A car moving at 10 km/h is accelerating at 15 m/s^2. It is overtaken by another moving at constant speed of 45 m/s. How much distance should the two travel before the first gets ahead.

Expert Answers

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The first car is moving at 10 km/h and is accelerating at 15 m/s^2. A second car moving at 45 m/s^2 overtakes the first. The initial speed of the first car at the point it is overtaken is 25/9 m/s.

Let the time taken by the first car to overtake the second be t s. In t seconds, the first car travels (25/9)*t + (1/2)*15*t^2 m. The second car travels 45*t m

Equating the two and solving for t:

(25/9)*t + (1/2)*15*t^2 = 45*t

=> 7.5t = 380/9

=> t = 152/27

The distance traveled in this duration is 760/3 m

The first car is able to overtake the second after the two travel for 760/3 m

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