First we have to calculate the acceleration of the car,
[tex]a =\frac{u-v}{t}[/tex]
Here, u is initial velocity of the car and its value is given 12.2 m/s and v is final velocity of the car and it comes to stop, so its value zero.
[tex]a=\frac{0-12.2 m/s}{t} =\frac{-12.2 \ m/s}{t}[/tex].
As during the braking the acceleration is constant, from the kinematic equation,
[tex]s=ut + \frac{1}{2} a t^2[/tex]
Here, s is the distance traveled by the car during braking and its value is given 36. 5 m.
Substituting all the values in kinematic equation, we get
[tex]36.5 m =(12.2 m/s) t + \frac{1}{2} (- \frac{12.2 m/s}{t}) t^2 \\\\ 36.5 \ m = (12.2 \ m/s) t -(6.1 \ m/s) t \\\\\ t = \frac{36.5 m}{6.1 m/s} = 5.98 s[/tex]
Therefore, car will stop after 5.98 s.
