We have that the distance traveled with breaks on is is mathematically given as
dt=44.38m
Question Parameters:
. He is traveling 18. 0m/s
His car is capable of a braking acceleration of 3. 65m/s^2
it takes him 0. 200 s to get the brakes on and he is 45. 0 m from the intersection when he sees the light. Will he be able to stop in time
Generally the equation for the Distance traveled is mathematically given as
d=v*t
Therefore
d=v*t
d=18*0.2
d=3.6m
Therefore
[tex]dt=\frac{V^2}{2a}\\\\dt=\frac{18^2}{2*3.65}[/tex]
dt=44.38
Hence, the distance traveled with breaks on is
dt=44.38m
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