Question

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

Answer 1

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.

Answer 2

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