# math

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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

• math -

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+(5225-2850)/38 = 212.5s

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