Acceleration and graphingWatch

Announcements
Thread starter 4 years ago
#1
Hello there

Please might anyone be able to advise me as to whether I've approached the following exercise correctly? i) 2/2 = 1 m/s

ii) 1/6 = 0.166 m/s

iii) -1/2 = - 0.5 m/s

iv) 2 seconds to 4 seconds and >6 seconds as no speeding up, slowing down or changing direction is occurring?

v) Gradient = (0-2)/(0-2) = 1. The same as i as dy/dx = gradient. 0-2 seconds is one increase of x which is the same as what the gradient shows. (I think this is a poor explanation).

vi) The values I got for dv/dt are:
0
1
0.5
0.166
0.125

The graph I plotted looks like this: Have I gone wrong? If anyone could please advise me, I'd be very appreciative.

Thank you very much.
0
4 years ago
#2
(Original post by Ggdf)
x
Your numerical answers are looking good, but your explanations might not be the ones they are looking for.

For iv), do you know any formulae for Force that might help you answer the question?

v) What does the gradient actually represent?

I think your graph should be a lot simpler than you're making out. If you think about any x value between 0 and 2, then y = x. So between 0 and 2, what is dy/dx for every single point? Does your current graph agree with this? What about the other points?
0
Thread starter 4 years ago
#3
(Original post by SeanFM)
Your numerical answers are looking good, but your explanations might not be the ones they are looking for.

For iv), do you know any formulae for Force that might help you answer the question?

v) What does the gradient actually represent?

I think your graph should be a lot simpler than you're making out. If you think about any x value between 0 and 2, then y = x. So between 0 and 2, what is dy/dx for every single point? Does your current graph agree with this? What about the other points?
Thank you very much for your reply.

Is force MLT-2, i.e. mass x acceleration?

Does force refer to a change in mass and length per 2 units of time? I can see what you're getting at, but am unable to explain why. Only that there is no change in length per 2 units of time during the time intervals I've stated. I'm however unable to factor in mass into my explanation at any point on the graph.

Does gradient represent the rate of change, i.e. how much the y axis changes for each unit of the x axis?

I've recalculated all the points as so:

0 = 0/0 = 0
1 = 1/1 =1
2 = 2/2 = 1
3 = 2/3 = 0.666
4 = 2/4 = 0.5
5 = 1.5 (unsure if this is the correct y value)/ 5 = 0.142
6 = 1/6 = 0.166
7 = 1/7 = 0.142
1/8 = 0.125

My graph now looks as follows: Best wishes
0
4 years ago
#4
(Original post by Ggdf)
Thank you very much for your reply.

Is force MLT-2, i.e. mass x acceleration?

Does force refer to a change in mass and length per 2 units of time? I can see what you're getting at, but am unable to explain why. Only that there is no change in length per 2 units of time during the time intervals I've stated. I'm however unable to factor in mass into my explanation at any point on the graph.

This is from a previous post of yours:

0
1
0.5
(and you're missing one more value for t=6 onwards)
What you've said about the 'no force' question is right. You don't need to worry about the mass, because the acceleration is enough to answer the question. So what does there being no acceleration mean for the force? (Hint: what's the value of a?)

You're right again, the gradient represents the rate of change. So in the context of this graph, what does dv/dt actually represent? Not just a rate of change, but the rate of change of.. which is...

The acceleration graph is just looking for instantaneous dv/dt, whereas you've calculated some averages. Again, going back to between 0 and 2, what's the value of dv/dt at x = 0.5? how about 0.01? So what does the graph of dv/dt look like between 0 and 2?

I think it may help to find some acceleration-time graphs in your textbook if you can, it may help.
0
Thread starter 4 years ago
#5
(Original post by SeanFM)
What you've said about the 'no force' question is right. You don't need to worry about the mass, because the acceleration is enough to answer the question. So what does there being no acceleration mean for the force? (Hint: what's the value of a?)

