A train starts from rest and accelerates uniformly, until it has traveled 3000 m and acquired a velocity of 24 m/s. The train then moves a constant velocity of 24 m/s for 430 s. The train then decelerates uniformly at 0.065 m/s², until it is brought to a halt.

Question

1. In Situation A, the acceleration during the first 3.0 km of travel is closest to
2. In Situation A, the distance traveled by the train during deceleration, in kilometers is closest to
3. In Situation A, the velocity of the train, when it has decelerated for 160 seconds, is closest to
4.In Situation A, the average velocity, during the first 8.0 km of travel, is closest to

Answer 1

Use your tried-and-true equations:

v = at
s = 1/2 at^2

#1.
3000 = 1/2 at^2
t = √(6000/a)

24 = a√(6000/a) = √6000a
a = 24^2/6000 = 0.096 m/s^2

and so on