A train which is moving with the uniform acceleration is observed to take 20 and 30 seconds to travel successive 0.4km. how much far then will it travel before coming to rest, if the acceleration remains uniform ?
Answers
Answer:
The train which is moving with uniform acceleration is observed to take 20 and 30 seconds to travel successive 0.4km.
Let - a denote the uniform negative acceleration (deceleration) with which the train is moving.
Let u1, u2 & u3 denote the velocities (in m/s) of the train at respectively beginning, middle & end of two successive (0.4 km = ) 400 m displacements.
So we get following kinematic relations,
u2 = u1 - 20*a ….. (1a)
u3 = u2 - 30*a ….. (1b)
400 = 20*u1 - a*400/2 = 20*u1 - 200*a ….. (1c)
400 = 30*u2 - a*900/2 = 30*u2 - 450*a
or 400 = 30*(u1 - 20*a) - 450*a [from (1a)]
or 400 = 30*u1 - 1050*a ….. (1d)
From (1c) & (1d) we get,
30*u1 - 1050*a = 20*u1 - 200*a
or 10*u1 = 850*a or u1 = 85*a ….. (2a)
From (1d) & (2a) we get,
30*85*a - 1050*a = 400 or 1500*a = 400 or a = 4/15 ….. (2b)
From (1a), (1b) & (2a), (2b) we get,
u2 = 85*a - 20*a = 65*a = 65*4/15 = 52/3
u3 = 65*a - 30*a = 35*a = 35*4/15 = 28/3 ….. (2c)
Let D & T denote the distance (in m) & the time (in s) respectively that the train will cover before coming to rest due to constant negative acceleration.
Hence we get following kinematic relations,
0 = u3 - a*T or T = u3/a = 35 (s) ….. (2d) [from (2b) & (2c)]
D = u3*T - (a/2)*T^2 = (28/3)*35 - (2/15)*35*35 [from (2b), (2c) & (2d)]
D = 35*(28/3 - 14/3) = 35*14/3 = 163.33 (m) [Ans]