Respuesta :
Answer:[tex]\theta =64.58^{\circ}[/tex]
Explanation:
Given
First you travel 4 mi North
position vector [tex]\vec{r_1}[/tex] for this displacement is given by
[tex]\vec{r_1}=4\hat{j}[/tex]
Now you travel 5.31 mi [tex]35^{\circ} east of North; Position vector [tex]\vec{r_2}[/tex] w.r.t [tex]r_1[/tex] is given by
[tex]\vec{r_{21}}=5.31(\sin(35)\hat{i}+\cos(35)\hat{j})[/tex]
Now you travel 1.1 mi in a direction [tex]15^{\circ}[/tex] North of east; Position vector [tex]\vec{r_3}[/tex] w.r.t [tex]r_2[/tex] is given by
[tex]\vec{r_{32}}=1.10(\sin(15)\hat{j}+\cos(15)\hat{i})[/tex]
Thus Position Vector of [tex]r_3[/tex] w.r.t starting Point
[tex]\vec{r_3}=\vec{r_{32}}+\vec{r_{21}}+\vec{r_1}[/tex][tex]\vec{r_3}=4\hat{j}+5.31(\sin(35)\hat{i}+\cos(35)\hat{j})+1.10(\sin(15)\hat{j}+\cos(15)\hat{i})[/tex]
[tex]\vec{r_3}=\hat{i}\left [ 5.31\sin 35+1.1\cos15\right ]+\hat{j}\left [ 4+5.31\cos 35+1.1\sin15\right ][/tex]
[tex]\vec{r_3}=4.10\hat{i}+8.63\hat{j}[/tex]
For angle
[tex]\tan \theta =\frac{8.63}{4.10}[/tex]
[tex]\tan \theta =2.10[/tex]
[tex]\theta =64.58^{\circ}\ North\ of\ east[/tex]
The magnitude of the hiker's displacement is 6.4 mi in a direction of 8.3⁰ north of east.
A sketch of the hiker's displacement
The sketch of the hiker's displacement will form a triangle and the resultant displacement of the triangle will be opposite to the largest angle of the triangle.
The largest angle of the triangle = 90 + 65 + 15 = 170⁰
Magnitude of the resulatant displacement
Apply Parallellogram law of vector addition
R² = a² + b² - 2bc(cosθ)
R² = (5.31)² + (1.1)² - 2(5.31)(1.1)cos(170)
R² = 40.9
R = √40.9
R = 6.4 mi
Direction of the displacement
Apply sine rule
R/sin170 = 5.31/sinθ
6.4/sin(170) = 5.31/sinθ
36.856 = 5.31/sinθ
sinθ = 5.31/36.856
sinθ = 0.144
θ = sin⁻¹(0.144)
θ = 8.3⁰
Thus, the magnitude of the hiker's displacement is 6.4 mi in a direction of 8.3⁰ north of east.
Learn more about resultant displacement here: https://brainly.com/question/13309193
