A train starts it 's journey from station P,accelerates at rate of 2 m/s² and reaches it 's maximum speed in 10 s .It maintains this speed for 30 min and retards uniformly to rest at the station Q after next 20 s .Calculate

1) the maximum speed of train ,

2) retardation

3) the distance between station Pand Q

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Solution

1.) u=0 , t = 10s, a= 2 m/s²
v=?
v = u + at
=> v = 0 + 210 = 20m/s
Maximum speed, v = 20m/s
2.) Here, initial velocity, u = 20m/s.
v=0,
t = 20s
a=?
v = u + at
=> 0 = 20 + a20
=> a = - 20/20
=> a = -1 m/s²
3.)
Case(i) s = ut + 0.5 at²
=> s = 010 + 0.521010
=> s = 100m
Case(ii)
s = vt
v = 20m/s
t = 30min = 30 60s = 1800s
s = 201800 = 36000m
Case(iii) s = ut + 0.5 at²
=> s = 2020 + 0.5(-1)20*20
=> s = 400 - 200 = 200m
So, distance between P and Q = 100+36000+200 = 36300m = 36.3km

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