An unmarked police car traveling a constant 95km/h is passed by a speeder. Precisely 2.00s after the speeder passes, the police officer steps on the acceleratorIf the police car accelerates uniformly at 3.00m/s2 and overtakes the speeder after accelerating for 7.00s , what was the speeder's speed?
They travel the same distance in the same time.
To solve this problem, we can use the kinematic equations of motion.
Let's break down the information given:
- Initial speed of the unmarked police car = 95 km/h
- Time elapsed after the speeder passes the police car before the police car accelerates = 2.00 s
- Acceleration of the police car = 3.00 m/s^2
- Time it takes for the police car to overtake the speeder = 7.00 s
To solve for the speeder's speed, we need to find the distance traveled by both the police car and the speeder.
First, we need to convert the initial speed of the police car from km/h to m/s. Since 1 km/h = 0.2778 m/s, the initial speed of the police car is 95 km/h * 0.2778 m/s = 26.67 m/s.
Next, we can find the distance traveled by the police car by using the equation:
distance = initial velocity * time + (1/2) * acceleration * time^2
Plugging in the values for the police car:
distance_police = 26.67 m/s * (2.00 s + 7.00 s) + (1/2) * 3.00 m/s^2 * (7.00 s)^2
Simplifying this equation:
distance_police = 26.67 m/s * 9.00 s + 1/2 * 3.00 m/s^2 * 49.00 s^2
distance_police = 240.03 m + 1/2 * 3.00 m/s^2 * 49.00 s^2
distance_police = 240.03 m + 73.50 m^2/s^2
distance_police = 313.53 m
Now, we need to find the distance traveled by the speeder. The speeder maintains a constant speed until the police car accelerates. So, the distance traveled by the speeder while the police car was accelerating is the same as the distance traveled by the police car.
Therefore, the speeder's speed is equal to the distance traveled by the speeder divided by the time it takes for the police car to overtake the speeder:
speed_speeder = distance_speeder / time_overtake
Since we already know the distance traveled by the police car is 313.53 m and the time it takes for the police car to overtake the speeder is 7.00 s, we can calculate the speeder's speed:
speed_speeder = 313.53 m / 7.00 s
speed_speeder = 44.79 m/s
Therefore, the speeder's speed is approximately 44.79 m/s.