Math Is Fun Forum

Ok so I didn't read the question.

So now I'm getting

3^(n-1)       (1/2)^(n-1)       (3/4)^(n-1)

When n=1 don't we want just one white triangle?

Bob

hi Phro,

I agree with your answers to a and c but not b.

When we start I'll take it that n = 0; we have a single white triangle side =1

At step 1 the white triangles have side = 1/2 = 1/(2^1)

At step 2 the white triangles have side = 1/4 = 1/(2^2)

.....................

If you spot a sequence and use it to get algebraic answers that seems ok to me. But all that's missing is to prove that the sequences are valid.  You can do that by, for instance, in each step the previous number of whites is made into four new smaller triangles, one of which is blue so 3/4 are still white. So each step increases the number of whites by a multiplier of times 3/4

Bob

I'm catching up fast,  On a weather map the region is the surface of the Earth (part of). Temperature is a scalar as it has just magnitude but not direction. Every point has a temperature so the 'field' of tempeatures matches the real world of the surface.

You can show wind simlarly but, as wind has magnitude and direction, this time it's a vector field.

Bob

Your acceleration calculations look ok. Just one thing bothers me. Does it make a difference if the force is applied for a measured amount of time. I've searched for an answer to this and made no progress.  Why ask this question?  If you apply a force of 200N for, say, 5 seconds then the impulse is 200 x 5 = 1000 N.sec and this gives that amount of momentum to the two objects.

Momentum before = 0

Momentum after = 1000

If I consider myself ( as the pusher) as at rest with the box moving away at velocity v after 5 seconds then

50 x v = 1000 so v = 1000/50 = 20 m/s

Using v = u + at     20 = 0 + 5a     so a = 4m/s^2

Oh joy! That's what you got before. So maybe all's well in the world of Newtonian mechanics.

Do you drive yet? Don't try to experience this 'box' effect by driving a car into a tree. The tree will definitely exert a big force on you!

Bob

Yes as it's 10 to the whole expression.

I'm of an age where there were no calculators to use for calculations. So logs were taught early on as you can make use of this property:

This converts a multiplication into an addition. Similarly divisions and roots.

Bob

Your calc leading to -50 is a good one. 

Yes to the frames question.

Speed is a scalar quantity so I wouldn't expect it to be negative. But velocity is a vector so direction is needed. When you start such a question decide which way is positive and if any calc leads to a negative it means it's going the other way.

eg. A ball is thrown vertically upwards at 27 m/s. Taking g as 10, what is its height after 5 seconds.

I'll take 'up' as positive.

g = -10; t = 5; u = 27

using s = ut + 0.5at^2

height = 27 x 5 - 0.5 x 10 x 5^2 = 135 - 125 = 10m (above the ground)

I'll take 'down' as positive

g = 10; t = 5; u = -27

height = -27 x 5 + 0.5 x 10 x 25 = -135 + 125 = -10m  (As down is positive this ball must be 10 m above the ground.

Suppose u = 7, and I throw the ball up leaning sightly over a cliff.

Take 'up' as positive.

height = 7 x 5 - 0.5 x 10 x 25 = 35 - 125 = -90m   The ball has gone up, come down and carried on downwards below my level down the cliff.

When was the ball again level with my position?

t = ?; u = 7; g = -10; s = 0

0 = 7t - 0.5 x 10 x t^2    =>       7t - 5t^2 = 0       t(7-5t) = 0 So we have two solutions, t = 0 and t = 7/5

We should expect the first as we know the ball was at that height when we threw it and started timing. The second is the one that answers the question.

Bob

The way I was taught to do this is to bring the 'WRT-object' to  rest by applying an equal and opposite speed to it, and apply the same speed to everything else.  As you are stationary that means the WRTs in the parts 3 and 4 are just the given information with no change needed.

For part 1 bring the train to rest by applying a speed of 30 m/s in the opposite direction.  Let's say, to avoid confusion, that the train is heading North, and so is the boarder.  So make the whole world go South at 30 m/s so that the train is brought to rest. The boarder is also given a speed South of 30 m/s so its speed North is now (5 - 30) m/s  ie. 25 m/s South.  If a person is on the train they think they're stationary and the whole world is travelling South at 30 m/s. Except the boarder of course who seems to be travelling South at 25 m/s.

You can try this yourself for part 2. You should get the exact opposite answer to part 1

So add the boarder's speed to the train's to get 30 + 5 m/s

Bob

OK, got it.

Are you familiar with matrix multiplication?  If not then there are two useful pages here:

https://www.mathsisfun.com/algebra/matr … lying.html

https://www.mathsisfun.com/algebra/matr … lator.html

In 2D transformation geometry 2 by 1 vectors (coordinates switched row and column) can be transformed by multiplying by a 2 by 2 matrix.

As any such matrix transforms (0,0) to (0,0) this only works when the transform leaves the origin invariant.  The one for the question does as y=x goes through the origin.

I'll call the matrix

y=x is invariant so (x,x) maps onto (x,x)

This gives us two equations

a + b = 1 so b = 1-a         and c + d = 1 so d = 1-c

So the matrix becomes

Now to fix which stretch by considering a single point under the transformation.

I'll start on the line at (2,2) and go one right and one down to (3,1) and again to (4,0)

That's a good point to consider as there's a zero in the calculation.

It's image is at one right and one down and the same again ie at (6,-2)

So (4,0) maps onto (6,-2)

This leads to a = 1.5 and c = -0.5

So the matrix for this transformation is

So what does this do to our points?

QED

Bob

hi rk_nithi

Welcome to you as a new member!

Easy bit first. This is a vector graphics program called Geometer's Sketchpad.  There is also a free program called Geogebra which works similarly.  Geo is more versatile but takes longer to learn.  Sketchpad costs money but I got it years ago (before Geo) and it still serves me well. I like the control it gives over thickness of lines and colours.

Transformations that leave the origin invariant can be represented by a matrix transformation.  So, in theory there must be one for this one but I'll have to do a bit of work to find it.  I'll come back to this later.

Bob

I've definitely been blown along by a wind on my back.  So you're correct.

The website should have said  "Every time the surface of a solid collides with an air molecule, there is force acting on the solid.  If the solid is trying to push through the air, this force will slow it down."

Bob

Not everything on the internet is true.

Physicists sometimes get things wrong; eg. saying there's a thing called centrifugal force.

Yes it is!!!  The forum is limping a bit.  You could try sending MIF a message.

Bob

Although I'm an administrator I cannot use code like that either.  Only MIF himself would be able to fiddle with the fundamental code underlying the forum.

Some BBCode commands are available, chiefly to enhance text. eg square brackets u for underlining, b for bold, huge for large text, color for colours.  Not all of the commands in BBCode work here I'm afraid. The forum moved to a new server several years ago and some commands which had worked now don't.

You can put in a link with url as I did in my previous post.  I think the format allows a 'title' for the link but I deliberately don't use it as I want members to be able to see where they're going when they click it.

http://www.mathisfunforum.com/viewtopic.php?id=30179

For smart maths formating you can use LaTex http://www.mathisfunforum.com/viewtopic.php?id=4397

This thread was made years ago, long before I joined.  You'll see the effect of the server change when you notice some code that clearly worked at the time, now doesn't.

Bob