An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 14km/h .Precisely 2.50s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.30m/s^2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?
15.5
Why would the police accelerate to catch the speeder who is already moving way slow then the police
To solve this problem, we need to find the time it takes for the police car to catch up to the speeder after the speeder passes.
Let's break down the information given:
- The initial speed of the police car is 95 km/h.
- The speeder is traveling at 14 km/h faster than the police car.
- The police car's acceleration is 2.30 m/s^2.
- The time between the speeder passing the police car and the police car accelerating is 2.50 seconds.
Let's start by converting the speeds of the police car and the speeder from km/h to m/s to keep the units consistent:
1 km/h = 1000 m / (60 min * 60 s) = 0.2778 m/s
So, the initial speed of the police car is:
95 km/h * 0.2778 m/s = 26.39 m/s
The speeder is going 14 km/h faster than the police car, so their speed is:
26.39 m/s + (14 km/h * 0.2778 m/s) = 26.39 m/s + 3.89 m/s = 30.28 m/s
Now let's find the distance between the police car and the speeder at the moment the police car starts accelerating. The distance traveled by the speeder during the 2.50 seconds is:
Distance = Speed * Time
Distance = 30.28 m/s * 2.50 s = 75.70 m
Since the speeder is traveling at a constant speed, we can assume the distance between the speeder and the police car remains constant during the time the police car accelerates.
Now, let's calculate the time it takes for the police car to catch up to the speeder. We can use the following equation of motion:
Distance = Initial velocity * Time + 0.5 * Acceleration * Time^2
Since the police car starts from rest, the initial velocity is 0. The distance is the same as the one calculated earlier (75.70 m), and the acceleration is given as 2.30 m/s^2.
75.70 m = 0 * Time + 0.5 * 2.30 m/s^2 * Time^2
Rearranging the equation:
1.15 * Time^2 = 75.70 m
Time^2 = 75.70 m / 1.15
Time^2 = 65.91 s^2
Taking the square root of both sides:
Time = √(65.91 s^2)
Time = 8.13 s (approximately, rounding to two decimal places)
Therefore, it takes approximately 8.13 seconds for the police car to catch up to the speeder after the speeder passes.