ANSWER

Step-by-step explanation
Parameters given:
Initial velocity, u = 19 m/s
Acceleration, a = -0.8m/s² (the train is slowing down)
Distance traveled, s = 175 m
To find the final velocity of the train, we have to apply one of Newton's equations of motion:

where v = final velocity
Substituting the given values into the equation, the final velocity of the train as its nose leaves the station is:
![\begin{gathered} v^2=19^2+(2\cdot-0.8\cdot175) \\ v^2=361-280 \\ v^2=81 \\ \Rightarrow v=\sqrt[]{81} \\ v=9m\/s \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/zi3rzvfleikmu6cwfktu8ft1s31n7aibz9.png)
That is the answer.