Two cars are moving with same velocity of $30 k m h^{- 1}$ maintaining a distance of $5 k m$ between them. Speed of third car moving in the opposite direction and meeting the two cars at an interval of $240 s$ is

45 k m h^{- 1}

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30 k m h^{- 1}

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55 k m h^{- 1}

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35 k m h^{- 1}

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Solution

The correct option is A $45 k m h^{- 1}$
Time interval,
$t = \frac{d i s t a n c e b e t w e e n c a r s}{r e l a t i v e v e l o c i t y}$
$= \frac{d}{v_{3} + 30}$
where $v_{3}$ is velocity of 3rd car and velocity of 1 or 2 is $30 k m h^{- 1}$
$⟹ \frac{240}{3600} = \frac{5}{v_{3} + 30}$
$v_{3} = (15 \times 5 - 30)$
$= 45 k m h^{- 1}$

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