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anonymous
 4 years ago
An unmarked police car traveling a constant 85km/hr is passed by a speeder traveling 140km/hr. About 3.00s after the speeder passes, the police officer steps on the accelerator, if the police car's acceleration is 1.80m/s^2 , how much time passes before the police car ovetakes the speeder after the speeder passes?(assumed moving at constant speed)
anonymous
 4 years ago
An unmarked police car traveling a constant 85km/hr is passed by a speeder traveling 140km/hr. About 3.00s after the speeder passes, the police officer steps on the accelerator, if the police car's acceleration is 1.80m/s^2 , how much time passes before the police car ovetakes the speeder after the speeder passes?(assumed moving at constant speed)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is a physics question involving kinematic equations. So far I've set up my problem as such: t = 3.00s, vspeeder140km/h, vp: 85km/h. ap: 1.80m/s^2. V stands for velocity of speeder and police car and a stands for acceleration.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm thinking that I would need to use this kinematics equation: d= vt+1/2at^2
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