Recall that
[tex]{v_f}^2-{v_0}^2=2a\Delta x[/tex]
where
- [tex]v_f[/tex] is the final velocity of the train; 0 in this case, because the train eventually stops
- [tex]v_0[/tex] is the initial velocity of the train; 1.0 m/s in this case, because this initial velocity refers to velocity it starts with after one of the wheels gets stuck, and the train is initially traveling at a constant speed of 1.0 m/s
- [tex]a[/tex] is the acceleration we want to find
- [tex]\Delta x[/tex] is the change in the train's position; 5.0 m in this case
So we have
[tex]-\left(1.0\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2a(5.0\,\mathrm m)[/tex]
[tex]\implies a=-0.10\,\dfrac{\mathrm m}{\mathrm s^2}[/tex]
