an automobile moving at constant velocity of 15m/sec passes a gasoline station. two seconds later another automobile leaves the gasoline station and accelerates at a constant rate of 2m/ sec squared. how soon will the second automobile overtake the first ?
They will pass when the distances traveled are equal.
15 t = (a/2)(t-2)^2 = (t-2)^2
Solve for t, which is measured from the time the 15 m/s car goes by the station.
t^2 -4t +4 = 15 t
t^2 -19t +4 = 0
t = (1/2)[19 +sqrt345) = 18.79 s
Both cars will have travelled 282 m at that time
To find out when the second automobile will overtake the first, we need to determine the time it takes for both automobiles to travel the same distance.
Let's break down the problem step by step:
1. We know that the first automobile is moving at a constant velocity of 15 m/sec.
2. Two seconds later, the second automobile starts from rest (zero velocity) at the gasoline station and accelerates at a constant rate of 2 m/sec².
First, let's find the distance traveled by the first automobile in the 2 seconds before the second automobile starts:
Distance = Velocity × Time
Distance = 15 m/sec × 2 sec
Distance = 30 meters
Now, both automobiles are at the same position. The second automobile starts from 0 meters and will accelerate until it catches up with the first.
To determine when the second automobile catches up with the first, we need to calculate the time it takes for the second automobile to cover a distance of 30 meters.
We can use the kinematic equation:
Distance = Initial Velocity × Time + (0.5) × Acceleration × Time²
Since the initial velocity is 0 (starting from rest), we have:
30 meters = 0 × Time + (0.5) × 2 m/sec² × Time²
Simplifying the equation:
30 meters = (1) × Time²
Time² = 30 meters
Time = √(30) meters
Calculating the square root of 30, we find:
Time ≈ 5.48 seconds
Therefore, it will take approximately 5.48 seconds for the second automobile to overtake the first.