To solve this problem, we can first calculate the distance between the police car and the speeder when the police officer starts accelerating. Then, we can find the time it takes for the police car to catch up to the speeder.
Let's break down the problem into steps:
Step 1: Calculate the distance between the police car and the speeder when the police officer starts accelerating.
The speeder is traveling at a constant speed of 14 km/h. In 2.50 seconds, the speeder covers a distance of:
Distance = Speed × Time
Distance = (14 km/h) × (2.50 s)
Distance = (14 km/h) × (2.50 s) × (1/3600 h/s) [Converting km/h to m/s]
Distance = (14 m/s) × (2.50 s)
Distance = 35 m
So, when the police officer starts accelerating, the speeder is 35 meters ahead.
Step 2: Calculate the time it takes for the police car to catch up to the speeder.
Now, we need to find how long it takes for the police car to cover the 35-meter distance and catch up to the speeder. We'll use the equations of motion to solve this.
We can use the equation of motion:
Distance = Initial Velocity × Time + (1/2) × Acceleration × Time^2
In this case, the initial velocity of the police car is 95 km/h, which we need to convert to m/s:
Initial Velocity = 95 km/h × (1/3600 h/s)
Initial Velocity = (95 m/s) × (1/3600 s)
Initial Velocity = 26.39 m/s
Using this information, we can plug in the values into the equation of motion and solve for time.
35 m = (26.39 m/s) × t + (1/2) × (2.30 m/s^2) × t^2
35 m = 26.39 m/s × t + 1.15 m/s^2 × t^2
0 = 1.15t^2 + 26.39t - 35
We solve this quadratic equation to find the time it takes for the police car to catch up to the speeder.
Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / 2a, we can substitute in the values:
t = [-(26.39) ± √((26.39)^2 - 4(1.15)(-35))] / (2 × 1.15)
Solving the equation gives us two possible solutions, but we discard the negative root since time cannot be negative:
t ≈ 1.19 s
Therefore, it will take approximately 1.19 seconds for the police car to overtake the speeder after the speeder passes.
Please note that in reality, the exact time may vary slightly due to factors such as the reaction time of the police officer.