Question
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?
Answers
oobleck
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!
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!
henry2,
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.
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.