veksetz

2021-12-30

A roller coaster moves 200 ft horizontally and the rises 135 ft at an angle of ${30}^{\circ }$ above the horizontal. Next, it travels 135 ft at an angle of ${40.0}^{\circ }$ below the horizontal. Use the graphical technique to find the roller cosater's displacement from its starting point at the end of this movement.

The x component of the total displacement vector is,
Substituting the values, to find vector ${r}_{x}$.
$\stackrel{\to }{{r}_{x}}={r}_{1}\stackrel{^}{i}+{r}_{2}\mathrm{cos}\theta \stackrel{^}{i}+{r}_{3}\mathrm{cos}\theta \stackrel{^}{i}$

The y component of the total displacement vector is,
Substituting the values, to find vector ${r}_{y}$.

The total displacement vector of the roller coaster is.
Substituting the values, to find vector r.
$\stackrel{\to }{r}=\stackrel{\to }{{r}_{x}}+\stackrel{\to }{{r}_{y}}$

The magnitude of the displacement of the roller coaster is,
$r=\sqrt{{\left(420.33ft\right)}^{2}+{\left(-19.28ft\right)}^{2}}$

Nick Camelot

Given:
Horizontal displacement of 200 ft: $x=200\phantom{\rule{0.167em}{0ex}}\text{ft}$
First segment:
Vertical displacement of 135 ft at an angle of 30 degrees above the horizontal: ${y}_{1}=135\phantom{\rule{0.167em}{0ex}}\text{ft}\phantom{\rule{0.167em}{0ex}}\mathrm{sin}\left({30}^{\circ }\right)$
Second segment:
Vertical displacement of 135 ft at an angle of 40 degrees below the horizontal: ${y}_{2}=-135\phantom{\rule{0.167em}{0ex}}\text{ft}\phantom{\rule{0.167em}{0ex}}\mathrm{sin}\left({40}^{\circ }\right)$
Now we can calculate the total vertical displacement $y$ as the sum of ${y}_{1}$ and ${y}_{2}$:
$y={y}_{1}+{y}_{2}$
Using the values above, we can substitute in the expressions for ${y}_{1}$ and ${y}_{2}$:
$y=135\phantom{\rule{0.167em}{0ex}}\text{ft}\phantom{\rule{0.167em}{0ex}}\mathrm{sin}\left({30}^{\circ }\right)-135\phantom{\rule{0.167em}{0ex}}\text{ft}\phantom{\rule{0.167em}{0ex}}\mathrm{sin}\left({40}^{\circ }\right)$
To find the displacement from the starting point, we can use the Pythagorean theorem:
$D=\sqrt{{x}^{2}+{y}^{2}}$
Substituting the given values:
$D=\sqrt{\left(200\phantom{\rule{0.167em}{0ex}}\text{ft}{\right)}^{2}+{\left(135\phantom{\rule{0.167em}{0ex}}\text{ft}\phantom{\rule{0.167em}{0ex}}\mathrm{sin}\left({30}^{\circ }\right)-135\phantom{\rule{0.167em}{0ex}}\text{ft}\phantom{\rule{0.167em}{0ex}}\mathrm{sin}\left({40}^{\circ }\right)\right)}^{2}}$
Simplifying the expression:
$D=\sqrt{40000\phantom{\rule{0.167em}{0ex}}{\text{ft}}^{2}+{\left(135\phantom{\rule{0.167em}{0ex}}\text{ft}\phantom{\rule{0.167em}{0ex}}\mathrm{sin}\left({30}^{\circ }\right)-135\phantom{\rule{0.167em}{0ex}}\text{ft}\phantom{\rule{0.167em}{0ex}}\mathrm{sin}\left({40}^{\circ }\right)\right)}^{2}}$
Evaluating the trigonometric functions:
$D=\sqrt{40000\phantom{\rule{0.167em}{0ex}}{\text{ft}}^{2}+{\left(135\phantom{\rule{0.167em}{0ex}}\text{ft}·\frac{1}{2}-135\phantom{\rule{0.167em}{0ex}}\text{ft}·\mathrm{sin}\left({40}^{\circ }\right)\right)}^{2}}$
Simplifying further:
$D=\sqrt{40000\phantom{\rule{0.167em}{0ex}}{\text{ft}}^{2}+{\left(67.5\phantom{\rule{0.167em}{0ex}}\text{ft}-135\phantom{\rule{0.167em}{0ex}}\text{ft}·\mathrm{sin}\left({40}^{\circ }\right)\right)}^{2}}$
Calculating the value inside the square root:
$D=\sqrt{40000\phantom{\rule{0.167em}{0ex}}{\text{ft}}^{2}+{\left(67.5\phantom{\rule{0.167em}{0ex}}\text{ft}-135\phantom{\rule{0.167em}{0ex}}\text{ft}·0.6428\right)}^{2}}$
Evaluating the expression inside the square root:
$D=\sqrt{40000\phantom{\rule{0.167em}{0ex}}{\text{ft}}^{2}+{\left(67.5\phantom{\rule{0.167em}{0ex}}\text{ft}-86.4975\phantom{\rule{0.167em}{0ex}}\text{ft}\right)}^{2}}$
Simplifying further:
$D=\sqrt{40000\phantom{\rule{0.167em}{0ex}}{\text{ft}}^{2}+\left(-18.9975\phantom{\rule{0.167em}{0ex}}\text{ft}{\right)}^{2}}$
Calculating the squared term:
$D=\sqrt{40000\phantom{\rule{0.167em}{0ex}}{\text{ft}}^{2}+360.9395\phantom{\rule{0.167em}{0ex}}{\text{ft}}^{2}}$
Simplifying the expression inside the square root:
$D=\sqrt{40360.9395\phantom{\rule{0.167em}{0ex}}{\text{ft}}^{2}}$
Taking the square root:
$D=200.901\phantom{\rule{0.167em}{0ex}}\text{ft}$
Therefore, the roller coaster's displacement from its starting point at the end of its movement is approximately 200.901 ft.

