An unmarked police car traveling a constant 90 km/h is passed by a speeder traveling 128 km/h. Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.80 m/s2 , how much time passes after the police car is passed by a speeder and before the police car overtakes the speeder (assumed moving at constant speed)?
90km/hr = 25m/s
128km/hr = 35.56m/s
Let t=0 be the moment the speeder passes
the cop. Then we want t when
35.56t = 25(t-1) + 1.40(t-1)^2
t = 11
To solve this problem, we need to determine the time it takes for the police car to overtake the speeder.
First, let's convert the speeds into m/s for consistency.
The speed of the police car is 90 km/h, which is equivalent to (90 * 1000) / 3600 = 25 m/s.
The speed of the speeder is 128 km/h, which is equivalent to (128 * 1000) / 3600 = 35.56 m/s.
The relative speed between the police car and the speeder is the difference between their speeds: 35.56 m/s - 25 m/s = 10.56 m/s.
Now, let's find out how far the speeder travels in the 1.00 second after passing the police car. The distance traveled can be calculated using the formula: distance = speed * time.
So, the distance traveled by the speeder in 1.00 second is 35.56 m/s * 1.00 s = 35.56 meters.
Now, let's calculate the time it takes for the police car to overtake the speeder. We can use the kinematic equation: displacement = initial velocity * time + (0.5 * acceleration * time^2).
The initial velocity of the police car is 25 m/s, the acceleration is 2.80 m/s^2, and the displacement is the distance traveled by the speeder after passing the police car (35.56 meters).
Let's rearrange the equation to solve for time: time = (-initial velocity + sqrt(initial velocity^2 + 2 * acceleration * displacement)) / acceleration.
Substituting the values into the equation:
time = (-25 + sqrt(25^2 + 2 * 2.80 * 35.56)) / 2.80
time = (-25 + sqrt(625 + 198.08)) / 2.80
time = (-25 + sqrt(823.08)) / 2.80
time = (-25 + 28.69) / 2.80
time = 3.69 / 2.80
time ≈ 1.32 seconds (rounded to two decimal places).
Therefore, approximately 1.32 seconds pass after the police car is passed by a speeder and before the police car overtakes the speeder.
To solve this problem, we need to find the time it takes for the police car to catch up to the speeder.
Step 1: Convert all the units to be consistent. We need to convert the speeds to meters per second.
The speed of the police car is 90 km/h = (90 * 1000) m / (60 * 60) s = 25 m/s.
The speed of the speeder is 128 km/h = (128 * 1000) m / (60 * 60) s = 35.56 m/s.
Step 2: Find the distance between the police car and the speeder when the police officer starts accelerating.
Since the police officer starts accelerating 1.00 s after the speeder passes, we need to find the distance the speeder travels during that time.
Distance = speed * time = 35.56 m/s * 1.00 s = 35.56 m.
Step 3: Set up the equations of motion for the police car.
Let t be the time it takes for the police car to catch up to the speeder. The distance the police car travels during this time is given by:
Distance = initial velocity * time + (1/2) * acceleration * time^2.
Since the initial velocity of the police car is 25 m/s and the acceleration is 2.80 m/s^2, we can write the equation as:
Distance = 25 m/s * t + (1/2) * 2.80 m/s^2 * t^2.
Step 4: Set up the equation for the distance the police car travels.
The distance the police car travels is equal to the distance the speeder travels plus the distance the police car travels after accelerating for time t.
Distance = 35.56 m + (25 m/s * t + (1/2) * 2.80 m/s^2 * t^2).
Step 5: Solve the equation to find the time.
Set up the equation: 35.56 m + (25 m/s * t + (1/2) * 2.80 m/s^2 * t^2) = 25 m/s * t + (1/2) * 2.80 m/s^2 * t^2.
Simplify and rearrange the equation: 10.56 m = 25 m/s * t.
Solve for t: t = 10.56 m / 25 m/s = 0.4224 s.
Therefore, the time it takes for the police car to overtake the speeder is approximately 0.4224 seconds.