Question
Mini essay: Reaction time and breaking deceleration areimportant when unexpected things happen while driving a car. Imagine that your reaction time is 200 milliseconds, and thatbreaking deceleration is less than one "gee"(9.8m/s2). What does this say about your abilityto stop if you're traveling at 70 mph (31m/s), and what rules for"distance behind the car ahead" might be useful to follow as aresult? Should these rules change if you are talking on thecell phone, and if so how?
If you answer in two parts.....I will rate both answers with themost possible points.
Explanation / Answer
Here are some ideas. First of all, based on the information given, there is a .2s(200millisecond) delay between when the driver wants to break andwhen the driver actually breaks. In this time, at 31m/s the carwould go a distance of 6.2m (d=v*t). Then the driver begins tobreak, causing the car to decelerate at a rate of less than9.8m/s2. Since that's a bit ambiguous some estimationwill be involved. There's a few ways you can do this. The easiestbeing just calling deceleration 9.8m/s2 or somenumber slightly less and plugging in. the other way would beto find the average value assuming the fastest acar can decelerate is 9.8m/s2 , but we probablywouldn't expect it do decelerate any slower than say 5m/s2. Apply the formulavf2=vo2+2ad. d =vf2/-2(-a) or d =vf2/2a and integrate to find the averagevalue which equals67.36m. This means that in the event a driver travelling at this speedwould need to come to a complete stop he or she should expect totravel a distance of approx 73.56m before being able to do so, thetime required to do so being around 2.4s . Therefore in order to be completely safe, adriver would have to maintain a distance between cars aheadproportional to their speed. The faster you go the more distancethere should be between cars. In reality, an ordinary "dangerous"situation would involve the car ahead of you slowing down and thenyou slowing down so not to hit the car ahead. If you are too closeyou may still hit them, but it will not be the exact distancepredicted by the above model. Nevertheless, the fact remains that asafe distance between cars is proportional to velocity (rules forthis you can probably find in a driver's manual.I don't drive). Finally, it is never a good idea to talk on a cellphone whiledriving, however if you are it would be a good idea to be evenfarther from cars ahead because you are not fully focused on theroad, which will increase the amount of time it takes to react to asituation, which will effectively increase the distance you wouldtravel before being able to come to a complete stop. *sorry for the shoddy grammar & typos, it was just a longtrain of thought. which equals67.36m. This means that in the event a driver travelling at this speedwould need to come to a complete stop he or she should expect totravel a distance of approx 73.56m before being able to do so, thetime required to do so being around 2.4s . Therefore in order to be completely safe, adriver would have to maintain a distance between cars aheadproportional to their speed. The faster you go the more distancethere should be between cars. In reality, an ordinary "dangerous"situation would involve the car ahead of you slowing down and thenyou slowing down so not to hit the car ahead. If you are too closeyou may still hit them, but it will not be the exact distancepredicted by the above model. Nevertheless, the fact remains that asafe distance between cars is proportional to velocity (rules forthis you can probably find in a driver's manual.I don't drive). Finally, it is never a good idea to talk on a cellphone whiledriving, however if you are it would be a good idea to be evenfarther from cars ahead because you are not fully focused on theroad, which will increase the amount of time it takes to react to asituation, which will effectively increase the distance you wouldtravel before being able to come to a complete stop. *sorry for the shoddy grammar & typos, it was just a longtrain of thought.