Race cars - Differential Equation

Reuel · 2011-03-09 04:36:58

I am still trying to work through some of these word problems but this one has me stuck.

"Leonardo had been in the lead ahead of his rival race car driver Michelangelo for some amount of time by a constant 3 miles. Just 2 miles from the finish, however, Leonardo's gas tank ran empty and his car decelerated ever after at a rate proportional to the square of his remaining speed. Within another mile, Leonardo's speed had halved exactly. If Michelangelo's speed remained constant, who won?"

Should this be set up as two different equations, one for each driver? The lacking of numbers for speeds is confusing. I dunno, that snowplow problem came after a little effort and this one seems as if it should be a lot alike, but I could use a little help with this one.

Thanks ahead of time.

bobbym · 2011-03-09 05:11:09

Hi Reuel;

I still do not have the required DE but a discrete analysis of it seems to yield that Michelangelo's car wins. Depending on the speed though.

Reuel · 2011-03-09 05:28:04

Hey bobbym, haven't talked to you in a few weeks.

Well, we know that the relevancy of the problem begins when Leonardo is 2 miles from the finish line and, at that instantaneous moment, when his car runs out of gas, Michelangelo is 3 miles behind him, so the total span is 5 miles.

When Leonardo's car begins to slow its deceleration is proportional to the square of his remaining speed. Is this a good way of stating it mathematically?

Negative because he is decelerating?

We also know that after one mile (which is half the distance) his speed had also halved. Does that mean his deceleration is actually linear?

Thanks.

bobbym · 2011-03-09 05:35:13

That is the DE I am playing with. I have solved it for the general case and am trying to determine the constants.

Reuel · 2011-03-09 05:43:29

I am very thankful for the help. I guess I am having an off-week.

More information: The difference in velocity between Leo and Mike before Leo ran out of gas was 0 which means that as Mike continues along, he does so at Leo's original velocity as Leo decelerates. That Leonardo reached half his speed in half the distance to the finish line, it sounds as though he might completely stop on the finish line and the whole time he is working up to doing so, Michelangelo has been closing the 3-mile gap between them at Leonardo's original speed.

Hmmmmmm...

bobbym · 2011-03-09 05:48:37

I am having trouble with determining the constants.

I am getting this but I do not understand the answer.

Solve these two simultaneously.

That is the point in which the two drivers finish at the same time.

Reuel · 2011-03-09 06:19:18

I don't understand the answer either. Or how you arrived at it.

I am working on it, too, but with none much luck thus far.

bobbym · 2011-03-09 06:20:19

Did you solve the DE?

Reuel · 2011-03-09 06:24:10

I solved

to get

For time zero I simply set it equal to some number s so that C=s.

bobbym · 2011-03-09 06:25:20

Hi;

I think you forgot the square.

Reuel · 2011-03-09 06:33:56

That could potentially pose as a problem.

I see that.

Might all of this be best put into terms of position rather than velocity? Could that in some way be made to work with the positions of the cars?

In your's, where did

come from?

Last edited by Reuel (2011-03-09 06:37:42)

bobbym · 2011-03-09 06:42:20

At

v(0) = v where v is the initial velocity. Which we do not know.

v(1) = v / 2

Now solve for k and C

Reuel · 2011-03-09 07:45:18

Are you still working on this? Because I am completely stuck. I thought I had a good idea but... no.

bobbym · 2011-03-09 07:47:05

Hi;

Yes, I am. I am not sure of the solution at all. What are you stuck on?

Reuel · 2011-03-09 07:48:47

I was going to try to find position functions for both of them but Leonardo is hard to do because of limiting information. I keep ending up with too many unknowns.

bobbym · 2011-03-09 08:00:32

I would expect that. What does yours look like?

Bob · 2011-03-09 08:48:13

hi Reuel and bobbym,

I think the following might get you out of trouble:

Let distance from the 2 mile point be s (in miles)

so

when s = 0 , v = U (the common velocity until the moment the fuel runs out)

when s = 1 v = U/2

Now you have your constants.

M has five miles to go at U mph so you can specify t when he finishes.

So where is L then?

Write v = ds/dt and get an equation linking s and t to answer this.

Bob

Math Is Fun Forum

#1 2011-03-09 04:36:58

Race cars - Differential Equation

#2 2011-03-09 05:11:09

Re: Race cars - Differential Equation

#3 2011-03-09 05:28:04

Re: Race cars - Differential Equation

#4 2011-03-09 05:35:13

Re: Race cars - Differential Equation

#5 2011-03-09 05:43:29

Re: Race cars - Differential Equation

#6 2011-03-09 05:48:37

Re: Race cars - Differential Equation

#7 2011-03-09 06:19:18

Re: Race cars - Differential Equation

#8 2011-03-09 06:20:19

Re: Race cars - Differential Equation

#9 2011-03-09 06:24:10

Re: Race cars - Differential Equation

#10 2011-03-09 06:25:20

Re: Race cars - Differential Equation

#11 2011-03-09 06:33:56

Re: Race cars - Differential Equation

#12 2011-03-09 06:42:20

Re: Race cars - Differential Equation

#13 2011-03-09 07:45:18

Re: Race cars - Differential Equation

#14 2011-03-09 07:47:05

Re: Race cars - Differential Equation

#15 2011-03-09 07:48:47

Re: Race cars - Differential Equation

#16 2011-03-09 08:00:32

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#17 2011-03-09 08:48:13

Re: Race cars - Differential Equation

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