A speeder passes a parked police car at a constant speed of 27.3 m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.55 m/s2 .
a) How much time passes before the speeder is overtaken by the police car?
b) How far does the speeder travel before being overtaken by the police car?
2 answers
see related questions below.
s=2u^2/a
s=2*27.3^2/2.55
s=584.54
s=ut
t=s/u
t=584.54/27.3= 21.41sec
s=2*27.3^2/2.55
s=584.54
s=ut
t=s/u
t=584.54/27.3= 21.41sec