An automobile moving at a constant velocity of 25m/s passes a gasoline station. Two seconds later, another automobile leaves the gasoline station and accelerates at a constant rate of 4.0m/s^2. how soon will the second automobile overtake the first?
v (t+2) = (1/2) a t^2
To solve this problem, you need to find out how long it takes for the second automobile to catch up with the first.
First, let's determine the initial position of the second automobile. We know that it starts from rest because it leaves the gasoline station and then accelerates. Therefore, its initial velocity (u) is 0 m/s.
Next, let's find the position equation for each automobile based on their velocities and the time it takes to catch up.
For the first automobile:
Position1 = Initial Position1 + Velocity1 * Time
For the second automobile:
Position2 = Initial Position2 + Velocity2 * Time + (1/2) * Acceleration * Time^2
Since the first automobile is already moving at a constant velocity, its position equation is straightforward:
Position1 = Velocity1 * Time
For the second automobile, since its initial position is not given, we can assume it starts from the same position as the first automobile due to the information provided in the question. Therefore, the initial position for both automobiles is 0.
Now we can set up the equations equating both positions and solve for the time (T) at which they are equal:
Velocity1 * Time = 0 + 0 * Time + (1/2) * 4.0 * Time^2
Simplifying the equation:
25 * T = 0.5 * 4.0 * T^2
Rearranging the equation, we get a quadratic equation:
2 * T^2 - 25 * T = 0
Factoring out T:
T * (2 * T - 25) = 0
Solving for T, we have two possible solutions:
T = 0 (which is not relevant to the problem) or T = 25 / 2
Thus, T = 12.5 seconds.
Therefore, the second automobile will overtake the first automobile in 12.5 seconds.