A speeder passes a parked police car at a constant speed of 24.8 m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.45 m/s2.
How much time passes before the speeder is overtaken by the police car?
To determine the time it takes for the police car to overtake the speeder, we need to find the time it takes for both vehicles to travel the same distance.
We can start by calculating the distance traveled by the speeder. Since the speeder is traveling at a constant speed of 24.8 m/s, we can use the equation:
distance = speed × time
Since the speeder passes the police car at that instant, we can assume they have both traveled the same distance when the police car catches up.
The equation for the distance traveled by the police car can be expressed as:
distance = initial velocity × time + 0.5 × acceleration × time^2
For the police car, the initial velocity is 0 m/s as it starts from rest, and the acceleration is 2.45 m/s^2.
Now we can set up the two distance equations and solve for time:
Distance traveled by speeder = Distance traveled by police car
speeder_distance = police_car_distance
speeder_speed × time = 0 × time + 0.5 × acceleration × time^2
24.8 × time = 0.5 × 2.45 × time^2
Now we can solve for time by canceling out the "time" term on both sides of the equation and rearranging:
24.8 = 1.225 × time
time = 24.8 / 1.225
time ≈ 20.24 seconds
Therefore, it takes approximately 20.24 seconds for the police car to overtake the speeder.