Chapter 3, Problem 49AP

As she picks up her riders, a bus driver traverses four successive displacements represented by the expression

$\begin{array}{l}(-6.30\text{b})\hat{\text{i}}-(4.00\text{b}\text{cos}\text{40}°)\hat{\text{i}}-(4.00\text{b}\text{sin}\text{40}°)\hat{\text{j}}\\ +(3.00\text{b}\text{cos}\text{50}°\text{)}\hat{\text{i}}-(3.00\text{b}\text{sin}\text{50}°\text{)}\hat{\text{j}}-(5.00\text{b) }\hat{\text{j}}\end{array}$

Here b represents one city block, a convenient unit of distance of uniform size; $\hat{\text{i}}$ is east; and $\hat{\text{j}}$ is north. The displacements at 40° and 50° represent travel on roadways in the city that are at these angles to the main east–west and north–south streets. (a) Draw a map of the successive displacements. (b) What total distance did she travel? (c) Compute the magnitude and direction of her total displacement. The logical structure of this problem and of several problems in later chapters was suggested by Alan Van Heuvelen and David Maloney, American Journal of Physics 67(3) 252–256, March 1999.

We note that − $\hat{\text{i}}$ = west and − $\hat{\text{j}}$ = south. The given mathematical representation of the trip can be written as 6.30 b west + 4.00 b at 40° south of west +3.00 b at 50° south of east +5.00 b south.

1. (a) The figure on the right shows a map of the successive displacements that the bus undergoes.
2. (b) The total odometer distance is the sum of the magnitudes of the four displacements:

   $6.30+4.00\text{b}\text{+}\text{3}\text{.00}\text{b}\text{+}\text{5}\text{.00}\text{b}\text{=}\text{18}\text{.3}\text{b}$

3. (c) $\begin{array}{l}\stackrel{\to }{\text{R}}\text{=}\text{(}-\text{6}\text{.30}-3.06+1.93)\text{b}\hat{\text{i}}+(-2.57-2.30-5.00)\text{b}\hat{\text{j}}\\ =-7.44\text{b}\hat{\text{i}}-9.87\text{b}\hat{\text{j}}\\ =\sqrt{{(7.44\text{b)}}^2+{(9.87\text{b)}}^{\text{2}}}\text{at}{\text{tan}}^{-1}\left(\frac{9.87}{7.44}\right)\text{south}\text{of}\text{west}\\ \text{=}\text{12}\text{.4}\text{b}\text{at}\text{53}\text{.0}°\text{south}\text{of}\text{west}\\ \text{=}\text{12}\text{.4}\text{b at 233}°\text{counterclockwise}\text{from}\text{east}\end{array}$

ANS. FIG. P3.49