Multiple step velocity/acceleration problem

[hannah2329]

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.

 

[Stonebridge]

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) ]

 

[hannah2329]

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.

 

[Stonebridge]

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.