As we know that reaction time will be
[tex]t = 0.50 s[/tex]
so the distance moved by car in reaction time
[tex]d = vt[/tex]
[tex]d = 20 \times 0.50[/tex]
[tex]d = 10 m[/tex]
now the distance remain after that from intersection point is given by
[tex]d = 110 - 10 = 100 m[/tex]
So our distance from the intersection will be 100 m when we apply brakes
now this distance should be covered till the car will stop
so here we will have
[tex]v_f = 0[/tex]
[tex]v_i = 20 m/s[/tex]
now from kinematics equation we will have
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - 20^2 = 2(a)100[/tex]
[tex]a = \frac{-400}{200} = -2 m/s^2[/tex]
so the acceleration required by brakes is -2 m/s/s
Now total time taken to stop the car after applying brakes will be given as
[tex]v_f - v_i = at[/tex]
[tex]0 - 20 = -2 (t)[/tex]
[tex]t = 10 s[/tex]
total time to stop the car is given as
[tex]T = 10 s + 0.5 s = 10.5 s[/tex]