A roller coaster moves 200 ft horizontally and the rises 135 ft at an

veksetz

veksetz

Answered question

2021-12-30

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

Answer & Explanation

Pademagk71

Pademagk71

Beginner2021-12-31Added 34 answers

The x component of the total displacement vector is,
Substituting the values, to find vector rx.
rx=r1i^+r2cosθi^+r3cosθi^
rx=(200 ft+(135 ft)cos(30)i^+(135 ft)cos(40)i^
=(420.33 ft)i^
The y component of the total displacement vector is,
Substituting the values, to find vector ry.
ry=(0 ft)j^+(135 ft)sin(30)j^(135 ft)sin(40)j
=(19.28 ft)j^
The total displacement vector of the roller coaster is.
Substituting the values, to find vector r.
r=rx+ry
r=(420.33 ft)i^(19.28 ft)j^
The magnitude of the displacement of the roller coaster is,
r=(420.33ft)2+(19.28ft)2
=420.8 ft

Nick Camelot

Nick Camelot

Skilled2023-05-19Added 164 answers

Given:
Horizontal displacement of 200 ft: x=200ft
First segment:
Vertical displacement of 135 ft at an angle of 30 degrees above the horizontal: y1=135ftsin(30)
Second segment:
Vertical displacement of 135 ft at an angle of 40 degrees below the horizontal: y2=135ftsin(40)
Now we can calculate the total vertical displacement y as the sum of y1 and y2:
y=y1+y2
Using the values above, we can substitute in the expressions for y1 and y2:
y=135ftsin(30)135ftsin(40)
To find the displacement from the starting point, we can use the Pythagorean theorem:
D=x2+y2
Substituting the given values:
D=(200ft)2+(135ftsin(30)135ftsin(40))2
Simplifying the expression:
D=40000ft2+(135ftsin(30)135ftsin(40))2
Evaluating the trigonometric functions:
D=40000ft2+(135ft·12135ft·sin(40))2
Simplifying further:
D=40000ft2+(67.5ft135ft·sin(40))2
Calculating the value inside the square root:
D=40000ft2+(67.5ft135ft·0.6428)2
Evaluating the expression inside the square root:
D=40000ft2+(67.5ft86.4975ft)2
Simplifying further:
D=40000ft2+(18.9975ft)2
Calculating the squared term:
D=40000ft2+360.9395ft2
Simplifying the expression inside the square root:
D=40360.9395ft2
Taking the square root:
D=200.901ft
Therefore, the roller coaster's displacement from its starting point at the end of its movement is approximately 200.901 ft.
Mr Solver

Mr Solver

Skilled2023-05-19Added 147 answers

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=200ft+135ftcos(40)
The vertical displacement is given by the sum of the vertical components of the two movements:
y=135ftsin(30)135ftsin(40)
To find the roller coaster's displacement from its starting point at the end of the movement, we can use the Pythagorean theorem:
D=x2+y2
Substituting the values, we have:
D=(200ft+135ftcos(40))2+(135ftsin(30)135ftsin(40))2
Evaluating this expression, we get:
D420.8ft
Therefore, the roller coaster's displacement from its starting point at the end of this movement is approximately 420.8 ft.
madeleinejames20

madeleinejames20

Skilled2023-05-19Added 165 answers

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 dx and the vertical distance as dy. The given information can be summarized as follows:
dx=200ft (horizontal distance)
dy1=135ft (vertical distance, first part)
θ1=30 (angle above the horizontal, first part)
dy2=135ft (vertical distance, second part)
θ2=40 (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:
Δx=dx
Δy=dy1+dy2
We can calculate the horizontal and vertical displacements separately:
Δx=200ft
Δy=135ft·sin(30)+135ft·sin(40)
Step 3:
Now let's calculate the values:
Δx=200ft
Δy=135ft·sin(30)+135ft·sin(40)
Calculating the values:
Δx=200ft
Δy=135ft·sin(30)+135ft·sin(40)68.91ft
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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