Question

Two trains, which are moving along different tracks in opposite directions are put on the same track by mistake. On noticing the mistake, when the trains are $300\, m$ apart the drivers start slowing down the trains. The graphs given below show decrease in their velocities as function of time. The separation between the trains when both have stopped is

• A. 120 m
• B. 20 m
• C. 60 m
• D. 280 m

Answer 1

Answer: (B)

For train I,
$ u_{1}=40 \,m / s$
As, $ v_{1}=u_{1}+a_{1} t $
$\Rightarrow 0=40+a_{1}(10) $
$\Rightarrow a_{1}=\frac{-40}{10}=-4 \,m / s ^{2}$
For train II
$u_{2}=-20 \,m / s$
As,$ v_{2}=u_{2}+a_{2} t $
$\Rightarrow 0=-20+a_{2}(8)$
$\Rightarrow a_{2}=\frac{20}{8}=\frac{10}{4} \,m / s ^{2}$
For train I,
$ v_{1}^{2} =u_{1}^{2}-2 a_{1} \,s_{1} $
$ \Rightarrow 0 =u_{1}^{2}-2 a_{1} \,s_{1} $
$ \Rightarrow s_{1} =\frac{u_{1}^{2}}{2 a_{1}}=\frac{(40)^{2}}{2 \times 4} $
$=\frac{1600}{8}=200 \,m $
For train II,
$v_{2}^{2}=u_{2}^{2}-2 a_{2} \, s_{2}$
$\Rightarrow 0=u_{2}^{2}-2 a_{2} \, S_{2}$
$\Rightarrow s_{2}=\frac{u_{2}^{2}}{2 a}$
or, $s_{7}=\frac{(-20)^{2}}{2\left(\frac{10}{4}\right)}$
or, $s_{2}=\frac{400 \times 2}{10}$
or, $s_{2}=80 \, m$
Total distance covered by both trains,
$s=s_{1}+s_{2}$
$s=200+80$
$s=280 \,m$
$\therefore $Separation between the trains
$=$ Remaining distance
$=300-280=20\,m$