Two trains A and B each of length 400 m are moving on two parallel tracks with a uniform speed 72 km in the same direction with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m . If after 50 s, the guard of B just brushes past A

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 (in m) between them?

Smart Achievers · Accepted Answer

(3)

For train A,

$u_{A} = 72 {km h}^{- 1} = 72 \times \frac{5}{18} {m s}^{- 1} = 20 {m s}^{- 1}, a_{A} = 0, t = 50 s,$

$∴$ $S_{A} = u_{A} t = (20) (50) = 1000 m$

For train B,

$u_{B} = 72 {km h}^{- 1} = 72 \times \frac{5}{18} {m s}^{- 1} = 20 {m s}^{- 1},$ $a_{B} = 1 {m s}^{- 2}, t = 50 s$

$∴$ $S_{B} = u_{B} t + \frac{1}{2} a_{B} t^{2} = (20 \times 50) + \frac{1}{2} \times 1 \times {(50)}^{2} = 2250 m$

Original distance between A and B $= S_{B} - S_{A}$

$= 2250 m - 1000 m = 1250 m$

Solution

1 750

2 1000

3 1250

4 2250

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