an automobile moving at a constant velocity of 50 ft/sec passes a gasoline station. 3 sec later, another automobile leaves the gasoline station and accelerates at the constant rate of 6.5 ft/sec2. how soon will the second automobile overtake the first?
Let t be the time after the constant-velocity car passes the station.
Require that they travel the same distance at time t
50 t = (1/2)*(6.5)(t-3)^2
Solve the resulting quadratic equation for t, which must exceed 3 s.
To find out how soon the second automobile will overtake the first, we need to determine the time it takes for the second automobile to cover the distance between the gasoline station and catch up with the first automobile.
We can start by finding the initial position of the second automobile, which is the distance covered by the first automobile during the 3-second head start. The distance covered by an object moving at a constant velocity can be calculated using the formula:
Distance = Velocity * Time
The distance covered by the first automobile during the 3-second head start is:
Distance1 = Velocity1 * Time1
Distance1 = 50 ft/sec * 3 sec
Distance1 = 150 ft
So, the initial position of the second automobile is 150 feet behind.
Next, we need to determine the time it takes for the second automobile to catch up with the first automobile. We can use the kinematic equation:
Distance = Initial Velocity * Time + (1/2) * Acceleration * Time^2
Since the second automobile starts from rest, its initial velocity is 0 ft/sec.
Substituting the values into the equation, we have:
150 ft = 0 ft/sec * Time + (1/2) * 6.5 ft/sec^2 * Time^2
Rearranging the equation and simplifying:
6.5 ft/sec^2 * Time^2 = 150 ft
Using algebra, we isolate Time^2:
Time^2 = 150 ft / (6.5 ft/sec^2)
Time^2 = 23.077 sec^2
Taking the square root of both sides, we find:
Time = √(23.077 sec^2)
Time ≈ 4.8 sec
Therefore, the second automobile will overtake the first automobile approximately 4.8 seconds after it leaves the gasoline station.