Question

Two trains A and B of length 400 m 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 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

Answer 1

Step 1: Given that:

Length of train A = Length of train B = $400m$

Initial velocity($u_A$) of train A = Initial speed of train B($u_B$)

$u_A$ = $u_B$ = $72km{h}^{-1}$ = $72\times \frac{5}{18}m{s}^{-1}=20m{s}^{-1}$

Train A is ahead of train B.

Acceleration in train B($a_B$) = $1m{s}^{-2}$

Given time(t) = $50sec$

Step 2: Calculation of the original distance between train A and train B:

The relative velocity of train B with respect to train A is given by;

$u_{BA}=u_B-u_A$

$u_{BA}=20m{s}^{-1}-20m{s}^{-1}$

$u_{BA}=0$

Distance covered in the given time by train B with respect to A is given by the second equation of motion;

$s=ut+\frac{1}{2}a{t}^2$

Let,

$s_{BA}$ be the distance that is to be covered by train B to just cross train B.

Putting the values, we get;

$s_{BA}=u_{BA}\times t+\frac{1}{2}{a}_B{t}^2$

$s_{BA}=0\times 50sec+\frac{1}{2}\times 1m{s}^{-2}\times {\left(50sec\right)}^2$

$s_{BA}=25\times 50$

$s_{BA}=1250m$

Here,

The original distance between the train = The distance covered by train B with respect to train A to just cross train A($s_{BA}$)

= $1250m$

Thus,

The original distance between them is = $1250m$.