Calculating Distance for a Train on a 1% Grade

[A_Sushi]

Can anyone help with the following question, please?
1.
A 60,000kg train is being pulled up a 1 percent grade (it rises 1m for each horizontal 100m). The friction force opposing the motion of the train is 4,000N. Initial speed of the train is 12 m/s. Through what horizontal distance will the train move before its speed is reduced to 9m/s?
3.
The answer suppose to be 191.2 m. I got 472.5 m as follows:
F_d=m*(v²-v²)/(2*distance)
4 000 = (60 000*(9²-12²)) / (2*distance) = 472.5 m

 

[MarkFL]

Use the kinematic equation:

$\Delta x=\dfrac{v_f^2-v_i^2}{2a}$

You are given the initial and final velocities...what is the acceleration? How does $\Delta x$ relate to the horizontal distance?

 

[fresh_42]

@A_Sushi Please do not delete our template, but use it instead. It is important for many reasons. Also we request you to show us some effort on your own, to see how we could help you best. Different people might have different difficulties. Thank you.

 

[A_Sushi]

Use the kinematic equation:

$\Delta x=\dfrac{v_f^2-v_i^2}{2a}$

You are given the initial and final velocities...what is the acceleration? How does $\Delta x$ relate to the horizontal distance?

The acceleration is not given. I do not know how to calculate the acceleration

 

[haruspex]

how to calculate the acceleration

Even though it says the train is being pulled up, I think you have to assume there is no engine attached, it is just coasting.
What are the forces on the train? What equation can you write relating those to acceleration?