Calculating the Speed on a Roller Coaster

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SUMMARY

The discussion focuses on calculating the speed of riders on a roller coaster after descending 75 meters of track. The initial calculations used the heights derived from the angles of descent, 17 degrees and 26 degrees, resulting in speeds of 16.6 m/s and 15.3 m/s respectively. However, the correct approach involves using the total change in height, which is 26 meters, leading to a final speed of approximately 22.8 m/s. This method leverages the principle of mechanical energy conservation, emphasizing that the specifics of the path do not affect the final speed as long as the height change is consistent.

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  • Understanding of basic physics concepts, particularly gravitational potential energy
  • Familiarity with trigonometric functions, specifically sine
  • Knowledge of the conservation of mechanical energy principle
  • Ability to perform calculations involving square roots and basic algebra
NEXT STEPS
  • Study the conservation of mechanical energy in physics
  • Learn about trigonometric functions and their applications in physics problems
  • Explore the effects of different angles on roller coaster design and speed
  • Practice solving problems involving gravitational potential energy and kinetic energy
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Students studying physics, particularly those interested in mechanics and energy conservation, as well as roller coaster designers and engineers looking to understand the dynamics of roller coaster speeds.

ilovemynny
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Homework Statement


While most people like high speeds on roller coasters, it is actually the change in acceleration that thrills them. To add more thrills to a ride, a roller coaster designer has decided to make the downward descent from his first hill bumpy. The first 49 m of the track will be at an angle of 17 degrees , while the next 28 m will be at a 26 degrees angle. (Both angles measured from the horizontal.) Assuming that the riders started with a speed of 0 m/s at the top of the hill, calculate their speed after they have ridden the first 75m of track?


Homework Equations


H x sin = y


The Attempt at a Solution


H x sin = y
Y = 49 x sin 17
Y = 14 m
V = 2 x g x h, square root
V = 2 x 9.81 x 14, square root
V = 274.68, square root
V = 16.6 m/s

H x sin = y
Y = 28 x sin 26
Y = 12 m
V = 2 x g x h, square root
V = 2 x 9.81 x 12, square root
V = 235.44, square root
V = 15.3 m/s

15.3m/s + 16.6m/s = 31.9m/s
Final Speed: 31.9m/s

I was wondering if I did this problem right, and if not how am I supposed to solve it?:confused:

p.s. when i write, square root that means there is a square root in the equation :-p
 
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You did not do it right. The expression

v=\sqrt{2gh}

works only if the initial speed vo is zero. So the first part of the calculation is correct but not the second. If I were you, I would use the overall change in height h in that expression and do it in one step. As long as the height change is the same, the specifics of the path do not matter. That's the beauty of mechanical energy conservation.
 
so then would it be like this:
75 x 2x 9.81, square root
1471.5, square root
and the final velocity is 38 m/s
 
ilovemynny said:
so then would it be like this:
75 x 2x 9.81, square root
1471.5, square root
and the final velocity is 38 m/s
Where did 75 come from in your expression?
 
it says calculate their speed after they have ridden the first 75m of track, and then you said it would be better to use the overall height
 
I said change in height, not distance traveled along the track. You need to calculate the overall change in height which you have already done. What do 14 m and 12 m represent?
 
ohhh okay 14 and 12 m represented the height
so then:
14 +12 = 26
so then 26 x 2 x 9.81 = 510
square root of 510 = 23m/s

so is 23 m/s the final velocity?
 
That looks about right, I got 22.8 m/s.
 
YES! thank you so much!
 

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