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Time to stop a train

  • Thread starter rogetz
  • Start date
  • #1
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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 dont 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)
 

Answers and Replies

  • #2
diazona
Homework Helper
2,175
6
There's another kinematic equation that might help you:
[tex]v^2 = u^2 + 2ad[/tex]

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

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