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?
Let t=0 when car passes police. We want the distance after that to be the same, so
49t = 19(t-.6) + 1/2 * 3 * (t-.6)^2
t = 21.54 s
To find out how long it takes for the police car to catch up with the speeder, we'll start by determining the initial distance between them.
Given:
Velocity of the police car (v1) = 19.0 m/s due north
Velocity of the speeder (v2) = 49.0 m/s due north
Reaction time (t) = 0.600 s
Acceleration of the police car (a) = 6.00 m/s^2
Since both vehicles are traveling due north, their motion is taking place in one dimension, and we can consider their velocities positive.
During the reaction time (t), the police car moves a distance given by:
d1 = v1 * t
Substituting the values:
d1 = (19.0 m/s) * (0.600 s)
Next, let's determine the distance covered by the speeder during the reaction time:
d2 = v2 * t
Substituting the values:
d2 = (49.0 m/s) * (0.600 s)
Now, we need to calculate the relative distance between the police car and the speeder:
d_rel = d2 - d1
Taking the difference:
d_rel = (49.0 m/s * 0.600 s) - (19.0 m/s * 0.600 s)
Now, we need to find out the time it takes for the police car to catch up with the speeder using this relative distance and the acceleration of the police car.
Using the equation of motion:
d_rel = (1/2) * a * t^2
Solving for t:
t^2 = (2 * d_rel) / a
Substituting the values:
t^2 = (2 * d_rel) / 6.00 m/s^2
Finally, we can find the time by taking the square root of t^2:
t = sqrt((2 * d_rel) / 6.00 m/s^2)
Evaluating this expression will give us the time it takes for the police car to catch up with the speeder, including the reaction time.