Mr Solver

Result:
$420.8$ ft
Solution:
Let's denote the horizontal displacement as $x$ and the vertical displacement as $y$.
The horizontal displacement is given by the sum of the horizontal components of the two movements:
$x=200\phantom{\rule{0.167em}{0ex}}\text{ft}+135\phantom{\rule{0.167em}{0ex}}\text{ft}\mathrm{cos}\left({40}^{\circ }\right)$
The vertical displacement is given by the sum of the vertical components of the two movements:
$y=135\phantom{\rule{0.167em}{0ex}}\text{ft}\mathrm{sin}\left({30}^{\circ }\right)-135\phantom{\rule{0.167em}{0ex}}\text{ft}\mathrm{sin}\left({40}^{\circ }\right)$
To find the roller coaster's displacement from its starting point at the end of the movement, we can use the Pythagorean theorem:
$D=\sqrt{{x}^{2}+{y}^{2}}$
Substituting the values, we have:
$D=\sqrt{{\left(200\phantom{\rule{0.167em}{0ex}}\text{ft}+135\phantom{\rule{0.167em}{0ex}}\text{ft}\mathrm{cos}\left({40}^{\circ }\right)\right)}^{2}+{\left(135\phantom{\rule{0.167em}{0ex}}\text{ft}\mathrm{sin}\left({30}^{\circ }\right)-135\phantom{\rule{0.167em}{0ex}}\text{ft}\mathrm{sin}\left({40}^{\circ }\right)\right)}^{2}}$
Evaluating this expression, we get:
$D\approx 420.8\phantom{\rule{0.167em}{0ex}}\text{ft}$
Therefore, the roller coaster's displacement from its starting point at the end of this movement is approximately $420.8$ ft.

Step 1:
To solve this problem using the graphical technique, we need to break down the motion into its horizontal and vertical components.
Let's denote the horizontal distance as ${d}_{x}$ and the vertical distance as ${d}_{y}$. The given information can be summarized as follows:
${d}_{x}=200\phantom{\rule{0.167em}{0ex}}\text{ft}$ (horizontal distance)
${d}_{y1}=135\phantom{\rule{0.167em}{0ex}}\text{ft}$ (vertical distance, first part)
${\theta }_{1}={30}^{\circ }$ (angle above the horizontal, first part)
${d}_{y2}=135\phantom{\rule{0.167em}{0ex}}\text{ft}$ (vertical distance, second part)
${\theta }_{2}={40}^{\circ }$ (angle below the horizontal, second part)
Step 2:
To find the roller coaster's displacement from its starting point at the end of this movement, we can use the following equations:
$\Delta x={d}_{x}$
$\Delta y={d}_{y1}+{d}_{y2}$
We can calculate the horizontal and vertical displacements separately:
$\Delta x=200\phantom{\rule{0.167em}{0ex}}\text{ft}$
$\Delta y=135\phantom{\rule{0.167em}{0ex}}\text{ft}·\mathrm{sin}\left({30}^{\circ }\right)+135\phantom{\rule{0.167em}{0ex}}\text{ft}·\mathrm{sin}\left(-{40}^{\circ }\right)$
Step 3:
Now let's calculate the values:
$\Delta x=200\phantom{\rule{0.167em}{0ex}}\text{ft}$
$\Delta y=135\phantom{\rule{0.167em}{0ex}}\text{ft}·\mathrm{sin}\left({30}^{\circ }\right)+135\phantom{\rule{0.167em}{0ex}}\text{ft}·\mathrm{sin}\left(-{40}^{\circ }\right)$
Calculating the values:
$\Delta x=200\phantom{\rule{0.167em}{0ex}}\text{ft}$
$\Delta y=135\phantom{\rule{0.167em}{0ex}}\text{ft}·\mathrm{sin}\left({30}^{\circ }\right)+135\phantom{\rule{0.167em}{0ex}}\text{ft}·\mathrm{sin}\left(-{40}^{\circ }\right)\approx 68.91\phantom{\rule{0.167em}{0ex}}\text{ft}$
Therefore, the roller coaster's displacement from its starting point at the end of this movement is approximately 68.91 ft horizontally and 68.91 ft vertically.

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