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paulb203

Member

Registered: 2023-02-24

Posts: 374

Position-time graphs; acceleration

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.

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KerimF

Member

From: Aleppo-Syria

Registered: 2018-08-10

Posts: 280

Re: Position-time graphs; acceleration

I am a bit busy now so I can just give a hint:
Try to draw the distance travelled by a falling object (a=g=10m/s2) versus time (assume that the downwards direction of the movement is positive). I don't think the graph will be a straight line.

Kerim

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Every living thing has no choice but to execute its pre-programmed instructions embedded in it (known as instincts).
But only a human may have the freedom and ability to oppose his natural robotic nature.
But, by opposing it, such a human becomes no more of this world.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,701

Re: Position-time graphs; acceleration

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

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

paulb203

Member

Registered: 2023-02-24

Posts: 374

Re: Position-time graphs; acceleration

Thanks, Kerim F

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Prioritise. Persevere. No pain, no gain.

paulb203

Member

Registered: 2023-02-24

Posts: 374

Re: Position-time graphs; acceleration

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.

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paulb203

Member

Registered: 2023-02-24

Posts: 374

Re: Position-time graphs; acceleration

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.

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Bob

Administrator

Registered: 2010-06-20

Posts: 10,701

Re: Position-time graphs; acceleration

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.

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

paulb203

Member

Registered: 2023-02-24

Posts: 374

Re: Position-time graphs; acceleration

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.

Thanks a lot, Bob.

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

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

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Prioritise. Persevere. No pain, no gain.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,701

Re: Position-time graphs; acceleration

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

In any area of maths you have to be careful not to assume things that you haven't been given.  So I would advise caution.

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?

Just that. The graph is all you have and all you need.

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?

Yes. You'd have to be told something about the rate of acceleration and for how long.  If a is constant, then the V-T graph is a straight line and the D-T graph is a quadratic. This comes from calculus but you can also get it from s = ut + ½at^2

If a isn't constant that's a one term course at first year Uni.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

paulb203

Member

Registered: 2023-02-24

Posts: 374

Re: Position-time graphs; acceleration

Thanks a lot, Bob

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Prioritise. Persevere. No pain, no gain.