You're right again, the gradient represents the rate of change. So in the context of this graph, what does dv/dt actually represent? Not just a rate of change, but the rate of change of.. which is...

The acceleration graph is just looking for instantaneous dv/dt, whereas you've calculated some averages. Again, going back to between 0 and 2, what's the value of dv/dt at x = 0.5? how about 0.01? So what does the graph of dv/dt look like between 0 and 2?

I think it may help to find some acceleration-time graphs in your textbook if you can, it may help.
Thank you so much again. I really appreciate your help!

Is the value of a 0? There is no acceleration, therefore no LT-2/acceleration in the force equation, therefore no force?

Does the gradient represent the rate of change of velocity with time? Which is 1? Would I say that between 0 and 2 seconds is the same as the gradient because this is where the line cuts through the origin, and where you can see the change in y with an increase in x? I'm not too sure how to describe the relationship between the gradient and 1 to 2 seconds.

Would x0.5 be y0.5 and x0.01 =y0.01. So dv/dt for 0.5 = 1 and 0.01 = 1?
Between those points, it looks like a straight line.
I'm not sure what I should do to calculate dv/dt for point between 2 seconds and 8 seconds?
0
4 years ago
#6
(Original post by Ggdf)
Thank you so much again. I really appreciate your help!

1. Is the value of a 0? There is no acceleration, therefore no LT-2/acceleration in the force equation, therefore no force?

2. Does the gradient represent the rate of change of velocity with time? Which is 1? Would I say that between 0 and 2 seconds is the same as the gradient because this is where the line cuts through the origin, and where you can see the change in y with an increase in x? I'm not too sure how to describe the relationship between the gradient and 1 to 2 seconds.

3. Would x0.5 be y0.5 and x0.01 =y0.01. So dv/dt for 0.5 = 1 and 0.01 = 1?
Between those points, it looks like a straight line.
I'm not sure what I should do to calculate dv/dt for point between 2 seconds and 8 seconds?

1. Yes, you're right. You could use less words and say F = ma = m*0 = 0, so F = 0. I think you can see why taht is.

2. Yes, the gradient represents that. But what is the rate of change of velocity? (Yes, it is 1). Let's think of something else to help you, you know that the rate of change of displacement (i.e distance over time) is velocity, right? So what is velocity over time? Why is that related to q(i)? It may help to look at q(i) and what is asks, that may give you the answer.

3. Yes, you're right about the straight line from 0 to 1. Knowing that I think you can do the rest . What's dv/dt between 2 and 4? What will this look like on your graph? How about the other two intervals? What's different about 4 to 6, and why do you need to be careful?
0
Thread starter 4 years ago
#7
(Original post by SeanFM)

1. Yes, you're right. You could use less words and say F = ma = m*0 = 0, so F = 0. I think you can see why taht is.

2. Yes, the gradient represents that. But what is the rate of change of velocity? (Yes, it is 1). Let's think of something else to help you, you know that the rate of change of displacement (i.e distance over time) is velocity, right? So what is velocity over time? Why is that related to q(i)? It may help to look at q(i) and what is asks, that may give you the answer.

3. Yes, you're right about the straight line from 0 to 1. Knowing that I think you can do the rest . What's dv/dt between 2 and 4? What will this look like on your graph? How about the other two intervals? What's different about 4 to 6, and why do you need to be careful?
2. Is it that velocity over time is acceleration? I see that we have to calculate the acceleration between 0 and 2 seconds. I'm not sure how I relate the gradient to this, only again that it is a straight line.

3. I'm really not sure how to calculate between 2 and 4 without getting the same answers from before. My only guess is that no change occurs at the points during this time interval, so all the points are 0? Would 4 to 6 be minus figures e.g. minus 5 = -0.4?

I'm sorry to be so slow, I'm struggling with this a lot. Thank you for all your help .
0
4 years ago
#8
(Original post by Ggdf)
2. Is it that velocity over time is acceleration? I see that we have to calculate the acceleration between 0 and 2 seconds. I'm not sure how I relate the gradient to this, only again that it is a straight line.

