Asked by Mar

A speeder passes a parked police car at a
constant speed of 25.8 m/s. At that instant,
the police car starts from rest with a uniform
acceleration of 2.68 m/s^2.

How much time passes before the speeder
is overtaken by the police car?
Answer in units of s

How far does the speeder travel before being
overtaken by the police car?
Answer in units of m.
Answered by MathMate
The police car catches up 2.68 m/s for every elapsed second. So to catch up 25.8 m/s, it will take
t=25.8 m/s^2 / 2.68 m/s
= 9.63 s
Since the speeder travels at a uniform speed of 25.8 m/s, the distance travelled can be calculated by
Distance = speed * time.
Answered by Wale
s=2u^2/a
s=2*25.8^2/2.68
s=496.75
s=ut
t=s/u
t=496.75/25.8
t=19.25sec
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