An inattentive driver is travelling 18m/s. When he notice a red light ahead. His car is capable of a breaking acceleration of 3.65m/s^2. If it takes him 0.200 second to get the brakes on and he is 45m from the intersection when he sees the ligth, will he be able to stop in time?

Answer 1

just plug in your numbers into the equation of motion:
how long does it take to stop (v=0)?
v = 18 - 3.65t
t = 4.932
How far does he go?
s = vt + 1/2 at^2
s = 18t - 1.825t^2
time to stop: 4.932+0.200 = 5.132 seconds
s = 18*5.132 - 1.825*5.132^2 = 44.310
>whew< close call!

Answer 2

V^2 = Vo^2 + 2a*d = 0.
324 + (-7.30)d = 0,
d = 44.4 m = stopping distance.

45 - 18m/s * 0.2s = 41.4 m. = Required stopping distance.
No, he will not be able to stop in time.