3. I'm really not sure how to calculate between 2 and 4 without getting the same answers from before. My only guess is that no change occurs at the points during this time interval, so all the points are 0? Would 4 to 6 be minus figures e.g. minus 5 = -0.4?

I'm sorry to be so slow, I'm struggling with this a lot. Thank you for all your help .
2. Yes, velocity over time is acceleration which is the gradient of the graph. So you've found the gradient and the acceleration from q(i) so..

3. Yep, 0 for some bits (flat) and then negative. Well done 0
Thread starter 4 years ago
#9
(Original post by SeanFM)
2. Yes, velocity over time is acceleration which is the gradient of the graph. So you've found the gradient and the acceleration from q(i) so..

3. Yep, 0 for some bits (flat) and then negative. Well done Thank you .

2. So the gradient is the same as the acceleration between 0 and 2 because this is where there is a straight line and we see an increase of y with x? Sorry, I see what you're getting at, but am unable to summarise it.

3. My new values look like this:
x y
0 0
1 1
2 1
3 0
4 0
5 -0.3
6 -0.166
7 0
8 0

And my graph like this: 0
4 years ago
#10
(Original post by Ggdf)
Thank you .

2. So the gradient is the same as the acceleration between 0 and 2 because this is where there is a straight line and we see an increase of y with x? Sorry, I see what you're getting at, but am unable to summarise it.

3. My new values look like this:
x y
0 0
1 1
2 1
3 0
4 0
5 -0.3
6 -0.166
7 0
8 0
2. The gradient is not only the same as acceleration, but it *is* the acceleration - the rate of change of velocity. I think for the question you just need to say that and then it's obvious that acceleration = acceleration.

3. You're almost there with the graph - remember that when there's a straight line on a velocity time graph the acceleration is constant (even when there's 0). So I'm not sure how you've got the values for the bit where the acceleration is negative, but think of it as it being really similar to between 0 and 2, just negative and a different gradient. So just a straight line on the acceleration graph with a certain value of a, which you can find.

I think they also want you to start the diagram at (0,1) instead of the origin as you have done in your graph. It's only because the program you're using makes it look like acceleration is increasing when really it's just a straight line. The acceleration/time graphs I have seen in Physics are generally straight lines horizontally, and when the acceleration changes at a certain point (let's say acceleration is 5 from 0 to 10, 0 from 10 to 20, -5 from 20 to 30), then there's a straight line y=5 from 0 to 10, at 10 the line goes almost vertically down to 0 at 10 and is a straight line to 20, then goes almost vertically down again from 0 to -5, and is a straight line to 30.)

But well done on solving the problem. I would recommend practicing drawing acceleration/time graphs, if you can find any questions on those.
0
Thread starter 4 years ago
#11
(Original post by SeanFM)
2. The gradient is not only the same as acceleration, but it *is* the acceleration - the rate of change of velocity. I think for the question you just need to say that and then it's obvious that acceleration = acceleration.

3. You're almost there with the graph - remember that when there's a straight line on a velocity time graph the acceleration is constant (even when there's 0). So I'm not sure how you've got the values for the bit where the acceleration is negative, but think of it as it being really similar to between 0 and 2, just negative and a different gradient. So just a straight line on the acceleration graph with a certain value of a, which you can find.

I think they also want you to start the diagram at (0,1) instead of the origin as you have done in your graph. It's only because the program you're using makes it look like acceleration is increasing when really it's just a straight line. The acceleration/time graphs I have seen in Physics are generally straight lines horizontally, and when the acceleration changes at a certain point (let's say acceleration is 5 from 0 to 10, 0 from 10 to 20, -5 from 20 to 30), then there's a straight line y=5 from 0 to 10, at 10 the line goes almost vertically down to 0 at 10 and is a straight line to 20, then goes almost vertically down again from 0 to -5, and is a straight line to 30.)

But well done on solving the problem. I would recommend practicing drawing acceleration/time graphs, if you can find any questions on those.
Thank you so much for your help, it's really kind of you!

