Posted by parul on Monday, May 28, 2012 at 3:02am.
a train starts from rest and accelerates uniformly to achieve a velocity of 20m/sec in 10 sec.then train maintains this speed for next 200 sec.the breaks are then applied and train comes to rest in next 50 sec.calculate
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 8:12pm
a. (VVo)/t = (200)/10 = 2 m/s^2.
b. a=(VVo)/t = (020)/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.