A speeder passes a parked police car at a constant speed of 39.9 m/s. At that in

Question

A speeder passes a parked police car at a constant speed of 39.9 m/s. At that instant,
the police car starts from rest with a uniform acceleration of 2.42 m/s 2 . How much time passes is overtaken by the police Answer in units of s.

How far does the car travel being overtaken by the police car? Answer in units of m.

Explanation / Answer

(a) For Speeder: Vi=39.9, Vf=39.9, ==> D=Vavg * t D= 39.9 * t For Police: Vi = 0, a =2.42, D = Vi * t + 0.5 * a * t^2 = (0)*t + ½ (2.42)* t^2 = 1.21* t^2 when overtaken by police car we have speeder = police 39.9 t = 1.21*t^2 39.9 = 1.21 t Therefore t = 39.9 / 1.21 = 32.97 s b) D =39.9 * t = 39.9 (32.97) = 1315.7 m

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