A speeding car is traveling at a constant 30.0 m/s when it passes a stopped police car. The police car accelerates at 7.0 m/s^2. a)how far will the police travel to catch up to the speeder? b) how fast will it be going when it catches up?
To find the answers to these questions, we need to consider the motion of both the speeding car and the police car. We can use the equations of motion to calculate the distance and final speed of the police car.
a) To find the distance the police car will travel to catch up to the speeder, we can use the equation:
distance = (initial velocity x time) + (1/2 x acceleration x time^2)
The initial velocity of the police car is 0 m/s since it is stationary, and the acceleration is 7.0 m/s^2. We need to find the time it takes for the police car to catch up to the speeder.
The relative velocity of the police car with respect to the speeding car is 30.0 m/s (speed of the car) + 0 m/s (initial speed of the police car) = 30.0 m/s.
Using the equation of relative velocity:
relative velocity = speed of the car - initial speed of the police car
30.0 m/s = 0 m/s - (-7.0 m/s^2) x time
30.0 m/s = 7.0 m/s^2 x time
time = 30.0 m/s / 7.0 m/s^2
time ≈ 4.29 s
Substituting the values back into the distance equation:
distance = (0 m/s x 4.29 s) + (1/2 x 7.0 m/s^2 x (4.29 s)^2)
distance ≈ (0 m/s x 4.29 s) + (1/2 x 7.0 m/s^2 x 18.3841 s^2)
distance ≈ 0 m + 64.305 m
distance ≈ 64.3 m
Therefore, the police car will travel approximately 64.3 meters to catch up to the speeder.
b) To find the final speed of the police car when it catches up, we can use the equation:
final velocity = initial velocity + (acceleration x time)
The initial velocity of the police car is 0 m/s, and the acceleration is 7.0 m/s^2.
Substituting the values into the equation:
final velocity = 0 m/s + (7.0 m/s^2 x 4.29 s)
final velocity = 0 m/s + 30.03 m/s
final velocity ≈ 30.0 m/s
Therefore, the police car will be traveling at approximately 30.0 m/s when it catches up to the speeder.