The driver of a train moving at 72 km h- 1 sights another train moving at 4 m s- 1 on the same track and in the same direction. He instantly applies brakes to produce a retardation of 1 m s- 2 . The minimum distance between the trains so that no collision occurs is

Question

The driver of a train moving at 72 km h- 1 sights another train moving at 4 m s- 1 on the same track and in the same direction. He instantly applies brakes to produce a retardation of 1 m s- 2 . The minimum distance between the trains so that no collision occurs is (A) 32 m (B) 64 m (C) 128 m (D) 256 m

Tardigrade Admin · Accepted Answer

Relative initial velocity =72× (5/18)-4=16 m s- 1 Now, 02-16× 16=2(- 1)S Or S=128 m

Q. The driver of a train moving at $72 km h^{- 1}$ sights another train moving at $4 m s^{- 1}$ on the same track and in the same direction. He instantly applies brakes to produce a retardation of $1 m s^{- 2}$ . The minimum distance between the trains so that no collision occurs is

32 m

64 m

128 m

256 m

Relative initial velocity $= 72 \times \frac{5}{18} - 4 = 16 m s^{- 1}$
Now, $0^{2} - 16 \times 16 = 2 (- 1) S$
Or $S = 128 m$

Q. The driver of a train moving at 72kmh−1 sights another train moving at 4ms−1 on the same track and in the same direction. He instantly applies brakes to produce a retardation of 1ms−2 . The minimum distance between the trains so that no collision occurs is