Question

A bird watcher meanders through the woods, walking 0.983 km due east, 0.116 km due south, and 4.01 km in a direction 34.3 ° north of west. The time required for this trip is 1.825 h. Determine the magnitudes of the bird watcher's (a) displacement and (b) average velocity

Answer 1

A bird watcher meanders through the woods, walking $0.983\mathrm{km}$ due east, $0.116\mathrm{km}$ due south, and $4.01\mathrm{km}$ in a direction $34.3{}^{\circ}$ , north of west. The time required for this trip is 1.825 hour. Determine the magnitudes of the bird watcher's (a) displacement and (b) average velocity.

Solution.

(a) $S - ?$

The displacement of the bird watcher's by the diagram:

$$\vec {S} = \vec {S} _ {1} + \vec {S} _ {2} + \vec {S} _ {3}.$$

From the right triangle OAB with the right angle located at $A$ :

$$\vec {S} _ {1 2} = \vec {S} _ {1} + \vec {S} _ {2}.$$

By Pythagorean Theorem $S_{12}$ is:

$$S _ {1 2} = \sqrt {S _ {1} ^ {2} + S _ {2} ^ {2}}.$$ $$S _ {1 2} = \sqrt {(0 . 9 8 3 k m) ^ {2} + (0 . 1 1 6 k m) ^ {2}} = 0. 9 8 9 k m.$$

From the triangle OBC:

$$\vec {S} = \vec {S} _ {1 2} + \vec {S} _ {3}.$$

The magnitude of the bird watcher's displacement $S$ from triangle OBC by Law of cosines:

$$S ^ {2} = S _ {1 2} ^ {2} + S _ {3} ^ {2} - 2 S _ {1 2} S _ {3} \cos \varphi_ {3}.$$

By the diagram:

$$\varphi_ {3} = \varphi_ {1} - \varphi_ {2};$$ $$\varphi_ {1} = 34.3{}^{\circ}.$$

$\varphi_{2} = \varphi_{4}$ as the pairs of angles, when parallel lines get crossed by another line.

An angle $\varphi_4$ from the right triangle $OAB$.

$$\tan \varphi_4 = \frac {S _ {2}}{S _ {1}}.$$ $$\tan \varphi_4 = \frac {0.116\,km}{0.983\,km} = 0.118.$$ $$\varphi_ {4} = \arctan (0.118) = 6.73{}^{\circ}.$$ $$\varphi_ {2} = \varphi_ {4} = 6.73{}^{\circ}.$$ $$\varphi_ {3} = 34.3{}^{\circ} - 6.73{}^{\circ} = 27.57{}^{\circ}.$$

The magnitude of the bird watcher's displacement is:

$$S = \sqrt {S _ {12}^2 + S _ {3}^2 - 2 S _ {12} S _ {3} \cos \varphi_ {3}}.$$ $$S = \sqrt {(0.989\,km)^2 + (4.01\,km)^2 - 2 \cdot 0.989\,km \cdot 4.01\,km \cdot \cos (27.57{}^{\circ})} = 3.166\,km.$$

(b) $v - ?$

The magnitude of the bird watcher's average velocity is:

$$v = \frac {S}{t}.$$ $$v = \frac {3.166\,km}{1.825\,hour} = 1.734\, \frac{km}{hour}.$$

Answer:

(a) The magnitude of the bird watcher's displacement is $S = 3.166\,km$.

(b) The magnitude of the bird watcher's average velocity is $v = 1.734\, \frac{km}{hour}$.