Question

Two trains $\text{A}$ and $\text{B}$ moving with speed of $36\text{km/hr}$ and $72\text{km/hr}$ respectively in opposite direction. A man moving in train $\text{A}$ with speed of $1.8\text{km/hr}$ opposite to the direction of train $\text{A}$. Find velocity of man as seen from train $\text{B}$ (in $\text{m/s}$).

• A. $32\text{m/s}$
• B. $29.5\text{m/s}$
• C. $32.5\text{m/s}$
• D. $28\text{m/s}$

Answer 1

The correct option is B $29.5\text{m/s}$


Here, $V_A=36\text{km/hr}=36\times \frac{1000}{3600}\text{m/s}=10\text{m/s}$

$V_B=-72\text{km/hr}=-72\times \frac{1000}{3600}\text{m/s}=-20\text{m/s}$

$V_{\text{Man, A}}=-1.8\text{km/hr}=-0.5\text{m/s}$

Velocity of man with respect to train $B$,

$V_{\text{man, B}}=V_{\text{man, A}}+V_{\text{A B}}$

$=V_{\text{man, A}}+(V_A-V_B)$
$\text{= -0.5 + 10 - ( -20)}$
$\text{= -0.5 + 30}\text{= 29.5 m/s}$

Hence, $\left(B\right)$ is the correct answer.