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
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=(vf²-vi²)/2a
=(0²-11²)/(2*(-8))
= 7.56 m
Total stopping distance
= 6.49+7.56 m
= 14.1 m
To calculate the stopping distance of the car, we need to determine the final velocity (Vf) of the car after applying the brakes.
First, let's calculate the final velocity using the equation: Vf = Vi + (acceleration x time)
Given:
Initial velocity (Vi) = 11.0 m/s
Acceleration (a) = -8 m/s^2 (negative because it's deceleration)
Reaction time (t) = 0.59 s
Vf = 11.0 + (-8) x 0.59
Vf = 6.28 m/s
Now, to find the stopping distance (x), we can use the equation: x = (1/2) x (Vi + Vf) x t
Plugging in the given values:
x = (1/2) x (11.0 + 6.28) x 0.59
x = 5.0976 meters
Therefore, the stopping distance of the car, as measured from the point where the driver first sees the red light, is approximately 5.1 meters.