Question

Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 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 the driver of A, what was the original distance between them ?

Answer 1

Given,

• $\sf u_A=72 \:km/h=20\:m/s$

• $\sf u_B=72 \:km/h=20\:m/s$

• $\sf Time=50\:s$

• $\sf Acceleration=1\:m/s^2$

We know,

$\huge\boxed{\sf s=ut+\dfrac{1}{2}at^2}}$

Let,

• $\sf {S_A \:be \:the \:distance\:covered\:by\:train\:A}$

• $\sf{ S_B\:be \:the \:distance\:covered\:by\:train\:B}$

$\boxed{\sf S_B=u_Bt+\dfrac{1}{2}at^2}$

$\implies\sf {S_B=(20\times50)+\dfrac{1}{2}(1\times(50)^2)}$

$\implies\sf S_B=1000+\dfrac{1}{2}(2500)$

$\implies\sf S_B=1000+1250$

$\implies\sf S_B=2250\:m$

$\boxed{\sf S_A=u_A\times t}$

$\implies\sf S_A=20\times50$

$\implies\sf S_A=1000\:m$

Therefore,

$\sf Original\:distance\:between\:trains\:A\:and\:B=S_B-S_A$

$\implies\sf Original\:distance\:between\:trains\:A\:and\:B=2250-1000$

$\implies\sf Original\:distance\:between\:trains\:A\:and\:B=1250\:m$

$\huge\boxed{\sf Original\:distance\:between\:trains\:A\:and\:B=1250\:m}$

Additional Information

→ To convert km/h into m/s we have to multiply the given value by  5/18.

Answer 2

Given Question?

Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 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 the driver of A, what was the original distance between them ?

Required Answer:-

Original distance between them is 1250m.

Explanation:-

$\large { \underline{ \rm{ \purple{Given:- }}}}$

• Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 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 the driver of A.

$\large { \underline{ \rm{ \purple{To \: Find:- }}}}$

• What was the original distance between them ?

$\large { \underline{ \rm{ \purple{Directions:-}}}}$

$\implies \mathtt{u_{a} = u_{b} = {72kmh}^{ - 1} }$

$\implies \mathtt{72 \times \frac{5}{18} = {20ms}^{ - 1} }$

$\large { \orange{ \bigstar{ \green { \boxed{ \rm{ Formula \: Used:-{ \mathtt {ut + \frac{1}{2} {at}^{2} }}}}}}}}$

$\implies \mathtt{S_b = u_{b}t + \frac{1}{2} {at}^{2} }$

$\implies \mathtt{20 \times 50 + \frac{1}{2} \times 1 \times {50}^{2} }$

$\implies \mathtt{S_b = 100 + 1250 = 2250}$

$\implies{\boxed{ \mathtt{S_b = 2250m}}}$

$\rm {Let \: S_a \: be \: the \: distance \: covered}$

$\rm{ by \: the \: train \: 'A' \: then}$

$\implies \mathtt{S_a=U_a \times t}$

$\implies { \boxed{ \mathtt{20 \times 50 = 1000m}}}$

$\rm{Original \: distance \: between \: the \: two \: trains = }$

$\implies \mathtt{S_b-S_a}$

$\implies \mathtt{2250 - 1000 = 1250m}$

$\pink{ \boxed{ \sf{{Hence, the \: original \: distance \: between \: them \: is \: 1250m. }}}}$