Question

Train $A$ travelling at $60\text{km/h}$ overtakes another train $B$ travelling at $42\text{km/h}$. Assuming each train to be $50\text{m}$ long and initially train $A$ to be just behind the train $B$, the total road distance used for the overtake is closest to

• A. $180\text{m}$
• B. $383\text{m}$
• C. $343\text{m}$
• D. $250\text{m}$

Answer 1

The correct option is B $383\text{m}$

Velocity of train $A$ $=60\text{km/h}=60\times \frac{5}{18}{\text{ms}}^{-1}=\frac{50}{3}{\text{ms}}^{-1}$
Velocity of train $B$ $=42\text{km/h}=42\times \frac{5}{18}{\text{ms}}^{-1}=\frac{35}{3}{\text{ms}}^{-1}$
Relative velocity of train $A$ w.r.t $B$ $=\frac{50}{3}-\frac{35}{3}=5{\text{ms}}^{-1}$
Train $A$ overtakes train $B$ when distance covered relatively by train A is $100\text{m}$ i.e. the sum of the length of the trains,
Using $S_{rel}=u_{rel}t+\frac{1}{2}{a}_{rel}{t}^2$
$\Rightarrow 100=5t+0$
$\therefore t=20\text{s}$
So, the road distance required to overtake would be displacement of faster train which is A + Length of train A
$⟹$ Road distance required $=v_{A}t+50=\frac{50}{3}\times 20+50\approx 383\text{m}$