3. What values might you put into the equation for points 5 and 6?
0
4 years ago
#12
(Original post by Ggdf)
Thank you so much for your help, it's really kind of you!

3. What values might you put into the equation for points 5 and 6?
dv/dx is constant between 4 and 6. Can you see why? Can you find it between 4 and 6?
0
Thread starter 4 years ago
#13
(Original post by SeanFM)
dv/dx is constant between 4 and 6. Can you see why? Can you find it between 4 and 6?
Is it because there is no change in v with change in x? I can see a negative change in v from 4 to 6, but am unsure what numbers to put in for v for points 5 and 6. I only seem to be able to find the ones I used. I think I'm still a little confused.
0
4 years ago
#14
(Original post by Ggdf)
Is it because there is no change in v with change in x? I can see a negative change in v from 4 to 6, but am unsure what numbers to put in for v for points 5 and 6. I only seem to be able to find the ones I used. I think I'm still a little confused.
Not quite - that would mean that the acceleration is 0. An acceleration of 0 is still constant acceleration (I think, as it's always 0).

I keep referring back to it, but it's really similar to what you had for 0 and 2 (the straight line). Remember how dv/dt was 1 for any point between 0 and 2? Whether it was 0.01, 0.5, or anything along the line. And how did you calculate it? Well, you took the two end points of that straight line - (0,0) and (2,2). It's practically the same for 4 and 6, and the points at 5 and 6 are the same, as is the point for 4, just like it was for 0 and 2. So what are the end points of the line?

I guess another way of thinking about it is 'what is the acceleration between 4 and 6'.
0
Thread starter 4 years ago
#15
(Original post by SeanFM)
Not quite - that would mean that the acceleration is 0. An acceleration of 0 is still constant acceleration (I think, as it's always 0).

I keep referring back to it, but it's really similar to what you had for 0 and 2 (the straight line). Remember how dv/dt was 1 for any point between 0 and 2? Whether it was 0.01, 0.5, or anything along the line. And how did you calculate it? Well, you took the two end points of that straight line - (0,0) and (2,2). It's practically the same for 4 and 6, and the points at 5 and 6 are the same, as is the point for 4, just like it was for 0 and 2. So what are the end points of the line?

I guess another way of thinking about it is 'what is the acceleration between 4 and 6'.
Thank you for explaining to me so well. I sadly still don't understand.

Is the acceleration not 0 between 4 and 6?

To calculate the straight line between 0 and 2 seconds, I took the y coordinate and divided it by the x coordinate. In the case of 0.01, I could see that the y and x coordinates were both roughly 0.01, so I divided y0.01 by x0.01 to get 1. The same was the case for 2 - the y and x coordinates both looked to be 2, so I divided y2 by x2 to get 1. However, I can't see how to do this between 4 and 6. It's hard to read the y values on the graph, but for 6 (for example), y looks to be about 1, therefore I divided y1 by x6 to get 0.166. I'm also unsure how to calculate between 2 and 4 - I only assumed the values at these points to be 0. I'm still very confused!
0
4 years ago
#16
(Original post by Ggdf)
Thank you for explaining to me so well. I sadly still don't understand.

Is the acceleration not 0 between 4 and 6?

To calculate the straight line between 0 and 2 seconds, I took the y coordinate and divided it by the x coordinate. In the case of 0.01, I could see that the y and x coordinates were both roughly 0.01, so I divided y0.01 by x0.01 to get 1. The same was the case for 2 - the y and x coordinates both looked to be 2, so I divided y2 by x2 to get 1. However, I can't see how to do this between 4 and 6. It's hard to read the y values on the graph, but for 6 (for example), y looks to be about 1, therefore I divided y1 by x6 to get 0.166. I'm also unsure how to calculate between 2 and 4 - I only assumed the values at these points to be 0. I'm still very confused!
Okay, no worries. If you're calculating dv/dt you take two values of v and two values of t (just like you might do (y2-y1)/(x2-x1) to find dy/dx between (x1,y1) and (x2,y2).

