Two cars start from the same point simultaneously and in the same direction. The first car travels at $40$ km/h, and the speed of the second car is $125$ per cent of the speed of the first car. Thirty minutes later a third car starts from the same point and in the same direction. It overtakes the second car $1.5$ h later than it overtook the first car. What is the speed of the third car?

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Solution

Speed of Second car is $⟹ \frac{125}{100} \times 40 = 50$
Let velocity of Third car be $x$
Let $t$ be Time taken to overtake first car
Distance covered is $x t = 40 t + 20$
$⟹ x = \frac{40 t + 20}{t}$
Distance covered to overtake second car is $x (t + 1.5) = 50 (t + 1.5) + 25$
$⟹ \frac{40 t + 20}{t} (t + 1.5) = 50 t + 100$
$⟹ t^{2} + 2 t - 3 = 0$
$⟹ t = 1, - 3$
Time cannot be negative So $t = 1 ⟹ x = 60$

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