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

The correct option is A

$29.5m{s}^{-1}$

Step 1: Given Data

Train $A$ and train $B$ are in opposite directions.

Velocity of train $A,v_A=36km/hr$

Velocity of train $B,v_B=-72km/hr$

A person is walking on train $A$ in the direction opposite to its motion.

The velocity of the person $v_{PA}=-1.8km/hr$

Let the velocity of the person observing from train $B$ be $v_{PB}$.

Step 2: Diagram

Step 3: Calculate the required velocity

Therefore,

The velocity of the person $v_P=$ velocity of the person with respect to $A,v_{PA}+$ velocity of $A,v_A$

Similarly,

The velocity of the person with respect to $B,v_{PB}$ $=$ The velocity of the person $v_P-$ velocity of $B,v_B$

$$\begin{gathered} \Rightarrow v_{PB}=v_{PA}+v_A-v_B \\ =-1.8+36-\left(-72\right) \\ =106.2km/hr \end{gathered}$$

Step 4: Convert Velocity into $m{s}^{-1}$

Now,

$$\begin{gathered} 1km/hr=\frac{5}{18}m/s \\ \therefore 106.2km/hr=\frac{5}{18}\times 106.2 \\ =29.5m/s \end{gathered}$$

Hence, the correct answer is option (A).