A police car is traveling at a velocity of 19.0 m/s due north, when a car zooms by at a constant velocity of 49.0 m/s due north. After a reaction time 0.600 s the policeman begins to pursue the speeder with an acceleration of 6.00 m/s^2. Including the reaction time, how long does it take for the police car to catch up with the speeder?
19t + 0.5a*t^2 = 49t + 49*0.6
19t + 0.5*6*t^2 = 49t + 29.4
3t^2 - 30t - 29.4 = 0
t = 10.9 s.
To find the time it takes for the police car to catch up with the speeder, we need to calculate the time it takes for the police car to cover the distance traveled by the speeder during the reaction time.
First, let's calculate the distance traveled by the speeder during the reaction time:
Distance = Velocity × Time
Distance = 49.0 m/s × 0.600 s
Distance = 29.4 m
Now let's calculate the time it takes for the police car to cover this distance:
Acceleration = (Final Velocity - Initial Velocity) / Time
6.00 m/s^2 = (Final Velocity - 19.0 m/s) / Time
Rearranging the formula, we can solve for time:
Time = (Final Velocity - Initial Velocity) / Acceleration
Time = (49.0 m/s - 19.0 m/s) / 6.00 m/s^2
Time = 30.0 m/s / 6.00 m/s^2
Time = 5.00 s
Adding the reaction time, the total time it takes for the police car to catch up with the speeder is:
Total Time = Reaction Time + Time to cover the distance
Total Time = 0.600 s + 5.00 s
Total Time = 5.60 s
Therefore, it takes the police car 5.60 seconds to catch up with the speeder.
To find the time it takes for the police car to catch up with the speeder, we need to determine the distance between them and divide it by the relative speed at which the police car is closing in on the speeder.
1. Calculate the distance traveled by the speeder during the policeman's reaction time:
distance_speeder = velocity_speeder * reaction_time
distance_speeder = 49.0 m/s * 0.600 s
distance_speeder = 29.4 m
2. Next, determine the relative velocity between the police car and the speeder:
relative_velocity = velocity_speeder - velocity_police_car
relative_velocity = 49.0 m/s - 19.0 m/s
relative_velocity = 30.0 m/s
3. Using the equation s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, t is the time, and a is the acceleration, we can find the time it takes for the police car to catch up with the speeder:
Let s be the distance between the police car and the speeder.
Let u be the relative velocity of the police car with respect to the speeder, and a be the acceleration of the police car.
s = ut + (1/2)at^2
s = 30.0 m/s * t + (1/2) * 6.00 m/s^2 * t^2
s = 30.0t + 3.00t^2
4. Substitute the distance traveled by the speeder during the reaction time into the equation:
29.4 m = 30.0t + 3.00t^2
5. Rearrange the equation to form a quadratic equation:
3.00t^2 + 30.0t - 29.4 = 0
6. Solve the quadratic equation using the quadratic formula or a calculator to find the values of t. In this case, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
For the quadratic equation 3.00t^2 + 30.0t - 29.4 = 0,
a = 3.00, b = 30.0, c = -29.4.
t = (-30.0 ± √(30.0^2 - 4 * 3.00 * -29.4)) / (2 * 3.00)
t = (-30.0 ± √(900.0 + 352.8)) / 6.00
t = (-30.0 ± √(1252.8)) / 6.00
t = (-30.0 ± 35.39) / 6.00
Solving for t, we get two values:
t = 1.26 s
t = -4.60 s
Since time cannot be negative, we discard t = -4.60 s.
7. Therefore, the police car takes approximately 1.26 seconds (including the reaction time) to catch up with the speeder.