A speeder passes a parked police car at a constant speed of 23.3 m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.75 m/s 2. How much time passes before the speeder is overtaken by the police car? Answer in units of s.

t = 16.94 s

Explanation:

t is the time passes before police catch the speeder

speed of speeder Vo = V = 23.3 m/s

T = t

Police Info

Vo = 0 m/s

a = 2.75 m/s^2

t = t

Now,

displacement of the police car = displacement of the speeder.

x_{police} = Vo * t + 1/2 at^2

since Vo = 0

x police = 1/2 at^2

x police = 1/2 (2.75) (t) ^2

Now the displacement of speeder is

x_{speeder} = Vt

x_{speeder} = 23.3 t

x_{speeder} = x_{police}

23.3 t = 1/2 * 2.75 t^2

23.3 t = 1.375 t^2

t = 23.3/1.375

t = 16.94

t = 16.94 s