Math Is Fun Forum

Wrong! It doesn't. The bot is confusing actual velocity with how much we want to alter the velocity.  It takes more effort if we want to alter velocity a lot, but Newton's law just says velocity stays constant until we apply an external force.  The amount of force determines how much the velocity changes but the start velocity is irrelevant.  That's why we have frames of reference; otherwise we've got to take account of all the possible velocities (orbits, movement of galaxies etc.)

It takes the same force to change velocity from say 10 m/s to 5 m/s whether the start velocity is 10 m/s or 10 million m/s.

Conclusion: Don't trust bots to tell you anything.

Bob

hi polarisgleam

Welcome to the forum.

If you look back through this thread you'll see that it's not just a case of adding 13. That's too easy for this forum.  Members try to find a calculation (the more interesting, the better) to make the next number.

Such as 2 x 2 x 7 x 13 x 83 = 30212.

Bob

It has to be like that because of all the other motions that we're all involved in.  eg. the Earth is hurtling through space (could estimate the speed but hopefully my point stands without that).  If you had to stop that speed too you'd need a lot of force just to take the dog for a walk!

Bob

There is a formula for the sum of n terms.  The sum to infinity will be infinite so the formula for that is

Do you know the finite sum formula? Would you like me to derive it?

Bob

I did! That looks very good.  I cannot install code; only MIF can do that.  Try emailing him at funmath@gmail.com

Bob

That's exactly what I meant.

Bob

I've made up some constraints to illustrate what I mean.

I'll define a large order as amount > 600
Medium as 250< amount < 600 including these endpoints
And small as amount < 250

If everything were to be reduced in the same proportion
then k = target amount / actual amount

But we want k for small to be 1, k for large to be sufficient to reach a total of 3000  and k for medium to be less than k for large.

I'll make k for medium to be 60 % of k for large.

This gives me a way to calculate k for large

The total for new large plus new medium is 3000 -200-50

So L.k + M - M(1-k)x0.6 = 2750

Rearranging k = (2750 - 0.4 x 500)/(4400 + 0.6x500)
= 0.54255...

Using this the new values are
1898,  488,  362,  200,  50 giving a total of 2998.

This is because I've rounded every calculation down.

So apply WHILE total < 3000 medium = old medium + 1 so this loops until 3000 is reached.

Bob

Oh yes,certainly.   I just used those as examples.  There's no limit to the number of possible frames.

Bob

Have I told you where to look for a long list of these?

Go to help and you'll find it in the third thread.

The thread existed long before I joined but it was hidden. I stumbled on it one day and copied it into the post in help.me
The whereabouts of the original is a mystery but I've added to my version so that's the place to go.

Do I know them all? Certainly not!  I don't fill my head with stuff like that

No idea  why they are given those names. Someone thought it was a good name I guess.

Bob

Italics means sloping

Emphasis means bold

Monster?  Nay! Admin have super powers to disintegrate unwanted characters.  And a character has two meanings; both of which are vulnerable.
B

Latex commands are 'called' by inserting them between the opening and closing command math

It used to be possible to suppress the action be using another bbcode command, code, but it seems to no longer work, probably happened when the forum switched to a new server.  So I'll leave off the first open bracket and the last close bracket to avoid the code interpreter trying to action the code.  Then I'll repeat but with all the brackets, so you can see what the command does.

math] commands here [/math

So to make a simple fraction

math] \frac{3}{4} [/math

Here's an integral

math] \int_0^\infty x^{x+2} + \alpha x .dx [/math

Years ago the member Dross wrote a tutorial to get people started https://www.mathisfunforum.com/viewtopic.php?id=4397

and others added to the thread.  I find the code commands are blacked out now and some other commands no longer work but there's still a lot of useful stuff here.  I go there if I want something I haven't used before eg. the infinity in the line above.

You can also do an internet search for the right command but this can be frustrating as not all are implemented here.

^ for to the power of works simply if followed by a single character eg

math] x^7 [/math

but if you want a more complicated power the power must be enclosed in {}

math] x^{y + 7} [/math

Otherwise this happens

If you see some Latex and want to see the underlying commands just click on it.

Hope that gets you started.

Bob

Usually I can answer your questions based on what I was taught and passed on in my own teaching.  This one is the exception. In all the Newtonian mechanics and physics questions I have encountered, inertia has never occurred.

I had to go up in my loft and get down all my old texts on applied maths and physics.  Some just refer to Newton and  basically say bodies possess the property of inertia, that is they do not accelerate unless acted upon by an external force. 

I also tried googling the units for inertia. I got two answers:

Mass is a measure of an object's inertia. The more mass an object has, the more inertia it has.

The SI unit for inertia is kilogram-meter squared (kg⋅m²). While the term "inertia" is often used to refer to an object's resistance to changes in motion, the actual concept of resistance to change is often referred to as moment of inertia. The moment of inertia is a measure of how an object resists rotational acceleration.

The first is inertia as it applies to linear motion and the second to rotating objects. 

The measure of linear kinetic energy is 1/5 . m . v^2 and for rotating objects 1/2 . I / w^2 where I is the moment of inertia and w the angular velocity.

We're not the only ones confused. Have a look at this thread :

https://physics.stackexchange.com/quest … n-stopping

So there we are. Everyone is confused and there's no one answer.

LATER EDIT: I've given this some over-night thought.  For an object with zero or uniform velocity it's inertia is because of its mass.  More mass, more inertia.  It's actual velocity is irrelevant.  The Earth rotates on its axis, and around the Sun, which in turn, rotates around the centre of the Galaxy, which is moving relative to other galaxies.  That's why we have the concept of frames of reference.

If you consider that trolley going at various speeds, then, yes, the amount of force to reduce that velocity to zero will vary, but that doesn't alter its inertia. It just gives you the means to do the calculation.  If you set up a frame of reference that travels at 6 m/s in the second case, then the trolley is only travelling at 6 m/s relative to that frame and so can be 'brought to rest' in the frame with the same effort as the first case.

Momentum gives you a way to measure the force required, but I showed in a previous post that impulse is just a way of applying v = u + at to determine F.

Bob