Friday

July 25, 2014

July 25, 2014

Posted by **Anonymous** on Saturday, June 2, 2012 at 5:10pm.

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 -
**Steve**, Saturday, June 2, 2012 at 5:50pmsince 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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