1. Relative Speed (Same Direction):
\[
\text{Speed of Train} = 15 \, \text{m/s}
\]
\[
\text{Time to Overtake} = 10 \, \text{s}
\]
\[
\text{Distance} = \text{Length of Train} = 100 \, \text{m}
\]
\[
\text{Relative Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{100}{10} = 10 \, \text{m/s}
\]
\[
\text{Speed of Man} = 15 - 10 = 5 \, \text{m/s}
\]

2. Relative Speed (Opposite Direction):
\[
\text{Relative Speed} = 15 + 5 = 20 \, \text{m/s}
\]

3. Time to Cross Man (Opposite Direction):
\[
\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}} = \frac{100}{20} = 5 \, \text{s}
\]

Thus, the train will take 5 seconds to cross the man if he is running in the opposite direction.

When objects move in opposite directions relative velocity= v1+ v2= 20 m/s

When moving in same direction relative velocity = V ₁ -V ₂= 10-20 = -10 m/s

Velocity of A w.r.t B = V _A - V _B
Velocity of B w.r.t A = V _B - V _A

Relative Velocity = V _A - V _B
10= 40 - V _B
V _B = 30 m/s

* Is 50 m/s a possible answer ?

As, 54 km/hr = 15 m/s both have same velocity in same direction their relative speed/velocity is zero

As both A and B have same velocity their relative speed is zero. so both have same slope in position time graph.

Let the length of the escalator be \(L\). Speed of person is \(v_p = L/45\). Speed of escalator is \(v_e = L/60\). When both move, their effective speed is \(v_{eff} = v_p + v_e = L/45 + L/60 = (4L+3L)/180 = 7L/180\). The time taken \(T = L/v_{eff} = L / (7L/180) = 180/7 \approx 25.71\) s.

Position of B: \(y_B(t) = H - (1/2)gt^2\). Meeting at \(H/2\): \(H/2 = H - (1/2)gt^2 \Rightarrow (1/2)gt^2 = H/2 \Rightarrow t = \sqrt{H/g}\). Position of A: \(y_A(t) = v_0t - (1/2)gt^2\). At meeting: \(H/2 = v_0t - H/2 \Rightarrow H = v_0t\). Substitute \(t\): \(H = v_0 \sqrt{H/g}\). Squaring both sides gives \(H^2 = v_0^2 (H/g) \Rightarrow v_0^2 = gH\).

Let the velocities be \(v_1\) and \(v_2\). Distance to cross is \(100 + 100 = 200\text{ m}\). For same direction: \(v_1 - v_2 = 200/20 = 10\text{ m/s}\). For opposite direction: \(v_1 + v_2 = 200/10 = 20\text{ m/s}\). Solving these gives \(v_1 = 15\text{ m/s}\) and \(v_2 = 5\text{ m/s}\).