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#2 Re: Help Me ! » Position-time graphs; acceleration » 2025-01-30 07:45:48

Bob wrote:

That's why you need to be careful believing everything you read off the internet; especially AI.*

If you're not told the initial velocity you must not assume it is any value at all.  For some problems an object might be starting from rest but, in this problem we know it isn't.

For the whole of the 3 seconds it's travelling at a constant velocity; there's no such thing as instantaneous acceleration; so it's starting velocity must be 5 m/s.

Bob

* note: I am not a robot.  I've had to prove this loads of times by ticking boxes, even when I'm asked to tick all the squares containing a bike and one square has a tiny, tiny bit of a bike wheel trying to trick me.  I have a grade A (when A* didn't exist) in applied maths ie. mechanics and I've taught it for 37 years. Doing v = u + at problems is something I'm pretty good at. smile

Thanks a lot, Bob.

I believe you! I've not quite grasped it yet, but I believe you smile

Just to clarify;

1. Was Google AI correct when they said 0m from the origin TYPICALLY means the object at rest? Or is even that incorrect?

2. What's the first thing that tells us, in this example, that there is no acceleration; is it the straight line slope? What else tells us?

3. What would the graph (the line) look like for an object starting at rest, accelerating from 0m/s to 5m/s, then maintaining 5m/s for several seconds; would it be curved for that first second, then straight for the remaining time?

**Respect for teaching for 37 years! smile  Stephen Fry (and me!) get annoyed at the lack of respect (from some) for teachers. **

#3 Re: Help Me ! » Position-time graphs; acceleration » 2025-01-29 23:56:56

From Google AI Overview;

Yes, in the context of physics, "0m from the origin" typically means an object is at rest, as it indicates the object is located at the starting point and not currently moving anywhere, with a position of zero displacement from the origin.

#4 Re: Help Me ! » Position-time graphs; acceleration » 2025-01-29 23:50:44

Bob wrote:

Constant velocity definitely means no acceleration.  The given information does not state that u=0 and I think that's why you're confused.  If you take u = v = 5

then v = u + at     => a = (v-u)/t = (5-5)/3 = 0

Bob

Thanks, Bob.

What did you mean when you said, "...the given information does not state that u=0..."

And why did you suggest we take the initial velocity to be 5m/s? I get that if we looked at it from 1s to 3s then u = 5. But my example was from 0s to 3s.

I said that, "xf-xi = 15m-0m = 15m"

So it's position at the outset was 0m from the origin, i.e, it was at rest. Is that correct; 0m from the orgin means at rest? I suppose I could leave my house to run round the block a few times and, when I passed my home, I would be 0m from my origin but still in motion (?) In that case, how do we know what is meant by 0m from the origin?

Can we ever assume the object is at rest, when it's 0m from its origin? And if so, it has to accelerate from 0m/s to 5m/s, before it can then maintain the constant v of 5m/s (?)

NB. I'm thinking of position-time graphs with position axis going from 0m to xm, and time axis going from 0m to xs, and the line being a straight diagonal from the origin at 0m, 0s.

#6 Help Me ! » Position-time graphs; acceleration » 2025-01-28 21:36:14

paulb203
Replies: 9

I’m told that a straight line on a position-time graph represents constant velocity and therefore no acceleration (the line is curved when there is acceleration).

But take the following scenario.

An object's change in position over 3 seconds is as follows;

xf-xi = 15m-0m = 15m

So, s(displacement); 15m
t; 3s
v=s/t
v=15m/3s
v avg=5m/s

And,

a=v-u/(t)
a=5m/s-0m/s/(3s)
a=5m/s/(3s)
a=5/3m/s^2

So we have an acceleration of 5/3m/s^2. Yet we have a straight line (which is supposed to represent constant velocity and therefore no acceleration (?).

**

So what am I missing?
Acceleration = change in velocity over time
Yes, there is no change in velocity between 1s and 2s. And between 2s and 3s. But there is a change in v between 0s and 1s (the object’s v changes from 0m/s to 5m/s). And we work out avg a over the full duration (3s).

