Due to slow reaction, the distance he had to travel to get the brakes on is; 44.4 m
We are given;
Initial velocity; u = 0 m/s
Final Velocity; v = 18 m/s
Acceleration; a = 3.65 m/s²
Time; t = 0.2 s
To find the stopping distance, we will use newton's equation of motion to get;
v² = u² + 2a*d
Plugging in the relevant values gives us;
18² = 0² + 2(3.65d)
324 = 7.3d
d = 324/7.3
Stopping Distance; d = 44.4 m
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