A police car is at rest beside a road, looking for speeders. A car passes going at a constant speed of 30.0 m/s. The police car starts up 4 seconds later and accelerates at 2 m/s sq. until it reaches a speed of 40m. If the police car maintains this speed until it catches up wit the speeding car, calculate the total distance in which he catches up with the speeder.
40 m is not a speed.
Probably a typo meant to say 40 m/s.
To calculate the total distance the police car catches up with the speeder, we can use the equation of motion.
Let's break down the problem into different parts:
1. The time it takes for the police car to catch up with the speeder.
2. The distance covered by the speeder during this time.
3. The distance covered by the police car during this time.
First, let's find the time it takes for the police car to catch up with the speeder. We can use the equation of motion:
s = ut + (1/2)at^2
Here, s represents the distance covered, u represents the initial velocity, a represents the acceleration, and t represents the time.
For the police car:
Initial velocity, u = 0 (since it starts from rest)
Acceleration, a = 2 m/s^2
Let's calculate the time it takes for the police car to reach a speed of 40 m/s:
v = u + at
40 = 0 + 2t
t = 40/2
t = 20 seconds
Now, let's find the distance covered by the speeder during this time:
Distance = Speed x Time
Distance = 30 m/s x 20 s
Distance = 600 meters
Finally, let's find the distance covered by the police car during this time:
Distance = (Initial velocity x Time) + (1/2)(Acceleration x Time^2)
Distance = (0 x 20) + (1/2)(2 x (20^2))
Distance = 0 + 1/2 x 2 x 400
Distance = 0 + 400
Distance = 400 meters
Therefore, the total distance in which the police car catches up with the speeding car is 600 meters + 400 meters = 1000 meters.