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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

  • physics -

    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.

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