A man 'a' goes from his home to somewhere with the speed of 40km/hr. After 15 min, second man 'b' also starts his journey from 'a' house with the speed of 45km/hr and he decided to overtake 'a'. At what distance from 'a' house he travels to overtake 'a' and what time he takes to overtake him

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Solution

Initial distance between the two men = $40 k m / h r \times 0.25 h r = 10 k m$
Relative velocity of two men = $45 k m / h r - 40 k m / h r = 5 k m / h r$
Therefore, time taken to cover the distance = $\frac{10 k m}{5 k m / h r} = 2 h r$
Therefore, b overtakes a 2 hours after b started. Also, the distance from the house is:
$45 k m / h r \times 2 h r = 90 k m$

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