Problems on Trains

Using v^2 = u^2 + 2as, let L be the length of the train. v1^2 = 0 + 2a(x) and v2^2 = 0 + 2a(x + L). The midpoint is at x + L/2. v_mid^2 = 2a(x + L/2) = 2ax + aL = v1^2 + (v2^2 - v1^2)/2 = (v1^2 + v2^2)/2.

Using the kinematic equation v^2 = u^2 + 2as, the velocity v_m at the midpoint s/2 is v_m^2 = u^2 + 2a(s/2) = u^2 + as. Since v^2 = u^2 + 2as, we have as = (v^2 - u^2)/2. Substituting this, v_m^2 = u^2 + (v^2 - u^2)/2 = (u^2 + v^2)/2. Thus, v_m = sqrt((u^2 + v^2)/2).

Let train length be L and speeds be v1, v2. (v1+v2) = 2L/3. If v1 becomes 1.5v1, (1.5v1+v2) = 2L/2.5 = 0.8L. Solving these equations for the same direction case (v1-v2) leads to 15 seconds.

For a train with constant acceleration a and length L, the velocity v at any point x is given by v^2 = u^2 + 2ax. At the front, x=0 and v=u. At the back, x=L and v=v. For the middle point, x=L/2. Thus, v_mid^2 = u^2 + 2a(L/2) = u^2 + aL. Since v^2 = u^2 + 2aL, we have aL = (v^2 - u^2)/2. Substituting this, v_mid^2 = u^2 + (v^2 - u^2)/2 = (u^2 + v^2)/2.

The train moves North at 10 m/s and the parrot moves South at 5 m/s. Their relative speed is 10 + 5 = 15 m/s. Time = Distance / Relative Speed = 150 / 15 = 10 seconds.

Relative speed = 10 + 15 = 25 m/s. Total distance to cover = 50 + 50 = 100 m. Time = Distance / Speed = 100 / 25 = 4 seconds.

For the train to be entirely in the tunnel, the tunnel must be at least as long as the train. However, the tunnel is 80 m and the train is 100 m. In classical physics, this is impossible. In special relativity, length contraction occurs: L = L0 * sqrt(1 - v^2/c^2). We need the contracted length of the train to be 80 m. 80 = 100 * sqrt(1 - v^2/c^2). 0.8 = sqrt(1 - v^2/c^2). 0.64 = 1 - v^2/c^2. v^2/c^2 = 0.36, so v = 0.6c.

When two trains travel in opposite directions, their relative speed is the sum of their individual speeds: 10 + 15 = 25 m/s. The total distance to cover is the sum of their lengths: 50 + 50 = 100 m. Time = Distance / Speed = 100 / 25 = 4 seconds.

Since the train and the parrot are moving in opposite directions, their relative speed is the sum of their individual speeds, which is 10 m/s + 5 m/s = 15 m/s. The distance to be covered to completely cross the train is equal to the length of the train, which is 150 m. Dividing the distance by the relative speed gives a time of 10 seconds.

This is a relativistic length contraction problem. The train (100m) must fit inside the tunnel (80m). Since the train is longer than the tunnel, it cannot fit in the tunnel in the rest frame. However, in a frame moving at speed v, the train's length contracts to L' = L * sqrt(1 - v^2/c^2). We need L' <= 80. 100 * sqrt(1 - v^2/c^2) <= 80 => sqrt(1 - v^2/c^2) <= 0.8 => 1 - v^2/c^2 <= 0.64 => v^2/c^2 >= 0.36 => v >= 0.6c.

The relative speed of the first train with respect to the second is 20 - 10 = 10 m/s. To pass the second train, the first must cover the sum of their lengths, which is 100 + 200 = 300 m. Time = Distance / Relative Speed = 300 / 10 = 30 seconds.

When the man runs in the same direction, the relative speed is 100/10 = 10 m/s, so his speed is 15 - 10 = 5 m/s. Running in the opposite direction gives a relative speed of 15 + 5 = 20 m/s, so the crossing time is 100/20 = 5 seconds.

Let train length be L and speed be V (m/s). (L + 96)/V = 12 and (L + 141)/V = 15. L + 96 = 12V and L + 141 = 15V. Subtracting gives 3V = 45, so V = 15 m/s. 15 m/s = 54 km/h. L = 12 * 15 - 96 = 180 - 96 = 84 meters.