**

Should I be focusing on the first second, where there is a change in v?
a=v-u/(t)
a=5m/s=0m/s/(1s)
a=5m/s/(1s)
a=5m/s^2
Hmmm..?
If I’ve calculated correctly the a over the 3s is 5/3m/s^2, and the a over just the first second is 5/m/s^2. Yet a straight line is supposed to represent zero acceleration.

#8 Re: Help Me ! » Velocity Time graphs » 2025-01-18 23:45:31

Thanks, Bob.

And, sorry, I didn't think I had submitted this one. I previewed it, saw all the code, and realised I don't know how to post an image on here (so I posted a thread on how to post images/photos) and thought I'd abandoned this post. Doh!

Anyway. Here is the question without all the code;

At what time does the mouse have the same position as t=0s?

I worked it out, eventually (3.5s). Did you get 4.5s?

It was the part, “...where rightwards is the positive velocity direction,” that had me puzzled.
Just to double check I’ve now got that;
Does it mean; when the velocity is positive, the displacement is to the right (and therefore, when the v is negative, the displacement is to the left)?

#9 Help Me ! » Velocity Time graphs » 2025-01-17 23:58:27

paulb203
Replies: 4

A mouse is trying to cross the street. Its velocity
\[v\] as a function of time
\[t\] is given in the graph below where rightwards is the positive velocity direction.
A set of black coordinate axes are given with the vertical axis labeled "v (m/s)" and the horizontal axes labeled "t (s)".  A curve that relates v to t is shown in blue.  It begins with a straight line of endpoints (0,0) and (1,5). This first line is connected to a second line with endpoints (1,5) and (3,-5). This second line is then connected to a third line of endpoints (3,-5) and (6,-5). 
\[\small{1}\]
\[\small{2}\]
\[\small{3}\]
\[\small{4}\]
\[\small{5}\]
\[\small{1}\]
\[\small{2}\]
\[\small{3}\]
\[\small{4}\]
\[\small{5}\]
\[\small{\llap{-}2}\]
\[\small{\llap{-}3}\]
\[\small{\llap{-}4}\]
\[\small{\llap{-}5}\]
\[v~(\text{m/s})\]
\[t~(\text s)\]
At what time does the mouse have the same position as
\[t = 0\,\text s\]?
Assume the motion is restricted to one dimension. Answer with two significant digits.

#10 Help Me ! » Large Numbers (quadrillion) » 2025-01-14 23:24:29

paulb203
Replies: 1

And Small Numbers (quadrillionth)

*

The electron has a mass of approximately 9x10^-31kg, apparently.

Is 10^31 10 quadrillion quadrillion?

Is 10^-30 one quadrillion quadrillionth?
Or is 10^-30 one quadrillionth quadrillionth?

And is 10^-31 one tenth of a quadrillion quadrillionth?
Or is 10^-31 one tenth of a quadrillionth quadrillionth?

And, finally, is 9x10^-31 9 tenths of a quadrillionth quadrillionth

#11 Re: Help Me ! » Vector quantities; direction » 2025-01-07 23:23:39

Thanks, Bob.

I'm too young for bakelite records, and listening to the Goons on the radio, but we were taught On The Ning Nang Nong at school smile

#14 Re: Help Me ! » Vector quantities; direction » 2025-01-06 00:38:44

KerimF wrote:

An object is thrown upwards. Its initial velocity is +20 m/s.
Find its velocity at t=5 sec. (Assume the absolute value of g is 10 m/s2)

Thanks, KerimF

I had to use Google to get the formula 

v=u+at

v=+20m/s +(-10m/s^2)(5s)

v= +20m/s + -50m/s

v= +20m/s -50m/s

v= -30m/s

or;

v= 30m/s downwards

Is that correct?

