Question

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

• A. $750\,m$
• B. $1000\,m$
• C. $1250\,m$
• D. $2250\,m$

Answer 1

Answer: (C)

$1250\,m$

Answer 2

For train $A$,
$u_{A}=72\,km\,h^{-1}=72\times\frac{5}{18}\,m\,s^{-1}=20\,m\,s^{-1}$, $a_{A}=0$, $t=50\,s$
$\therefore S_{A}=u_{A}\,t=\left(20\right)\left(50\right)=1000\,m$
For train $B$,
$u_{B}=72\,km\,h^{-1}=72\times\frac{5}{18}\,m\,s^{-1}=20\,m\,s^{-1}$
$a_{B}=\,m\,s^{-2}$, $t=50\,s$
$\therefore S_{B}=u_{B}t+\frac{1}{2}\,a_{B}t^{2}$
$=\left(20\times50\right)+\frac{1}{2}\times1\times\left(50\right)^{2}$
$=2250\,m$
Original distance between $A$ and $B = S_B - S_A$
$= 2250\, m - 1000 \,m = 1250 \,m$