An unmarked police car traveling a constant 80 km/h is passed by a speeder traveling 125 km/h. Precisely 3.00 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.40 m/s^2, how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

Xf(police)=Xf(car)

Xf(police)= (22.22)(3) + (22.22)t + 0.5(2.40)t^2..(Xf=Xi+Vt+0.5at^2)

Xf(car)= (34.72)(3) + 34.72t..(Xf=Xi+Vt)-constant velocity

To find the time it takes for the police car to overtake the speeder, we'll need to find the distance that the speeder travels during the 3.00 seconds before the police officer steps on the accelerator.

First, let's convert the speeds to meters per second (m/s):
Speeder's speed = 125 km/h = (125 * 1000) / (60 * 60) = 34.72 m/s
Police car's speed = 80 km/h = (80 * 1000) / (60 * 60) = 22.22 m/s

During the 3.00 seconds, the speeder will have traveled a distance of:
Distance = Speed * Time = 34.72 m/s * 3.00 s = 104.16 m

Now, let's find the time it takes for the police car to catch up with the speeder after accelerating.

We can use the equation of motion: s = ut + (1/2)at^2
Where:
s is the distance covered
u is the initial velocity
a is the acceleration
t is the time taken

Since the initial velocity of the police car is 22.22 m/s and we want to find the time, we rearrange the equation:
t = (sqrt((2s)/a)) - (u/a)

Substituting the known values:
t = (sqrt((2 * 104.16 m) / (2.40 m/s^2)) - (22.22 m/s) / (2.40 m/s^2)

Calculating this, we find:
t ≈ 14.96 s

Therefore, it takes approximately 14.96 seconds for the police car to overtake the speeder after the speeder passes.

To solve this problem, we can break it down into two parts: the time it takes for the police car to catch up with the speeder, and the time it takes for both vehicles to travel that distance.

First, let's find the time it takes for the police car to catch up with the speeder after the speeder passes.
The speeder is traveling at 125 km/h, which is equivalent to 125,000 m/3600 s or approximately 34.72 m/s.

Now, let's find the distance between the police car and the speeder when the police car starts accelerating. This can be calculated using the formula:
distance = speed × time.

Since the speeder has a head start of 34.72 m/s for 3.00 seconds before the police car starts accelerating, the distance the speeder travels in those 3.00 seconds is:
distance = speed × time = 34.72 m/s × 3.00 s = 104.16 m.

After 3.00 seconds, the police car starts accelerating. We can use the formula of motion to find the time it takes for the police car to catch up with the speeder. The formula is:
distance = initial velocity × time + (1/2) × acceleration × time^2.

We are given the initial velocity of the police car, which is 80 km/h, or approximately 22.22 m/s, and the acceleration of the police car, which is 2.40 m/s^2. We don't know the time it takes for the police car to catch up, so let's call it "t" for now.

The distance the police car travels during that time is the sum of the initial distance between the police car and the speeder (104.16 m) and the distance covered while accelerating. We can set up the following equation:

104.16 m + 22.22 m/s × t + (1/2) × (2.40 m/s^2) × t^2 = 34.72 m/s × t

Simplifying the equation:

104.16 m + 22.22 m/s × t + 1.20 m/s^2 × t^2 = 34.72 m/s × t

Rearranging the equation to bring all terms to one side:

1.20 m/s^2 × t^2 + (22.22 m/s - 34.72 m/s) × t + (104.16 m - 0) = 0

Now, we can solve this quadratic equation to find the value of "t," the time it takes for the police car to catch up with the speeder after the speeder passes.

Once we find the value of "t," we add the initial 3.00 seconds to get the total time passed before the police car overtakes the speeder after the speeder passes.