For the first 5.0 seconds of its motion, the train covers a distance of
[tex]\left(75\,\dfrac{\mathrm m}{\mathrm s}\right)(5.0\,\mathrm s)+\dfrac12\left(-2.5\,\dfrac{\mathrm m}{\mathrm s}\right)(5.0\,\mathrm s)^2=344\,\mathrm m[/tex]
After the first 5.0 seconds, the train's velocity is
[tex]75\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(5.0\,\mathrm s)=62.5\,\dfrac{\mathrm m}{\mathrm s}[/tex]
For the next 100.0 seconds, the trains covers a distance of
[tex]\left(62.5\,\dfrac{\mathrm m}{\mathrm s}\right)(100.0\,\mathrm s)=6250\,\mathrm m[/tex]
and thus a total distance of 6594 meters.
After the first 105.0 seconds, the train's velocity [tex]v[/tex] at time [tex]t[/tex] (where [tex]t=0[/tex] corresponds to the 105.0 second mark) is given by
[tex]v(t)=62.5\,\dfrac{\mathrm m}{\mathrm s}+\left(-5\,\dfrac{\mathrm m}{\mathrm s^2}\right)t[/tex]
and its position [tex]x[/tex] at time [tex]t[/tex] (again, [tex]t=0[/tex] corresponds to the 105.0 second mark) by
[tex]x(t)=\left(62.5\,\dfrac{\mathrm m}{\mathrm s}\right)t+\dfrac12\left(-5\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2[/tex]
The time it takes for the train to stop is such that [tex]v(t)=0[/tex], for which we have
[tex]0=62.5\,\dfrac{\mathrm m}{\mathrm s}+\left(-5\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\implies t=12.5\,\mathrm s[/tex]
Then the distance covered in the final slow-down phase of the train is
[tex]x(12.5\,\mathrm s)=391\,\mathrm m[/tex]
giving a total distance covered of about 6990 meters.
