Answer to Question #155181 in Civil and Environmental Engineering for Shin

Question #155181

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
2021-01-19T02:31:23-0500

The driver's perception-reaction time is:


t=t1t2t=t_1-t_2 ,


where t1t_1 - the time from the moment, when the driver presses the brake, to the moment, when the car stops;

t2t_2 - the time from the moment, when the driver presses the brake, to the moment, when the car strikes the object at a speed of 30 kmph.

First, find t1t_1. The car speed depends on time as follows:


v=v0+atv=v_0+at ,


where v0v_0 - the initial speed;

aa - the acceleration.

In the first case v=0v=0, so


t1=v0at_1=-\frac{v_0}{a} .


The acceleration can be found from the Newton's second law:


ma=μmg,ma=-\mu mg,


a=μg.a=-\mu g.


Henсe, assuming that 50 kmph is 13,89 mps,


t1=v0μg=13.890.69.81=2.36 s.t_1=\frac{v_0}{\mu g}=\frac{13.89}{0.6\cdot 9.81}=2.36\space s.


Then, find t2t_2. In the second case v=30 kmphv=30\space kmph or 8.33 mps8.33\space mps so


t2=v0vμg=13.898.330.69.81=0.95 s.t_2=\frac{v_0-v}{\mu g}=\frac{13.89-8.33}{0.6\cdot 9.81}=0.95\space s.


t=t1t2=2.360.95=1.41 s.t=t_1-t_2=2.36-0.95=1.41\space s.


Answer: The driver's perception-reaction time is 1.41 seconds.



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