physics
An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 14km/h .Precisely 2.50s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.30m/s^2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?
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Using the kinematics equation, we can calculate the time it will take the police officer to overtake the speeder. We plug in the given values: V = 95 + 14 = 109 km/h a = 2.3 m/s^2 t = 2.50 s We use the equation ∆s = 1/2 a t^2 and solve for t: 109 km/h x 1000 (m/km) x (1 h/3600) = 30.25 m/s ∆s = 30.25^2/2 x 2.3 = 351.5 m t = √(2* 351.5/2.3) = 14.66 s Therefore, it takes 14.66 seconds for the police officer to overtake the speeder after the speeder passes.