#15 Re: Help Me ! » Vector quantities; direction » 2025-01-05 23:47:48

Ah, thanks, Bob, that does ring a bell. My memory is not what it... sorry, what were we talking about?

Also, if an object is shown to have moved in the negative direction could that mean either it turned around and headed back towards its origin; or, it could have went into reverse (if it was a vehicle; or, I suppose, a person, etc, like Spike Milligan Walking Backwards for Christmas, for example).

#16 Help Me ! » Vector quantities; direction » 2025-01-05 23:07:55

paulb203
Replies: 8

If the velocity of an object is -2m/s does the negative sign stand for direction? If so does that mean -2m/s is the complete answer, we don't need anything else such as North, South, Rightwards, Leftwards?

#17 Re: Help Me ! » When the y axis is labelled x » 2025-01-05 22:58:16

Thanks, Bob.

Although when I Googled abscissa and ordinate the first few results said those were names for the x and y co-ordinates, not the axes themselves.

#19 Help Me ! » When the y axis is labelled x » 2025-01-04 23:42:00

paulb203
Replies: 6

On a basic co-ordinate system is the horizontal axis always x, and the vertical axis always y?
And if they’re not labelled should we always think of them like that? Should they always be labelled? Is it considered slack when they’re not?
What about a position/time graph, where the y axis is labelled x for position? Is this often confusing for the novice?

#21 Re: Help Me ! » Is velocity ever a scalar quantity? » 2024-12-29 00:06:16

Thanks, Bob, thanks, Phil.

Having asked around I think I’ve got it now, and realise I should’ve included more information, and therefore context, with my question.

Khan was quite thorough with their notation. They showed various symbols and asked what each stands for and whether it was a vector or a scalar.
When they put x (for position) they had the little rightwards arrow above it, indicating a vector.
They also had an arrowed v, also indicating a vector.
And then they had a plain, unarrowed v, indicating, I think I see now, in this context, a scalar. Although someone pointed out that, strictly speaking, there should be vertical bars either side of this unarrowed v, to indicate absolute value, to indicate the magnitude alone of the velocity vector.

I’ve noticed in lots of maths questions (GCSE level) they keep it simple, e.g, no bar line above to indicate average rather than instantaneous; or no avg to indicate average; no arrows to indicate vectors; etc. I guess this makes sense given the context. It’s maths, not physics, even though it technically it is physics, if you know what I mean. And those letters (v, s, u, etc) could be any letters when the exercise is, for example, to show that you can change the subject of a formula.

Good points about the steering wheel and gas pedal, Phil smile

#22 Help Me ! » Is velocity ever a scalar quantity? » 2024-12-26 23:04:44

paulb203
Replies: 3

I thought velocity was always a vector quantity, one with both magnitude and direction.

When it came to suvat equations, where v = final velocity, and u = initial velocity, I thought both of those were vector quantities, e.g;

v (final velocity) 112km/hr North

u (initial velocity) 0km/hr (what do we put for direction when the object is initially stationary?)

But in a Khan Academy question they ask what does the letter v (lower case with no arrow above it, or anything else) stand for, and whether it’s a vector or a scalar.

I answered ‘velocity’ (it was multiple choice with no option for ‘final velocity’) and that it was a vector.

Their answer was;

“The symbol v represents speed, a scalar.”

I know speed is a scalar, but thought v stood for final velocity. I also know that final velocity IS speed, but thought that we had to include direction given the velocity part.

#25 Re: This is Cool » Teapot » 2024-12-22 00:41:29

"Chocolate teapot

A chocolate teapot is an analogy for any useless item; a teapot made from chocolate would melt, and be impossible to use.

Experimental researchers in 2001 did indeed fail to successfully use a chocolate teapot they had made. Later research, however, by The Naked Scientists in 2008, showed that such a teapot could be used to make tea, provided that the walls of the teapot were more than one centimetre thick."

Question. What  if you took your chocolate teapot up a mountain where water boils at less than 100 degrees C?

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