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]