An unmarked police car traveling a constant 85.0 km/h is passed by a speeder traveling 130 km/h .

Precisely 3.00 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.30 m/s2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

To solve this problem, we will first calculate the distance traveled by the speeder in the 3.00 seconds before the police officer starts accelerating.

Distance = Speed × Time
Distance = 130 km/h × (3.00 s / 3600 s/h) [Converting km/h to m/s]
Distance = (130 × 1000 m) / (3600 s) × 3.00 s
Distance = 108.33 m

Now, let's determine the time it takes for the police car to catch up to the speeder.

We'll use the equation of motion:
Distance = Initial velocity × Time + 0.5 × Acceleration × Time^2

Since the police car starts from rest, the initial velocity is 0 m/s.

108.33 m = 0 m/s × T + 0.5 × 2.30 m/s^2 × T^2
108.33 m = 1.15 m/s^2 × T^2

Simplifying the equation:
T^2 = 108.33 m / 1.15 m/s^2
T^2 = 94.24 s^2
T ≈ √94.24
T ≈ 9.71 s

Therefore, it takes approximately 9.71 seconds for the police car to overtake the speeder after the speeder passes.

To solve this problem, we need to find the time it takes for the police car to overtake the speeder after the speeder passes.

First, let's convert the speed of the police car from km/h to m/s:
85.0 km/h = (85.0 km/h) * (1000 m/km) * (1 h/3600 s) ≈ 23.61 m/s

Next, let's find the distance traveled by the speeder before the police officer steps on the accelerator. We can do this by multiplying the speed of the speeder (130 km/h) by the time it takes for the police officer to step on the accelerator (3.00 s):
Distance traveled by the speeder = (130 km/h) * (1000 m/km) * (1 h/3600 s) * (3.00 s) ≈ 108.33 m

Now, let's find the distance the police car needs to cover to catch up with the speeder. Since both vehicles are moving at a constant speed, the distance the police car needs to cover is equal to the distance traveled by the speeder before the police officer steps on the accelerator:
Distance the police car needs to cover = 108.33 m

Next, we'll use the formula for distance traveled under constant acceleration:
Distance = Initial velocity * Time + (1/2) * Acceleration * Time^2

Let's substitute the known values into the equation. The initial velocity of the police car is 23.61 m/s, and the acceleration is 2.30 m/s^2. We'll solve for the time it takes for the police car to cover the distance:
108.33 m = (23.61 m/s) * Time + (1/2) * (2.30 m/s^2) * Time^2

Simplifying the equation, we get:
0.95 * Time^2 + 23.61 * Time - 108.33 = 0

To solve this quadratic equation, we can use the quadratic formula:
Time = (-b ± √(b^2 - 4ac)) / (2a)

Where a = 0.95, b = 23.61, c = -108.33.

Solving for Time using the quadratic formula, we find two possible times: Time1 and Time2.

Plugging the values into the formula, we get:
Time1 ≈ 2.62 s
Time2 ≈ -33.52 s

Since we cannot have a negative time, we discard Time2. Therefore, the time it takes for the police car to overtake the speeder is approximately 2.62 seconds.

Vp = 85,000m/3600s = 23.61 m/s.

Vs = 130,000/3600 = 36.11 m/s.
Ds = 36.11m/s * 3s = 108.3 m. Head start

Vo*t + 0.5a*t^2 = Vs*t + 108.3
23.61t + 0.5*2.30*t^2 = 36.11t + 108.3.
-12.5t + 1.15t^2 = 108.3.
1.15t^2 - 12.5t - 108.3 = 0.
Use Quadratic Formula.
t = 16.6 s.