Posted by Anonymous on .
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
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.