A train on a long straight stretch of tracks starts from test and speeds up steadily to 19 m/s in 6.0 minutes?

Question

at which point it switches to slowing down with a constant acceleration of magnitude 0.12 m/s for 2.0 minutes. What was the average speed of the train during this time, that is, from starting at rest to the end of the 2.0 minutes of slowing down?


hint: think of it as two separate problems, one for the train speeding up, and one for the train slowing down; then put the two together to answer the question.

Answer 1

If the train has a negative acceleration of 0.12m/s^2 for 2.0 minutes, it actually slows to a stop and starts back the other way.

Leg1:
v(avg = 19/2 = 9.5m/s
D = rt = 9.5(6min)(60sec/min) = 3420m

Leg2:
Part1:[slows to stop]
(9.5m/s)/(0.12m/s^2) = 79.1667sec
v(avg) = [v(zero) - v(final)]/2 = 9.5/2 = 4.75m/s
D = rt = (4,75)(79.1667) = 376.042m
Part2:[accelerating back the other way]
120 - 79.1667 = 40.8333sec
(40.8333)(0.12) = 4.90000m/s [back the other way]
v(avg) = (v(zero) + v(final))/2 = 4.9/2 = 2.45m/s
D = rt = (2.45)(4..8333) = 100.042m

Total distance = 3420 + 376.042 + 100.042 = 3896.08 m
Total time = 6min + 2min = 8min(60sec/min) = 480sec
Speed = Total Distance/Total time = 3896.08/480 = 8.11683m/s