PHYSICS

1. Concepts and Principles

The components of vector $\overrightarrow{A}$ are the projections of the vector in the and the y axes. If 8 is defined as the anglebetween vector $\overrightarrow{A}$ and the +x-axis, the components of vector $\overrightarrow{A}$ are

$A_x=A\cos \theta$ (1)

$A_y=A\sin \theta$ (2)

The magnitude of vector $\overrightarrow{A}$ is found by using Pythagorean theorem:

$A=\overrightarrow{A}=\sqrt{A_x^2+A_y^2}$

It follows that the tangent of the angle $\theta$ is found by dividing $A_y$ by $A_x$:

$\tan \theta =\frac{A_y}{A_x}\to \theta ={\tan }^{-1}\frac{A_y}{A_x}$

2. Given Data

Let the first displacement be $\overrightarrow{A}$, the second displacement be $\overrightarrow{B}$, the third displacement be $\overrightarrow{C}$, and the resultant displacement be $\overrightarrow{R}$. Let the x-axis be toward east and the y-axis be toward north and the starting point be at the origin of coordinates

$\overrightarrow{A}$: 20.8 m, in the +y-direction (due north)

$\overrightarrow{B}$: 38 m, in the +x-direction (due east)

$\overrightarrow{C}$: 18 m, ${33}^{\circ }$ west of south

3. Required Data

We are asked to find the magnitude and direction of the resultant displacement $\overrightarrow{R}$, which is the straight line from the starting point to the hidden coin, shown in the figure below.

4. Solution

The x component of the resultat displacement is equal to the sum of the x components of vectors $\overrightarrow{A}$, $\overrightarrow{B}$ and $\overrightarrow{C}$:

$R_x=A_x+B_x+C_x$

Where $A_x=0$ since the vector $\overrightarrow{A}$ is completely in the +y-direction (due north), $B_x=38$ m since vector $\overrightarrow{B}$ is completely in the +x-direction, and $C_x=-C\sin {33}^{\circ }:$

$R_x=0+38-C\sin {33}^{\circ }=38=38(-18)(\sin {33}^{\circ })=28.2$ m

The y component of the resultant displacement is equal to the sum of the y components of vectors  $\overrightarrow{A}$, $\overrightarrow{B}$ and $\overrightarrow{C}$:

$R_y=A_y+B_y+C_y$

Where $A_y=20.8$ m, $B_y=0$, and $C_y=-C\cos {33}^{\circ }$ is found from Equation (2), and$B_y=46$ km:

$R_y=20.8+0-C\cos {33}^{\circ }=20.8-(18)(\cos {33}^{\circ })=5.7$ m

The magnitude of the resultant displacement (how far the person should run) is found from Equation (3):

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(28.2{)}^2+(5.7{)}^2}=28.8$ m

The angle made by the resultant displacement with the +z-axis is found from Equation (4):

$\theta ={\tan }^{-1}\frac{R_y}{R_x}={\tan }^{-1}(\frac{5.7}{28.2})={11.4}^{\circ }$

ANSWER: $R=28.8$ m and $\theta ={11.4}^{\circ }$