We use kinematic equations,
During reaction time,
[tex]S = ut + \frac{1}{2} at^2[/tex] (A)
During the deceleration,
[tex]v^2= u^2 + 2as[/tex]. (B)
Here, u is initial velocity, v is final velocity , a is acceleration and t is time.
There is no deceleration during his 0.8 sec reaction time so a=0, from equation (A),
[tex]S = 20 \ m/s (0.8 \ s) + 0 \times t= 16 \ m[/tex]
During the deceleration (a = -7.0 m/s^2) , final velocity will be zero i.e v = 0, therefore from equation (B),
[tex]0 = (20 m/s)^2 +2 (-7.0 m/s^2) \ s \\\\ s = \frac{-400}{-14} = 28. 57 m[/tex]
Thus, the distance traveled by car after he sees need to put brakes is
[tex]S + s = 16 \ m + 28.57 \ m =44 .57 \ m[/tex]
