Answer:
[tex]a=-0.001388\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of a train, u = 2.5 m/s
Finally, it reaches the station, v = 0 (at rest)
Time, t = 30 minutes = 1800 s
Acceleration is equal to the rate of change of velocity. So,
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{0-2.5}{1800}\\\\a=-0.001388\ m/s^2[/tex]
So, its acceleration is [tex]0.001388\ m/s^2[/tex].