A speeder passes a parked police car at a

constant speed of 20.2 m/s. At that instant,
the police car starts from rest with a uniform
acceleration of 2.86 m/s
2
.
How much time passes before the speeder
is overtaken by the police car?
Answer in units of s

d2 = d1

0.5a*t^2 = V1*t
1.43*t^2 = 20.2t
1.43t^2 - 20.2t = 0
Use Quadratic formula.
t = 14.1 s.

s=ut...(I)

s=1/2ut^2...(ii)
from (i)
t=s/u...(iii)
sub in (ii)
s[1-as/2u^2]=0
as/2u^2=1
s=2u^2/a
s=2*20.1^2/2.65=304.9m
since t=s/u
t=304.9/20.1= 15.17sec

To find the time it takes for the police car to catch up to the speeder, we can use kinematic equations. Here's how you can solve it step by step:

1. Determine the initial position and acceleration of the speeder and the police car.
- The speeder is moving at a constant speed, so its initial velocity (v₀) is 20.2 m/s. Its acceleration is 0 m/s² since it is not accelerating.
- The police car starts from rest, so its initial velocity (v₀) is 0 m/s, and its acceleration (a) is 2.86 m/s².

2. Determine the position-time equation for both the speeder and the police car.
- The position-time equation can be written as: x = v₀t + 1/2at², where x is the position, v₀ is the initial velocity, a is the acceleration, and t is the time.

3. Set up the equations for the speeder and the police car.
- For the speeder: x_speeder = 20.2t
- For the police car: x_police = 1/2 * 2.86t²

4. Set the positions of the speeder and the police car equal to each other to find the time it takes for the police car to catch up to the speeder.
- 20.2t = 1/2 * 2.86t²

5. Rearrange the equation and solve for t.
- 20.2t = 1.43t²
- 1.43t² - 20.2t = 0
- Factoring out t: t(1.43t - 20.2) = 0
- t = 0 (discard this solution since time cannot be zero) or t = 14.11 seconds

Therefore, it takes approximately 14.11 seconds for the police car to catch up to the speeder.