Question

Two trains $A$ and $B$ of length $400m$ each are moving on two parallel tracks with a uniform speed of $72km{h}^{-1}$ in the same direction, with $A$ ahead of $B$. The driver of $B$ decides to overtake $A$ and accelerates by $1m{s}^{-2}$. If after $50s$, the guard of $B$ just brushes past the driver of $A$, what was the original distance between them?

• A. $100m$
• B. $1150m$
• C. $1300m$
• D. $1250m$

Answer 1

The correct option is D $1250m$

Given that,
$u_A=u_B=72km{h}^{-1}=72\times \frac{5}{18}=20{ms}^{-1}$
Using the relations, $s=ut+\frac{1}{2}a{t}^2$, we get
$S_B=u_{B}t+\frac{1}{2}a{t}^2=20\times 50+\frac{1}{2}\times 1\times {\left(50\right)}^2$
$S_B=1000+1250=2250m$
Also, let $S_A$ be the distance covered by the train $A$, then $S_A=u_A\times t$
$=20\times 50=1000m$
Original distance between the two trains
$=S_B-S_A$
$=2250-1000=1250m$