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

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450 m

2250 m

1000 m

1250 m

answer is A.

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Detailed Solution

For train A,

$\begin{array}{l} u_{A} = 72 {kmh}^{- 1} = 72 \times \frac{5}{18} {ms}^{- 1} \\ = 20 {ms}^{- 1}, a_{A} - 0, t = 50 s \\ ∴ S_{A} = u_{A} t = (20) (50) = 1000 m \end{array}$

For train B,

$\begin{array}{l} u_{B} = 72 {kmh}^{- 1} = 72 \times \frac{5}{18} {ms}^{- 1} \\ = 20 {ms}^{- 1} \\ a_{B} = 1 {ms}^{- 2}, t = 50 s \\ ∴ S_{B} = u_{B} t + \frac{1}{2} a_{B} t^{2} = (20 \times 50) + \frac{1}{2} \\ \times 1 \times (50)^{2} = 2250 m \end{array}$

Original distance between A and B

$\begin{array}{l} = S_{B} - S_{A} - 2 l \\ = 2250 m - 1000 m - 2 \times 400 m = 450 m \end{array}$

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