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 between them?

• A. 750 m
• B. 1000 m
• C. 1250 m
• D. 2250 m

Answer 1

Answer: (C)

1250 m

Answer 2

For train A.

$u_{A} = 72kmh^{- 1} = 72 \times \frac{5}{18}ms^{- 1} = 20ms^{- 1}$

$a_{A} = 0,t = 50s.$

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

For train B,

$u_{B} = 72kmh^{- 1} = 72 \times \frac{5}{18}ms^{- 1} = 20ms^{- 1}$

$a_{B} = 1ms^{- 2},t = 50s$

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

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

= 2250 m – 1000 m = 1250m

Alternative solutions

$\mathrm { u } _ { \mathrm { BA } } = \mathrm { u } _ { \mathrm { B } } - \mathrm { u } _ { \mathrm { A } } = 2 \mathrm {~ms} ^ { - 1 } - 20 \mathrm {~ms} ^ { - 1 } = 0 \mathrm {~ms} ^ { - 1 }$ $a_{BA} = a_{B} - a_{A} = 1ms^{- 2} - 0ms^{- 2} = 1ms^{- 2}$

T = 50 s

$S_{BA} = u_{BA}t + \frac{1}{2}a_{BA}t^{2} = 0 \times 50 + \frac{1}{2} \times 1 \times (50)^{2} = 1250m$