Answer:
(A) after 12.94 sec speeder will be overtaken by police car
(B) speeder will travel 181.9377 m before coming to rest
Explanation:
We have given that speeder passes a parked police car with a constant speed of 28.1 m/sec
As the police car starts from rest so its initial velocity u = 0 m/sec
It is given that police car is travelling with a constant acceleration of [tex]2.17m/sec^2[/tex]
So acceleration a = [tex]2.17m/sc^2[/tex]
From first equation of motion v = u+at
So [tex]28.1=0+2.17\times t[/tex]
t = 12.94 sec
So after 12.94 sec speeder will be overtaken by police car
(b) From second equation of motion [tex]s=ut+\frac{1}{2}at^2[/tex]
[tex]s=0\times 12.94+\frac{1}{2}\times 2.17\times 12.94^2=181.9377m[/tex]
So speeder will travel 181.9377 m before coming to rest
