Stopping a Train: Calculating Time from Velocity

  • Thread starter Thread starter rogetz
  • Start date Start date
  • Tags Tags
    Time Train
AI Thread Summary
To determine the time it takes for a train to stop from a given initial velocity using a distance versus velocity graph, the user has derived the equation s=2.9u^2 + 29u for stopping distance. They attempted to invert the graph to express velocity in terms of distance, resulting in u=-2x10^-5s^2 + 0.09s. The user is confused about calculating deceleration (a) and its relationship to time, as they know that acceleration is the change in velocity over time. They are seeking clarification on how to apply kinematic equations, specifically v^2 = u^2 + 2ad, to solve for time. The discussion emphasizes the need for a clearer understanding of the relationship between velocity, distance, and time in the context of train braking dynamics.
rogetz
Messages
1
Reaction score
0

Homework Statement


I have a distance versus velocity graph that gives me the braking distance for a train (for known gradients of track - but gradient is irrevelant for my problem). the graph has distance on the vertical axis and velocity on the horizontal axis. What i want to know is that for a given initial velocity, how long - in time - does it take the train to stop?


Homework Equations


the graph equation i have worked out to be s=2.9u^2 + 29u (where s=distance, u=initial velocity)
(i was given several data points i.e. a train with an intial velocity of x km/h would take y m to stop)


The Attempt at a Solution


i thought i would invert the graph so to make velocity on the vertical and distance on the horizontal. doing this u=-2x10^-5s^2 + 0.09s

using Newtons 2nd (?) law, v=u+at but since v=0 (i.e stopped), then u=-at. what i don't know is a (accelaration - in my case decelleration). but if change in velocity divided by change in time (dv/dt) is accleration and velocity is change in distance divided by change in time (ds/dt) then...

this is where i get lost

please help - it should be easy (i think but i hope not)
 
Physics news on Phys.org
There's another kinematic equation that might help you:
v^2 = u^2 + 2ad

P.S. v = u + at isn't one of Newton's laws, it's really just math.
 
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