d = u t for car
d = 20.1 t
d = (1/2) a t^2 for cruiser
d = (1/2) 2.65 t^2 = 1.325 t^2
so
1.325 t^2 - 20.1 t = 0
t (1.325 t - 20.1) = 0
they are together at t = 0
and again at
t = 20.1/1.325 = 15.2 seconds
A speeder passes a parked police car at a
constant speed of 20.1 m/s. At that instant,
the police car starts from rest with a uniform
acceleration of 2.65 m/s
How much time passes before the speeder
is overtaken by the police car?
Answer in units of s
1 answer
1 answer