Newbie physics help? im suffering from not smart syndrome

[newphysicist]

Hi, I am new at physics, and i can't seem to get the right answer for these questions. Quick help would be appreciated.

q1: A hockey player shoots a 200 gram puck at 25m/s. the ice has a friction of 0.09 kinetic.
how far will the puck slide before it comes to a rest?

q2: A train has 3 cars, mc, mb, and ma. The acceleration of the train is 1.54 m/s^2. ma has a mass for 10kg, mb a mass of 35kg, mc has a mass of 20kg.The force of the train is 100 N forward.
Note: No friction in part a and b!
<mc-mb-ma> <-train going east
a) Calculate the tension force between ma and mb as well as the tension force between mb and mc.
b)Also calculate the braking force of the train if it is moving at 20 m/s when the brakes are suddenly applied. if it takes 23m to stop, calculate the braking force.
c) (with friction) If the coefficent of kinetic friction is equal to 0.10, calculate the engine force required to keep the train moving at a constant velocity. there are only frictional forces on car b and c.
d)Calculate the engine force that is required to accelerate the train at a rate of 1.0m/s^2[E]

Attempt at a solution:
For q2 I'm not too sure about. I did 45(35+10) * 1.54 for ma-mb, and the same format for the next part. both answers, however, are wrong. can someone please help me through the steps?
For q1, i figured out the normal force was 1962 (200g * 9.81). The Force of friction would be 176.58 (1962 x 0.09) -25 since Fn and Fg cancel out. my acceleration from this was 151.58/200=0.7579
i put it into the kinematic eq d=vf^2 - vi^2 / 2a and got 236.84m, or 200m with significant digits. The correct answer is 354m. I am not sure how they got that answer, please explain?

 

[AvroArrow]

For q2a, the tension force between ma and mb is 45(35+10)*1.54=822.3N, and the tension force between mb and mc is 45(20+35)*1.54=1133.5N. For q2b, the braking force of the train is the sum of the tensions (822.3+1133.5)=1955.8N. For q2c, the engine force required to keep the train moving at a constant velocity is the sum of the tensions plus the friction force on car b and c, which is 0.10*45(35+20)=81N. So, the engine force required is 1936.8N. For q2d, the engine force required to accelerate the train at 1.0m/s^2 is the sum of the tensions plus the acceleration force, which is 45(35+20)*1.0=1080N. So, the engine force required is 2113.5N.