Amusment park yesterday for physics class

  • Thread starter Thread starter Format
  • Start date Start date
  • Tags Tags
    Class Physics
AI Thread Summary
To calculate the acceleration of the "hellevator" ride during the first 5 meters, one must use the equation a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity (0 at the start), and t is the time taken to ascend. The final velocity can be determined using the free fall equation vf = vi + at, considering the acceleration due to gravity (9.8 m/s²). The height can be expressed as h = (1/2)at², allowing for the calculation of acceleration if the time to reach 5 meters is known. After the initial ascent, the ride's motion is influenced by gravity, requiring adjustments to the equations for maximum height calculations. Understanding these physics principles enhances the educational experience of amusement park rides.
Format
Messages
87
Reaction score
0
When to an amusment park yesterday for physics class. Went on a ride called the "hellevator" where you sit down and it shoots up a certain distance then let's you free fall. My question is if the acceleration upwards occurs in the first 5 meters how do you figure out that acceleration?
 
Physics news on Phys.org
You are going to need more information to find the acceleration.

Given an acceleration of a m/s2, the speed after t seconds is v= at and the height is h= (1/2)at2.

If you know the time, T, taken to get to that first 5 meters, you could solve
5= (1/2)a(T2): a= 10/T2.

After that first time, T, you continue upwards under the influence of grativity:
you speed at time t will be v= -g(t-T)+ aT and height will be h= 5+aTt- (1/2)g(t-T)2.

At the maximum height, v= -g(t-T)+ aT= 0 so you can solve for t (in terms of T). If you know that maximum height (usually easier to measure than time), you could solve the second equation for a.
 


To calculate the acceleration in the first 5 meters of the "hellevator" ride, you would need to use the equation a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time interval. In this case, the initial velocity is 0 because the ride starts from rest, and the final velocity can be calculated using the free fall equation vf = vi + at, where a is the acceleration due to gravity (9.8 m/s^2) and t is the time it takes for the ride to reach the top. Once you have the final velocity, you can plug it into the first equation to solve for the acceleration. Keep in mind that the acceleration will be negative since the ride is moving upwards, while gravity is pulling downwards. It is also important to note that this calculation assumes that the ride is moving at a constant acceleration, which may not be the case in reality. Other factors such as air resistance and mechanical limitations of the ride may affect the acceleration. Overall, understanding the physics behind amusement park rides can add a fun and educational element to the experience.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top