A speeder passes a parked police car at a constant speed of 25.1 m/s. At that instant,the police car starts from rest with a uniform acceleration of 2.86 m/s^2.

1) How much time passes before the speeder is overtaken by the police car?
Answer in units of s
2) How far does the speeder travel before being overtaken by the police car?
Answer in units of m

1. 25.1t = 0.5a*t^2.

1.43t^2 - 25.1t = 0
Use Quadratic Formula:
t = 17.55 s.

2. d = 25.1 + 17.55 = 440.5 m.

To find the answers to these questions, we can use kinematic equations that relate the initial velocity, final velocity, acceleration, time, and displacement of an object.

Let's solve each question step by step:

1) How much time passes before the speeder is overtaken by the police car?

To find the time it takes for the police car to overtake the speeder, we need to find the time it takes for both vehicles to reach the same position.

First, we need to determine the distance travelled by the police car in this time. We will use the equation:
s = ut + 1/2at^2

For the police car:
Initial velocity, u = 0 (since it starts from rest)
Acceleration, a = 2.86 m/s^2
Initial displacement, s = 0 (since it starts from rest and the reference point is where the speeder passes the parked car)

Let's find the time it takes for the police car to catch up to the speeder.

Using the equation:
s = ut + 1/2at^2

0 = 0 + 1/2 * 2.86 * t^2
0 = 1.43 * t^2
t^2 = 0
t = 0 seconds

This means that the police car catches up to the speeder at the same instant it starts moving. Therefore, no time passes before the speeder is overtaken by the police car.

2) How far does the speeder travel before being overtaken by the police car?

To find the distance traveled by the speeder before being overtaken, we need to find the distance traveled by the police car in the same time.

We can use the equation:
s = ut + 1/2at^2

For the speeder:
Velocity, u = 25.1 m/s (constant)
Time, t = 0 seconds (since the police car catches up to the speeder at the same instant)

Using the equation:
s = ut + 1/2at^2

s = 25.1 * 0 + 1/2 * 0 * 0^2
s = 0

This means that the speeder doesn't travel any distance before being overtaken by the police car.