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October 1, 2014

October 1, 2014

Posted by **parul** on Monday, May 28, 2012 at 3:02am.

a)acceleration in first 10 sec.

b)acceleration in last 50 sec.

c)the total distance traveled by the train in whole journey

d)average velocity of the trip.

- physics -
**Henry**, Wednesday, May 30, 2012 at 8:12pma. (V-Vo)/t = (20-0)/10 = 2 m/s^2.

b. a=(V-Vo)/t = (0-20)/50 = -0.4 m/s^2.

c. d1 = 0.5a*t^2 = 1*10^2 = 100 m.

d2 = 20m/s * 200s = 4000 m.

d3 = Vo*t + 0.5a*t^2

d3 = 20*50 + (-0.2)*50^2 = 500 m.

d1+d2+d3 = 100 + 4000 + 500 = 4600 m. = Total dist. traveled.

d. T = t1 + t2 + t3 = 10 + 200 + 50 = 260 s.

Va = (d1+d2+d3)/T = 4600m / 260 = 17.7

m/s. = Average velocity.

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