| Topic: |
Science > Physics |
| User: |
"quat" |
| Date: |
09 Apr 2006 08:25:37 PM |
| Object: |
race driving problem |
Going top speed, driver X leads Y steadily by 3 miles. Only 2 miles from
the end, X runs out of gas. Thereafter, x decelerates with time at a rate
proportional to the square of her instantateous speed and in the next mile
X's speed exactly halves. Who wins? (The race track is straight.)
So far I have "x decelerates with time at a rate proportional to the square
of her instantateous speed" implies:
x'' = - a*(x')^2
I also thought to start measuring time (t = 0) right when X runs out of gas,
so at this point, Y is exactly 3 miles behind X and X is two miles from the
end. At this time X has a velocity v0. When X is 1 mile from the end it
has a velocity v0/2.
I think v0 and v0/2 are info to be used as initial conditions? But I need
them in terms of time.
I guess my questions are, is my ODE setup correct or is it missing terms?
How to find the initial conditions in terms of time? How to determine the
proportionality constant a?
.
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| User: "Hexenmeister" |
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| Title: Re: race driving problem |
10 Apr 2006 01:22:13 AM |
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"quat" <spam@void.com> wrote in message =
news:Vti_f.177$Jk2.119@fed1read03...
| Going top speed, driver X leads Y steadily by 3 miles. Only 2 miles =
from=20
| the end, X runs out of gas. Thereafter, x decelerates with time at a =
rate=20
| proportional to the square of her instantateous speed and in the next =
mile=20
| X's speed exactly halves. Who wins? (The race track is straight.)
| So far I have "x decelerates with time at a rate proportional to the =
square=20
| of her instantateous speed" implies:
|
| x'' =3D - a*(x')^2
|=20
| I also thought to start measuring time (t =3D 0) right when X runs out =
of gas,=20
| so at this point, Y is exactly 3 miles behind X and X is two miles =
from the=20
| end. At this time X has a velocity v0. When X is 1 mile from the end =
it=20
| has a velocity v0/2.
|=20
| I think v0 and v0/2 are info to be used as initial conditions? But I =
need=20
| them in terms of time.
|=20
| I guess my questions are, is my ODE setup correct or is it missing =
terms?=20
| How to find the initial conditions in terms of time? How to determine =
the=20
| proportionality constant a?=20
The easy way to work problems like this is to think in reverse, time
running backwards.=20
Suppose you start the clock with X at zero mph, Y at constant
speed and both at the start (finish) line, Y moving.
Obviously Y will get ahead initially, and for X to catch up
he/she must accelerate first to Y's speed and then exceed it.
But X can only go as fast as Y, so the best X can do is be=20
3 miles behind Y.=20
Now you take into consideration that X's speed is v0
two miles from the start line. Hope that helps.
Androcles.
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| User: "ABarlow" |
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| Title: Re: race driving problem |
09 Apr 2006 10:19:30 PM |
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Your ODE is set up properly in terms of x and t; however, if you look
at how your initial conditions are set up, you really aren't looking
for x(t), you're looking for v(x). Consider your initial conditions,
you have:
v(0) = v0
v(1) = v0/2
Hence, we want to rewrite the equation in terms of velocity and
position, not position and time. You will want to make use of the chain
rule, dv/dt = (dv/dx)(dx/dt), for example, would probably be useful.
Once you have v(x), you can probably find x(t) explicitly if you want
it. Solving the second order ODE you have there directly will probably
be very, very ugly.
You should be able to find the constant a and your other integration
constants using your initial conditions, but you may need to introduce
a third one, probably using x(t) where x(0) = 0.
A.
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| User: "Henning Makholm" |
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| Title: Re: race driving problem |
10 Apr 2006 08:20:58 AM |
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Scripsit "ABarlow" <impartially_insane@hotmail.com>
Your ODE is set up properly in terms of x and t; however, if you look
at how your initial conditions are set up, you really aren't looking
for x(t), you're looking for v(x). Consider your initial conditions,
you have:
[...]
Solving the second order ODE you have there directly will probably
be very, very ugly.
It's not that bad, actually. It is really just a fairly easy
first-order ODE in v(t). The general solution is quickly integrated
to find x(t), after which it is not difficult to derive expressions
for t(x) and v(x) and find out what a must be. All the functions turn
out to have closed elementary expressions.
The tricky bit is that the actual speeds are underdetermined, because
all the given measures in the problem text are lengths. So after
having derived the various expressions one has to _imagine_ that for
example v(0) is known and fix the integration constants in terms of
it. Later this assumed v(0) ought to disappear from the expression
that tells how far the trailing car goes in the time until the leading
one passes the finish line.
--
Henning Makholm "Unmetered water, dear. Run it deep."
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