Two trains are moving with velocities v1= 10 ms and v2 = 20 ms on the same track in
opposite directions. After the application of brakes if their retarding rates are a1= 2 ms and
a2= 1 ms respectively, then the minimum distance of separation between the trains to avoid
collision is
(a) 150 m
(b) 225 m
(c) 450 m
(d) 300 m
Answers
b) 225 m
Explanation:
For first train,
2aS1 = Vf^2 – Vi^2
As, the train will be at rest in ending so, final velocity will be zero.
2 (-2) S1 = - (10)^2
S1 = 25 m
For first train,
2aS2 = Vf^2 – Vi^2
As, the train will be at rest in ending so, final velocity will be zero.
2 (-1) S2 = - (20)^2
S1 = 200 m
Hence,
Minimum separation = S1 + S2
Minimum Separation = 25 + 200 = 225
Answer:
225m is the minimum distance
Explanation:
V1=10m/s
V2=20m/s
S1=? & S2=?
S= S1+S2(Total distance covered by both trains)
How much distance is covered by first train for using Retarding factor a1=2
2aS1 = Vf² – Vi²
As, the train will be at rest in ending so, final velocity will be zero.
2 (-2) S1 = - (10)²
S1 = 25 m.
After covering 25m distance vf becomes 0 and train stops
2 (-1) S2 = - (20)²
S1 = 200 m
Hence,
S = S1 + S2 = 25 + 200 = 225(Total distance both train will cover)