A certain traffic light is yellow for 3 seconds before it turns red.
As you approach the intersection, you must decide whether you can
safely go through the yellow light before it turns red or whether you
have time to stop before reaching the intersection. Suppose your
reaction time is 0.5 seconds, you are initially moving at 20 m/s, and
you can decelerate at a maximum rate of 5 m/s^2 when you step on the
brakes and accelerate at a maximum rate of 3 m/s^2 when you step on
the gas.

(a) to come to a complete stop before reaching the intersection, what
is the minimum distance you can be from the light when it turns
yellow?

(b)To pass the light before it turns red, what is the maximum distance
you can be from the light when it turns yellow?

(c)Repeat the calculations assuming an initial speed of 35 m/s and
find the range of distance for which you can neither stop nor get
through the light in time.

a) Use the equation V2^2 = V1^2 + 2ax to solve for the distance needed
to stop (40m). You need 10m to react - 0.5s at 20m/s. Therfore you
have to be at least 50m away.

b) Use x = V1t + 0.5at^2 to solve for the distance covered in 2.5s
accelerating at 3m/s^2 (59.375m). Add 10m for reaction time yielding
69.375m.

c) Repeat part a and b and you will get min = 140m and max = 114.375m.
Therefore if you are between these two distances you will be unable to
stop in time or go through the light.

Best Regards
jpariag

I did try to add #s in, the result is:
a.-20.4m
b.11.9m

Is that all rite? plx response me back, thx...........

I'm not sure how you got those answers... but I'll try to offer some more help.

a) Using V2^2 = V1^2 + 2ax  where V2 = 0m/s (you have stopped), V1 = 20m/s,
a = -5m/s^2. Solve for x. You get:

0 = 20^2 + 2(-5)x
x = 40m

Therefore you will need 40m to come to a stop from 20m/s while
decelerating at 5m/s^2. You take 0.5s to react and this time you will
travel 10m (x = vt). Therefore you need 50m total to be able to stop
safely.

b) When the light turns yellow, you take 0.5s to react and you travel
10m in that time (x = vt). You then have 2.5s to accelerate past the
light.
Using x = V1t + 0.5at^2 where V1 = 20m/s and a=3m/s^2 and t = 2.5s solve for x.
You should get x = 59.375m. Add this to the 10m you took to react and
you get a total of 69.375m. That means you cannot be further than
69.375m away from the light if you want to make it through.

c) To solve this part, just repeat parts a and b but use V1 = 35m/s.
You should find that you need a minimum distance of 140m to stop
safely and you cannot be further than 114.375m to make it through the
light. That means that if you are inbetween those two distances (130m
for example), you cannot stop in time (you need 140m to do that) and
you cannot make it throught the light (you cannot be further than
114.375m away)

Hope that helps
jpariag