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.

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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&#039;s displacement from its starting point at the end of this movement.

Pademagk71 · Accepted Answer

The x component of the total displacement vector is, Substituting the values, to find vector rx. rx→=r1i^+r2cos⁡θi^+r3cos⁡θi^ 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 · Accepted Answer

Given:Horizontal displacement of 200 ft: x=200ftFirst 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+y2Using 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+y2Substituting the given values:D=(200ft)2+(135ftsin(30∘)−135ftsin(40∘))2Simplifying the expression:D=40000ft2+(135ftsin(30∘)−135ftsin(40∘))2Evaluating the trigonometric functions:D=40000ft2+(135ft·12−135ft·sin(40∘))2Simplifying further:D=40000ft2+(67.5ft−135ft·sin(40∘))2Calculating the value inside the square root:D=40000ft2+(67.5ft−135ft·0.6428)2Evaluating the expression inside the square root:D=40000ft2+(67.5ft−86.4975ft)2Simplifying further:D=40000ft2+(−18.9975ft)2Calculating the squared term:D=40000ft2+360.9395ft2Simplifying the expression inside the square root:D=40360.9395ft2Taking the square root:D=200.901ftTherefore, the roller coaster&#039;s displacement from its starting point at the end of its movement is approximately 200.901 ft.

Mr Solver · Accepted Answer

Result:420.8 ftSolution:Let&#039;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&#039;s displacement from its starting point at the end of the movement, we can use the Pythagorean theorem:D=x2+y2Substituting the values, we have:D=(200ft+135ftcos(40∘))2+(135ftsin(30∘)−135ftsin(40∘))2Evaluating this expression, we get:D≈420.8ftTherefore, the roller coaster&#039;s displacement from its starting point at the end of this movement is approximately 420.8 ft.

madeleinejames20 · Accepted Answer

Step 1:To solve this problem using the graphical technique, we need to break down the motion into its horizontal and vertical components.Let&#039;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&#039;s displacement from its starting point at the end of this movement, we can use the following equations:Δx=dxΔy=dy1+dy2We can calculate the horizontal and vertical displacements separately:Δx=200ftΔy=135ft·sin(30∘)+135ft·sin(−40∘)Step 3:Now let&#039;s calculate the values:Δx=200ftΔy=135ft·sin(30∘)+135ft·sin(−40∘)Calculating the values:Δx=200ftΔy=135ft·sin(30∘)+135ft·sin(−40∘)≈68.91ftTherefore, the roller coaster&#039;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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