Answer to Question #163089 in Physics for killua

Question

A hiker begins a trip by first walking 30 km southeast from her car. She stops and sets up her tent for the night. On the second day, she walks 40 km in a direction 70° north of east, at which point she discovers a forest ranger’s tower. (a) determine the components of the hiker’s displacement for each day. (b) determine the components of the hiker’s resultant displacement R for the trip (find an expression for R in terms of unit vector notation). (c) determine the magnitude and direction of the resultant.

Answer 1

a) According to the directions distribution below

the first displacement vector is given as follows:

$$\mathbf{d}_1 = 30(\cos45\degree, -\cos 45\degree) = (15,-15)$$

The second displacement is given as follows:

$$\mathbf{d}_2 = (40\cos70\degree, 40\sin70\degree)$$

b) The resultant displacement is:

$$\mathbf{R} = \mathbf{d}_1 + \mathbf{d}_2 = (15+40\cos 70\degree,-15+40\sin 70\degree)\\ \mathbf{R}\approx (28.68,22.59)$$

c) The magnitude is:

$$R = \sqrt{28.68^2+22.59^2} \approx 36.51m$$

The direction is:

$$\theta = \arctan\left( \dfrac{22.59}{28.68} \right) \approx 38\degree$$

which is $38\degree$ north of east.

Answer. a) $(15,-15)$, $(40\cos70\degree, 40\sin70\degree)$, b) $(28.68,22.59)$, c) $36.51$ m, $38\degree$ north of east.