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hey people.
need a little help here.
i need to get information on the amount of time we need to stop our car in time,like the mathematical equations and everything, and the factors that will make you a safe driver. im even sure where to start. anyone with any ideas are welcomed. thanks alot!
rach
Igor Khavkine
Sep14-04, 12:16 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nrach <heavens_gal@hotmail.com> wrote in message news:<rach.1c8a8g@physicsforums.com>...\n> hey people.\n> need a little help here.\n>\n> i need to get information on the amount of time we need to stop our car\n> in time,like the mathematical equations and everything, and the factors\n> that will make you a safe driver. im even sure where to start. anyone\n> with any ideas are welcomed. thanks alot!\n\nNewton\'s third law: a = F/m. a - acceleration, F - force, m - mass.\nKinetic energy: K = mv^2/2, v - velocity.\nWork done by a force: dW = F.dx, dx - infinitesimal displacement.\n\nAssuming that your car\'s velocity is v when you start breaking, and that\nthe force F of friction from the road is constant, the distance travelled\nbefore you come to a complete stop is\n\nx = mv^2/(2F),\n\nand the time taken to stop is obtained by solving the equaition\n\nx = vt - Ft^2/(2m).\n\nThe force F of the road on the car can be obtained as a friction force:\n\nF = mg mu_s, if the car is not silding, g - acceleration due to gravity,\nmu_s - coefficient of static friction between the road\nand the tires,\n\nor\n\nF = mg mu_k, if the car is sliding, mu_k - coefficient of static friction\nbetween the road and the tires.\n\nHowever, I believe that becoming a better driver can only be a matter\nof experience and good judgement.\n\nIgor\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>rach <heavens_gal@hotmail.com> wrote in message news:<rach.1c8a8g@physicsforums.com>...
> hey people.
> need a little help here.
>
> i need to get information on the amount of time we need to stop our car
> in time,like the mathematical equations and everything, and the factors
> that will make you a safe driver. im even sure where to start. anyone
> with any ideas are welcomed. thanks alot!
Newton's third law: a = F/m. a - acceleration, F - force, m - mass.
Kinetic energy: K = mv^2/2, v - velocity.
Work done by a force: dW = F.dx, dx - infinitesimal displacement.
Assuming that your car's velocity is v when you start breaking, and that
the force F of friction from the road is constant, the distance travelled
before you come to a complete stop is
x = mv^2/(2F),
and the time taken to stop is obtained by solving the equaition
x = vt - Ft^2/(2m).
The force F of the road on the car can be obtained as a friction force:
F = mg \mu_s, if the car is not silding, g - acceleration due to gravity,
\mu_s - coefficient of static friction between the road
and the tires,
or
F = mg \mu_k, if the car is sliding, \mu_k - coefficient of static friction
between the road and the tires.
However, I believe that becoming a better driver can only be a matter
of experience and good judgement.
Igor
Paul Draper
Sep17-04, 05:32 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nrach <heavens_gal@hotmail.com> wrote in message news:<rach.1c8a8g@physicsforums.com>...\n> hey people.\n> need a little help here.\n>\n> i need to get information on the amount of time we need to stop our car\n> in time,like the mathematical equations and everything, and the factors\n> that will make you a safe driver. im even sure where to start. anyone\n> with any ideas are welcomed. thanks alot!\n>\n> rach\n>\n\nWhat you are probably more interested in is the DISTANCE needed to\nstop the car, since most folks can\'t judge the time between now and\nwhen they\'ll collide with that rocking chair that just fell off the\npickup truck in front of them.\n\nRoughly, the distance required to come to a stop from a velocity v is\nd, where d = v*v / (2 * mu * g).\nHere g is the gravitational acceleration, a number that depends on\nwhat units you\'re using (9.8 m/s/s if working metric), and mu is a\nnumber that depends on the tires you bought and the road conditions\nand whether you have an anti-lock brake system. If you lock the brakes\nand skid, or if the road is wet, or if you have cold tires, mu goes\ndown and the distance to stop increases. Police have tables for mu,\nwhich is how they can look at the length of the skid marks and figure\nout how fast you were going when you panicked and hit the brakes. Note\nthat it will take you NINE times as far to stop from 120 kph as it\ndoes to stop from 40 kph.\n\nDriver safety classes say that you should maintain, for every 10mph or\n15kph, one car length between you and a MOVING car in front of you.\nYour homework assignment is to a) find out why this is a linear\nrelationship with speed and not proportional to the square of the\nspeed, and b) how you should modify this rule to avoid hitting\nstationary objects, like that rocking chair.\n\nPD\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>rach <heavens_gal@hotmail.com> wrote in message news:<rach.1c8a8g@physicsforums.com>...
> hey people.
> need a little help here.
>
> i need to get information on the amount of time we need to stop our car
> in time,like the mathematical equations and everything, and the factors
> that will make you a safe driver. im even sure where to start. anyone
> with any ideas are welcomed. thanks alot!
>
> rach
>
What you are probably more interested in is the DISTANCE needed to
stop the car, since most folks can't judge the time between now and
when they'll collide with that rocking chair that just fell off the
pickup truck in front of them.
Roughly, the distance required to come to a stop from a velocity v is
d, where d = v*v / (2 * \mu * g).
Here g is the gravitational acceleration, a number that depends on
what units you're using (9.8 m/s/s if working metric), and \mu is a
number that depends on the tires you bought and the road conditions
and whether you have an anti-lock brake system. If you lock the brakes
and skid, or if the road is wet, or if you have cold tires, \mu goes
down and the distance to stop increases. Police have tables for \mu,
which is how they can look at the length of the skid marks and figure
out how fast you were going when you panicked and hit the brakes. Note
that it will take you NINE times as far to stop from 120 kph as it
does to stop from 40 kph.
Driver safety classes say that you should maintain, for every 10mph or
15kph, one car length between you and a MOVING car in front of you.
Your homework assignment is to a) find out why this is a linear
relationship with speed and not proportional to the square of the
speed, and b) how you should modify this rule to avoid hitting
stationary objects, like that rocking chair.
PD
J. J. Lodder
Sep19-04, 06:56 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nPaul Draper <pdraper@yahoo.com> wrote:\n\n> Driver safety classes say that you should maintain, for every 10mph or\n> 15kph, one car length between you and a MOVING car in front of you.\n\nCar length is pretty irrelevant under the circumstances.\nA better rule is that you should keep at least 2 seconds distance.\nFor example 60 m when travelling at 30 m/s (= 108 km/h)\n\nExperience bears this out:\nThe chance of a chain collision occurring increases greatly\nwhen the flow becomes larger than about 2000 cars/hour.\n(As it too often is in these crowded days)\n\nBest,\n\nJan\n\nPS For those amused by the subject (and not afraid of equations)\nthe little book: Prigogine, Kinetic Theory of Vehicular Trafic\nmay be worth a look at, if your library has it.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Paul Draper <pdraper@yahoo.com> wrote:
> Driver safety classes say that you should maintain, for every 10mph or
> 15kph, one car length between you and a MOVING car in front of you.
Car length is pretty irrelevant under the circumstances.
A better rule is that you should keep at least 2 seconds distance.
For example 60 m when travelling at 30 m/s (= 108 km/h)
Experience bears this out:
The chance of a chain collision occurring increases greatly
when the flow becomes larger than about 2000 cars/hour.
(As it too often is in these crowded days)
Best,
Jan
PS For those amused by the subject (and not afraid of equations)
the little book: Prigogine, Kinetic Theory of Vehicular Trafic
may be worth a look at, if your library has it.
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