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Physics

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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 - ,

    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

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