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When two bodies approach each other uniformly, the distance between them decreases by 32 m every 20 s. If they move in the same direction, distance between them increases by 6 m for every 10 s. Then velocities of those two bodies are

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$3.3 m s^{- 1}, 1 m s^{- 1}$

$2.2 m s^{- 1}, 0.5 m s^{- 1}$

$1.1 m s^{- 1}, 0.5 m s^{- 1}$

$1.1 m s^{- 1}, 1 m s^{- 1}$

answer is C.

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Detailed Solution

$(v_{1} + v_{2}) = \frac{32}{20}$

$v_{1} - v_{2} = \frac{6}{10}$

Solving $v_{1} = 1.1 m / s, v_{2} = 0.5 m / s$

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