Math Problems

[Elanor's Angel]

Thank you Janny! Right now I only know it to the 60 dig. after the dec. but I'll still probably get to 100+ by the end of the year.(It's taken me 4 days so far to get from 0 dig to 60)
About the race problem, I live in America, so I was going by Miles per hour. So in order for him so get an average of 60mph with the first lap at 30 he would need a speed of 90 mps on the next lap.

If it's still wrong I'm blameing it on a fried brain, from memorizeing Hamlet's advise to the players, the legend of narsil (type thing), and 60 dig of pi in the last four dayes with almost no sleep.(plus haveing to do a TON of homework)

[Radagast The Brown]

The mph or kph doesn't matter.
Logically - let's say one lap is 120 mlies (it doesn't matter, it just makes my calculation easier). if he does the first lap in 60mph average, you'd do it in 2 hours. Meaning... two laps in 4 hours.
If you do the first lap in 30 mph average, you do it in 4 hours. You can see you have absolutely no time to do the next lap?

If you want a mathemtical explanation, check Joanthan's post. He's much better in this than me. .

[Elanor's Angel]

oh... I still don't get it. *opens head for all to see*
If say, the corce is 1 mile long, and he does it in 1 min. then for that lap his average mph is 60 mps... then say he does the second in 2 min, then his average on that lap is.... *stops abruptly*, 2 min...OOOOOOOOOOOOOH! I get it! Slams head on desk.
(i am so dumb.)

[Janny]

You got there.

The maths dep leant me a cool graphics calculator (cuz I'm taking further maths) so I drew it and you get this nice little asymptote to 60.
Incidentally, I believe 30 is the cut off point. A first lap of 30.00000001 would have a solution... I think.

[Jonathan]

Yes, you're quite right. If the driver had had a better speed than 30 in the first lap, he would have been able to get an average speed of 60.

[Janny]

Because 1/x = 0 when 1/30 + 1/x = 2/60, right?

Mind you, at 31 mph the second lap has to be something like 960 mph...

[Jonathan]

Well then it would still be theoretically possible to reach 60

[Elanor's Angel]

Yeah, I was just being an idiot.

[Janny]

Okay. As a result of watching the news, my father became curious about the bullet that are fired into the air (trival I know compared to what the news actually was).
I believe 180 m/s (400 mph) is a reasonable estimate for the speed of a bullet from a rifle, and I was going to model it as a 1D thing.
Using V^2 = U^2 + 2as and assuming gravity to be 9.8 m/s/s
=> 0 = 32400 - 19.6s
=> s ~= 1653 m (about a mile)

I worked out s first because I was curious how high they go. But is this right? Is it wrong to model it under constant acceleration of 9.8? How high do you have to go before gravity changes?

Disregarding this and assuming gravity does not change, it then comes back to earth at 180 m/s. I'm guessing that this is greater than the terminal velocit of a bullet... but how do you calculate that? I know there is a coefficient that I need... but I can't find it...

Of course, who knows, they may be firing blanks.

[Jonathan]

Originally Posted by Janny

I worked out s first because I was curious how high they go. But is this right? Is it wrong to model it under constant acceleration of 9.8? How high do you have to go before gravity changes?

I'd say it's right to say the acceleration is constant. You'd have shoot the bullet way higher than a mile up in the sky to notice any change in gravity.

Originally Posted by Janny

Disregarding this and assuming gravity does not change, it then comes back to earth at 180 m/s. I'm guessing that this is greater than the terminal velocit of a bullet... but how do you calculate that? I know there is a coefficient that I need... but I can't find it...

I'm not really following, do you mean an air resistance coefficient? Or a coefficient that tells you how the force of gravity decreases with increasing altitude? If you mean the latter, the force decreases by 3 millionths m/s² with every metre from the centre of the earth. A bullet will have a hard time escaping gravity

Originally Posted by Janny

Of course, who knows, they may be firing blanks.

Depends . If your father watched news from Falluja, I guess they used sharp ammunition.

[edit]Yay, found my physics formula book in my old school bag! Thought I had returned it to the teacher before graduation. I'm glad I didn't

[Janny]

Originally Posted by Jonathan

I'm not really following, do you mean an air resistance coefficient? Or a coefficient that tells you how the force of gravity decreases with increasing altitude? If you mean the latter, the force decreases by 3 millionths m/s² with every metre from the centre of the earth. A bullet will have a hard time escaping gravity

I think I mean the former. Basically, what would be the terminal velocity of a bullet coming down from 1.6 km in the air?
How would you work it out?

Thanks

[Jonathan]

Hmm, on its way down, the bullet would eventually fall at a constant speed of 9.8/k metres per second, where k is the coefficient of friction. I don't what the air's friction would be, but it would change with the altitude I suppose. The air is a bit thinner a mile up in the sky.

[Janny]

Yes, it was that k I wanted.

I don't suppose it matters that the air is thinner further up, given that objects travelling faster than terminal velocity slow down until they reach it. It would then be k just above the surface of the earth which is important.

Edit: If k is not less than 1 it's going to be limited to 9.8 m/s anyway...
which would still make a dent I suppose.

[Jonathan]

Well if you know the mass of the bullet, the air resistance will be mass · 9.8 Newton. Though that doesn't say much about the friction at all... Bah, it's midnight and I'm too tired to think

[Nurvingiel]

Isn't terminal velocity the fastest any object can fall given its mass? It would also need a certain amount of height to achieve this velocity - I believe a heavier object would take longer to reach it.

But actually I think Jonathan answered the question already... now I'm confused.

[Janny]

I think there's some such effect such that an object 9 times heavier would take three times longer (i.e. a square root).

"weight · 9.8"
Do you mean mass? As in F = ma.

[Jonathan]

Yes but let's not get into detail . It's enough to know that what goes up must come down

[Janny]

Sounds like you're ducking out of a calculation there!

[Mercutio]

i'm having trouble with this precalc problem:

the diagram:

 PROBLEM:  Prove all medians intersect at the same point (centroid)
using coordinate geometry (no theorems allowed except pythagorean, but I
don’t believe that’s useful here)

So, I just find the equations for the lines and then algebraically solve for
their common solution, right?

Well…First let’s find the slopes (m) ---  (y1-y2/x1-x2)

m RT = (b/2) / (a/2 – c)     m PU = (b/2) / (a – c/2)   m QS = (b) / (c/2 + a/2)

If it’s simpler, with the original coordinates I couldn’ve used P as (2a, 0 ) for
example, and then have less halves through the rest of the stuff.  Here’s the 
slopes that way:

m RT = b/(a-2c)     m PU = b/ (-c+2a)   m QS = 2b/(c-a)

^Ok so I may not have quite typed that in right, anyway, let’s keep going…

Now plug them into point slope form for lines:   y – y1 = m (x – x1) 

*Note when I did this, I checked it with real numbers for a, b, and c, and I 
got a single point as there intersection.  I assume than that the line 
equations are correct
Also, in my original sketch, point S was on the other side of the y axis and I 
used R as (-c , 0); I hope that won’t throw you off.

Equations:

RT  y = (bx+bc)/(a+2c)      QS  y = (2bx)/(c-a) + b       PU  y = (bx-ba)/(-c-2a)

Now, theoretically I should be able to solve two equations simaltaneously, and 
then two others, and then those two answers…yadda yadda yadda… and get 
(x, y) of the centroid in terms of a, b, and c.  But it ain’t happen’ baby.    Did 
I miss something along the way?

[Janny]

Love to help, dear Mercution, but the link is broke.