Multiple step velocity/acceleration problem

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The problem involves calculating the acceleration of a train moving from 26.1 m/s to 15.9 m/s over 12.7 seconds, resulting in an acceleration of -0.803 m/s². To determine the additional time needed to stop the train at this acceleration, it is calculated to be 19.8 seconds. The next step is to find the total distance required to bring the train to a complete stop, which involves using the average velocity for both segments of motion. The average velocity is calculated as the mean of the initial and final speeds for each segment, and the total distance is the sum of distances for both accelerations. Understanding the relationship between distance, average velocity, and time is crucial for solving this problem.
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Homework Statement


a train is moving at 26.1m/s. 12.7 sec later its speed is 15.9m/s. find the trains acceleration- -.803m/s^2

what additional time would be necessary to bring the train to a stop if it continues to accelerate at the same rate calculated?
-.803=0-15.9/t which is 19.8 seconds

this is the part i can't figure out. i only have two tries left.

find the total distance necessary to bring the train to a complete stop from the beginning initial velocity.




Homework Equations





The Attempt at a Solution


i tried setting it up 26.1=d/32.5 but that wasnt right and now i don't know what else to do.
 
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You need to add the two values of distance for the two separate accelerations.
Does the fact that distance traveled = average velocity x time help?
For both accelerations you have the initial and final velocity as well as the time.
[average velocity = 0.5 x (initial + final) ]
 
I understand that I'm just not sure which velocity I'm supposed to use on each step that's where I'm stuck because everytime I tried it I got the answer wrong.
 
a train is moving at 26.1m/s. 12.7 sec later its speed is 15.9m/s. find the trains acceleration- -.803m/s^2
1st step
velocity goes from 26.1 to 15.9m/s. What is the average?
Distance traveled = average velocity x time

what additional time would be necessary to bring the train to a stop if it continues to accelerate at the same rate calculated?
-.803=0-15.9/t which is 19.8 seconds
2nd step
Velocity goes from 15.9m/s (from part 1) to zero
What is the average velocity?
Distance is average velocity x time.
 

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