Posted by Sue on .
A car is traveling at 11.0 m/s, and the driver sees a traffic light turn red. After .59 s (the reaction time), the driver applies the brakes, and the car decelerates at 8 m/s^2. What is the stopping distance of the car, as measured from the point where the driver first sees the red light?
Vf= 11+(8m/s^2)(.59)
Vf=6.28
x=1/2(Vi +Vf)(t)
x=1/2(11+6.28)(.59)
x=5.0976

Physics 
MathMate,
During the reaction time of 0.59 s, car is still moving at 11 m/s, so distance to be added is 11 m/s * 0.59 s = 6.49 m.
The deceleration is
a=8 m/s²
vi=11 m/s
vf=0 m/s (trying to stop)
vf²=vi²+2ax
=>
x=(vf²vi²)/2a
=(0²11²)/(2*(8))
= 7.56 m
Total stopping distance
= 6.49+7.56 m
= 14.1 m