km}$ $45.0º$ north of west; then $4 . \text{70 km}$ $60.0º$ south of east; then $1.30 \text{ km}$ $25.0º$ south of west; then $5 . \text{10 km}$ straight east; then $1.70 \text{ km}$ $5.00º$ east of north; then $7 . \text{20 km}$ $55.0º$ south of west; and finally $2 . \text{80 km}$ $10.0º$ north of east. What is his final position relative to the island?