SOLUTION: this is physics:An automobile moving at a constant velocity of 45 ft/s passes a gasoline station. two seconds later, another automobile leaves the gasoline station and accelerates

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Question 1042754: this is physics:An automobile moving at a constant velocity of 45 ft/s passes a gasoline station. two seconds later, another automobile leaves the gasoline station and accelerates at the constant rate of 6 ft/s^2. how soon will the second automobile overtake the first?
Answer by ikleyn(52866)   (Show Source): You can put this solution on YOUR website!
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this is physics:An automobile moving at a constant velocity of 45 ft/s passes a gasoline station.
Two seconds later, another automobile leaves the gasoline station and accelerates at the constant rate of 6 ft/s^2.
How soon will the second automobile overtake the first?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let "t" be the time from the moment when the second car left the station.

Then you have an equation

 = .

The left side is the distance from the station to the overtaking point, covered by the first car.

The right side is the same distance, covered by the second care.

Thus you have this quadratic equation

 = ,   or

 = .

Solve it using the quadratic formula:

 =  = .

Only positive root is accepted: t = 15.94 seconds.

Ansqer: 15.94 seconds after the second car left the station.


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