Question #155243

a test of a driver's perception reaction time is being conducted on a special testing track with wet pavement and a driving speed of 50 kph. when the driver is sober, a stop can be made just in time to avoid hitting an object that is visible 40 m. ahead. after a few drinks of san miguel beer, under exactly the same condition, the driver fails to stop in time and strikes the object at a speed of 30 kph. determine the driver's perception-reaction time after he was drinking. assume coefficient of friction is 0.60.


1
Expert's answer
2021-01-19T05:19:41-0500

d=vd= v1 ×tr\times tr

d=d= distance travelled for perception

vv1 = initial velocity

trtr= perception time


trtr =dv1\frac{d}{v1} = 4050=0.8seconds\frac{40}{50}= 0.8 seconds

coefficient of friction μ\mu = forceappliednormalweight\frac{force applied }{normal weight }= mamg\frac{ma}{mg}


0.6=mgma=mam×9.81\frac{mg}{ma}= \frac{ma}{m\times 9.81}

a=9.81×0.6=5.886m/sa=9.81\times0.6=5.886 m/s

aa (acceleration/deceleration) = finalvelocityinitialvelocitytime2\frac{final velocity- initial velocity}{time^2}

= mgt\frac{mg}{t} =vvv- vvu/t22

5.886= 030t2=equivalent=30t2\frac{0-30}{t^2}=equivalent = \frac{30}{t^2}

t= 305.886=2.258seconds\sqrt[]{\frac{30}{5.886}} = 2.258 seconds


total perception time for the unconscious driver is therefore prolonged as below;

t= 0.8 + 2.258

= 3.058 seconds



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