Expert: Sherry Wallin - 7/7/2009

Question
A train travels from A to B,a distance of 20000m,taking 1000s.The journey has 3 stages.In the first stage the train starts from rest at A & accelerates uniformly until its speed is Vms^-1.In the second stage the train travels at constant speed Vms^-1 for 600s.During the third stage of the journey the train decelerates uniformly,coming to rest at B.Given that the acceleration of the train during the first stage of the journey is 0.15ms^-2,find the distance travelled by the train during the third stage of the journey.Many thanks

Answer
Hi Warm~
    Here is my reasoning, see if this makes sense to you:
The distance traveled in the first leg of the trip has to equal the distance traveled in the last leg of the trip since it is given that the train 'decelerates uniformly'. This means that the acceleration and the deceleration are at the same rate and thus the distance getting up to speed and the distance to a stop are equal. Since the total transit time is 1000s and the 2nd leg ot the trip is constant for 600s it stands to reason that the first leg and third leg is equal to
1000s - 600s = 400s and now divide that by 2 so that you know the time for the 3rd leg of the trip. 400s/2 = 200s which is the time on the third leg. Finally use for the distance traveled  the formula
d = v_i*t + (1/2)at^2 where v_i = 0 so v_i*t = 0 and
(1/2)(.15)(200)^2 = 3000m

Math Prof