Problems on Trains

Speed = 50 km/h = 50 * 5/18 m/s = 125/9 m/s. Time = Distance / Speed = 150 / (125/9) = 150 * 9 / 125 = 6 * 9 / 5 = 54 / 5 = 10.8 s.

Relative speed = (L1+L2) / Time = (100+200)m / 120s = 300/120 = 2.5 m/s. 2.5 m/s = 2.5 * 18/5 = 9 km/hr. Since they are in the same direction, V_x - V_y = 9. 120 - V_y = 9, so V_y = 111 km/hr.

Relative speed = 210m / 6s = 35 m/s. 35 m/s = 35 * 18/5 = 126 kmph. Since they are moving in opposite directions, relative speed = speed of train + speed of man. 126 = speed of train + 9, so speed of train = 117 kmph.

Train X travels from Meerut to Delhi in 1 hour (speed = D/1). Train Y travels from Delhi to Meerut in 1.5 hours (speed = D/1.5). Relative speed is D + D/1.5 = 2.5D/1.5 = 5D/3. They meet at time t = Distance / Relative Speed = D / (5D/3) = 3/5 hours = 36 minutes after 4 PM.

Relative speed = 32 + 3 = 35 km/h = 35 * (5/18) m/s = 175/18 m/s. Length = 350 m. Time = 350 / (175/18) = 350 * 18 / 175 = 2 * 18 = 36 seconds.

The train speed is 78 km/hr, which equals 78 * (5/18) = 65/3 m/s. In 1 minute (60 seconds), the train covers a total distance of (65/3) * 60 = 1300 meters. Since the train is 800 meters long, the tunnel length is 1300 - 800 = 500 meters.

Train speed v (m/s). Man speed 6 km/hr = 5/3 m/s. Relative speed = v + 5/3. Distance = 150m. Time = 6s. 150 / (v + 5/3) = 6 implies v + 5/3 = 25, so v = 25 - 1.66 = 23.33 m/s. Convert to km/hr: 23.33 * 18/5 = 84 km/hr.

Let d be the total distance and v be the speed. The original time is d/v. The accident happens at distance v (1 hour). Delay is 0.5 hours. Remaining distance (d-v) is covered at 3/4v. Total time = 1 + 0.5 + (d-v)/(0.75v) = d/v + 3.5. This simplifies to (d-v)/(0.75v) - (d-v)/v = 3. Solving for d-v gives 9v. Using the second condition (accident at v+90), we find the distance.

Let speed be v km/hr. v km/hr = v * (1000/3600) m/sec = v * (5/18) m/sec. Time per rail = 30 / (v * 5/18) = 30 * 18 / 5v = 108/v seconds. If the number of clicks is N, then time = N * (108/v). The question implies N = v, so time = v * (108/v) = 108 seconds.

Let train length be L and speed be v. Crossing post: L/v = 1, so L = v. Crossing platform: (L + 300) / v = 3. Substitute L = v: (v + 300) / v = 3. v + 300 = 3v, 2v = 300, v = 150. Since L = v, L = 150.

Relative speed = 60 + 48 = 108 km/hr = 30 m/s. Let train lengths be L and L/2. Total distance = 1.5L. 1.5L = 30 * 15 = 450, so L = 300. Train 1 speed = 60 km/hr = 50/3 m/s. Passing bridge: (300 + Bridge) = (50/3) * 51 = 850. Bridge = 550 m.

Relative speed = Distance / Time = 300 m / 33 s = 100/11 m/s. Convert to km/hr: (100/11) * (18/5) = 360/11 = 32.7272... km/hr. Since the man is moving in the same direction, Train Speed - Man Speed = 32.7272. Train Speed = 32.7272 + 3 = 35.7272 km/hr, which is 35 and 8/11 km/hr.

Convert speed to m/s: 54 km/hr * (5/18) = 15 m/s. Time = distance / speed = 225 / 15 = 15 seconds.