Physics, 16.07.2019 14:30 athenadailey8717

Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? 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?

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The representation of this problem is shown in Figure 1. So our goal is to find the vector $\overrightarrow{R}$ . From the figure we know that:

$\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}$

From geometry, we know that:

$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$

Then using vector decomposition into components:

$For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85$

Therefore:

$R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m$

So if you want to find out how far are you from your starting point you need to know the magnitude of the vector $\overrightarrow{R}$ , that is:

$\left | \overrightarrow{R} \right |=
\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}$

Finally, let's find the compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

$\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}$

Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0?

Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0?

Answer from: Cheyenne7327

Representation of the given information in vector form can be used to find

the distance and direction from the start to the final position.

The distance from the starting point is approximately 19.5 metersThe compass direction of the line connecting the starting point and final position is 4.65° South of West

Reasons:

Distance, w₁, walked in the direction 20° West of North = 12.0 m

Distance, w₂, walked in the direction 40.0° South of West = 20.0 m

Distance and direction from the starting location; Required

Solution:

The distances presented in vector form are;

$\overrightarrow{w_1}$ = -12.0×sin(20°)·i + 12.0×cos(20°)·j

$\overrightarrow{w_2}$ = -20.0×cos(40°)·i - 20.0×sin(40°)·j

Therefore, we have;

R = $\overrightarrow{w_1}$ + $\overrightarrow{w_2}$ = (-12×sin(20°) + -20×cos(40°))·i + (12×cos(20°) + - 20×sin(40°))·j

R = -19.43·i - 1.58·j

Therefore;

|R| = √((-19.43)² + (-1.58)²) ≈ 19.5

The distance from the starting point is |R| ≈ 19.5 m

The tangent of the angle θ formed by the line connecting your starting

point to your final position is given as follows;

$tan(\theta) = \dfrac{Opposite \ side \ length\ to \ \angle\theta}{Adjacent\ side \ length\ to \ \angle\theta}$

The opposite side length = The distance moved along the positive y-

direction = The j coefficient

The adjacent side length = The distance moved along the positive x-direction = The i coefficient

Therefore;

$tan(\theta) = \left(\dfrac{-1.58}{-19.43} \right)$

The direction is therefore;

$\theta = arctan \left(\dfrac{-1.58}{-19.43} \right) \approx 4.65^{\circ}$

Therefore, the compass direction of a line connecting the starting and final

position is 4.65° South of West.

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e/c, where e is its total energy

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i really dont know but i will try okay.

so i think u would just add all them up and get your answer..

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