Stopping a Train: Calculating Time from Velocity

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SUMMARY

The discussion focuses on calculating the time it takes for a train to stop based on its initial velocity using a distance versus velocity graph. The equation derived from the graph is s = 2.9u² + 29u, where 's' represents distance and 'u' is the initial velocity. The user attempts to invert the graph to express velocity as a function of distance, resulting in u = -2x10^-5s² + 0.09s. The user seeks clarification on determining acceleration (deceleration) to apply kinematic equations effectively.

PREREQUISITES
  • Understanding of kinematic equations, specifically v² = u² + 2ad
  • Familiarity with graph interpretation, particularly distance versus velocity graphs
  • Knowledge of Newton's laws of motion, especially the concept of acceleration
  • Basic algebra skills for manipulating equations
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  • Study the derivation and application of kinematic equations in physics
  • Learn how to interpret and manipulate distance versus velocity graphs
  • Research methods for calculating acceleration from velocity and time
  • Explore practical examples of braking distance calculations for trains
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Students studying physics, particularly those focusing on kinematics, as well as engineers and professionals involved in railway safety and design.

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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 don't know is a (accelaration - in my case decelleration). but if change in velocity divided by change in time (dv/dt) is acceleration 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)
 
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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.
 

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