A speeder passes a parked police car at a
constant speed of 28.4 m/s. At that instant,
the police car starts from rest with a uniform
acceleration of 2.93 m/s2.
How much time passes before the speeder
is overtaken by the police car?
Answer in units of s

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the speeder's location xs(t) is
xs(t) = 28.4t
Now what's the police car's position, xp(t) ?

(1/2)(2.93)t^2

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