math
posted by Anonymous on .
The diagram below shows the speed time graph for a train travelling between two stations. The train starts from rest and accelerates uniformly for 150 seconds. It then travels at a constant speed for 300 seconds and finally decelerates uniformly for 200 seconds.
Fig.
Given that the distance between the two stations is 10 450m, calculate the:
(a) maximum speed, in km/h, the train attained;
(b) acceleration;
(c) distance the train traveled during the last 100 seconds
(d) time the train takes to travel the first half of the journey

since the deceleration took 200s and the acceleration took only 150s, d = 3/4 a.
v after 150s = 150a
10450 = 1/2 a*150^2 + 300(150a) + 1/2 (3a/4)*200^2
(a) v = 38m/s
(b) a = 19/75 = .253m/s^2
(c) 950m
(d) 150+(52252850)/38 = 212.5s