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

Motion in A Straight Line PYQ — JEE Main 2020 · 2 Sep · Shift 1 · Q45 with Solution

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Correct answer: B

Let the direction of motion of train A be positive.

Speed of train A: $36\ \text{km/h}$
Speed of train B: $72\ \text{km/h}$ in the opposite direction, so $v_B=-72\ \text{km/h}$

The person walks inside train A opposite to the motion of train A with speed $1.8\ \text{km/h}$ .

So, relative to train A, $v_{P/A}=-1.8\ \text{km/h}$

Hence velocity of person relative to ground is $v_P=v_A+v_{P/A}=36-1.8=34.2\ \text{km/h}$

Relative velocity of person with respect to train B: $v_{P/B}=v_P-v_B$ $v_{P/B}=34.2-(-72)=106.2\ \text{km/h}$

Using $1\ \text{km/h}=\frac{5}{18}\ \text{m/s}$

So, $106.2\times \frac{5}{18}=29.5\ \text{m/s}$

$29.5\ \text{m/s}$ corresponds to Option B.

Final Answer: $29.5\ \text{m/s}$ , i.e. Option B.