Determine stopping distance of a train traveling at 20m/s

[Alexanddros81]

This is problem 12.81 from Pytels Dynamics 2nd edition

1. Homework Statement
A train traveling at 20m/s is brought to an emergency stop. During braking,
the acceleration is a=-(7/4)+(t/16) m/s^2, where t is the time in seconds measured
from when the brakes were applied. (a) Integrate the acceleration from t=0 to
t=16s using Euler's method with Δt=2s. (b) Use the results of the integration to
determine the stopping distance of the train and compare you answer with 138.7m,
the value found analytically.

Homework Equations

The Attempt at a Solution

[/B]
According to the results of the Euler method above the stopping distance is between 150.5-152 m
That is way above 138.7m. Am I correct?

 

[TSny]

Looks good to me.

 

[Alexanddros81]

Is the stopping distance between x₇ and x₈ where the velocity is 0?

Do I use linear interpolation to find the stopping distance?
for example:

Do I use something like this?

If yes how do I apply linear interpolation in above graph to find x?

 

[BvU]

You could do that, but as you see there is little gain in accuracy (*)
You would use something like $${x-150\over v_7} = {x-152\over v_8} \Rightarrow x = \left ( {150.5\over v_7}-{152\over v_8} \right )/ \left ({1\over v_7}-{1\over v
_8}\right ) $$ or 151.14.

(*) the error the Euler integration makes in the first step is already about 3.5 m
( $x = v_0 t -{1\over 2} a_0 t^2 \$ gives 36.5 m instead of 40 m, the exact formula $v_0 t - {1\over 2} {7\over 4} t^2 + {1\over 2}{1\over 3} {\displaystyle t^3\over 16}$ gives 36.6 m )