Train A and train B are running on parallel tracks in the opposite directions with speed of 36 km hour$$^{-1}$$ and 72 km hour$$^{-1}$$, respectively. A person is walking in train A in the direction opposite to its motion with a speed of 1.8 km hour$$^{-1}$$. Speed (in m s$$^{-1}$$) of this person as observed from train B will be close to: (take the distance between the tracks as negligible)

Let us choose the direction in which train A is moving as the positive direction.

Speed of train A with respect to the ground is given as $$+36\text{ km h}^{-1}.$$

The person is walking inside train A opposite to the motion of the train. Therefore, the speed of the person with respect to train A is

$$v_{\text{person, A}} = -1.8\text{ km h}^{-1}.$$

To find the speed of the person with respect to the ground, we use the relation

$$v_{\text{person, ground}} = v_{\text{train A, ground}} + v_{\text{person, A}}.$$

Substituting the known values, we obtain

$$v_{\text{person, ground}} = 36 + (-1.8) = 34.2\text{ km h}^{-1}.$$

Now, train B is moving in the opposite direction (negative direction) with speed

$$v_{\text{train B, ground}} = -72\text{ km h}^{-1}.$$

The relative speed of the person as observed from train B is given by the standard formula

$$v_{\text{person relative to B}} = v_{\text{person, ground}} - v_{\text{train B, ground}}.$$

Substituting the values, we have

$$v_{\text{person relative to B}} = 34.2 - (-72) = 34.2 + 72 = 106.2\text{ km h}^{-1}.$$

We must now convert this speed from kilometres per hour to metres per second using

$$1\text{ km h}^{-1} = \frac{5}{18}\text{ m s}^{-1}.$$

Hence,

$$v_{\text{person relative to B}} = 106.2 \times \frac{5}{18}\text{ m s}^{-1}.$$

First multiply:

$$106.2 \times 5 = 531.0.$$

Now divide by 18:

$$\frac{531.0}{18} = 29.5\text{ m s}^{-1}.$$

Therefore, the speed of the person as observed from train B is approximately $$29.5\text{ m s}^{-1}.$$

Hence, the correct answer is Option A.