So between 0 and 2 what you're actually doing is taking (0,0) and (2,2) - so it would be (2 - 0)/(2 -0), which is 1.

Between 2 and 4, the two endpoints are (2,2) and (4,2). So what is dv/dt?

What are the endpoints for 4 to 6? What's dv/dt?

If you're wondering why we can use these values to plot dv/dt for all the points between each time interval (eg 1 for 0 to 2), it's because when there's a straight line on the velocity-time graph means the acceleration is constant (the same between all points).
0
Thread starter 4 years ago
#17
(Original post by SeanFM)
Okay, no worries. If you're calculating dv/dt you take two values of v and two values of t (just like you might do (y2-y1)/(x2-x1) to find dy/dx between (x1,y1) and (x2,y2).

So between 0 and 2 what you're actually doing is taking (0,0) and (2,2) - so it would be (2 - 0)/(2 -0), which is 1.

Between 2 and 4, the two endpoints are (2,2) and (4,2). So what is dv/dt?

What are the endpoints for 4 to 6? What's dv/dt?

If you're wondering why we can use these values to plot dv/dt for all the points between each time interval (eg 1 for 0 to 2), it's because when there's a straight line on the velocity-time graph means the acceleration is constant (the same between all points).
I'm really sorry to have put you through all this explaining. My command of maths really is poor, so much so that I couldn't really recognise what exactly what it was I didn't understand (I'm embarrassed). I wasn't aware of how to calculate dv/dt. I can't thank you enough for explaining to me!

Am I correct to understand that to calculate dv/dt, we do (y1-y2)/(x1-x2)?

Therefore:

For 2-4 seconds: The endpoints are (2,2) and (4,2). The dv/dt equation is therefore (2-2)/(4-2) = 0 ?

For 4-6 seconds: The endpoints are (4,2) and (6,1). The dv/dt equation is therefore (1-2)/(6-4) = -0.5 ?

For 6-8 seconds: The endpoints are (6,1) and (8,1). The dv/dt equation is therefore (1-1)/(8-6) = 0 ?

My points therefore look like this:
x y
0 0
1 1
2 1
3 0
4 0
5 -0.5
6 -0.5
7 0
8 0

And my graph looks like this: Thank you very much for explaining to me about the intermediate points. That was hugely helpful and has clarified it a lot for me!
0
4 years ago
#18
(Original post by Ggdf)
X
Yes, all of the points are correct - well done . It's good that you know how to get there now.

I guess the way your graphing program works is that it joins up all the lines together, but if it was on paper you would want just straight lines like I've described in the previous post - so when the acceleration changes from 0 to -0.5 for example, it doesn't gradually get down to -0.5 over a period of time, right from after t=4 it becomes -0.5. So if you like you could have two points at the times where the acceleration changes to see what kind of shape they want you to draw - so (4,0) and (4,-0.5), and (6,-0.5) and (6,0), and also (2,1) and (2,0). (All points on the dv/dt graph, not velocity-time).

But otherwise good job 0
Thread starter 4 years ago
#19
(Original post by SeanFM)
Yes, all of the points are correct - well done . It's good that you know how to get there now.

I guess the way your graphing program works is that it joins up all the lines together, but if it was on paper you would want just straight lines like I've described in the previous post - so when the acceleration changes from 0 to -0.5 for example, it doesn't gradually get down to -0.5 over a period of time, right from after t=4 it becomes -0.5. So if you like you could have two points at the times where the acceleration changes to see what kind of shape they want you to draw - so (4,0) and (4,-0.5), and (6,-0.5) and (6,0), and also (2,1) and (2,0). (All points on the dv/dt graph, not velocity-time).

But otherwise good job You have been really fantastic . Thank you so much for guiding me through. I feel I've really learned something.

I tried changing my graph to a straight line one and got this: I then tried entering two points for 2, 4 and 6 and got a graph like this: The points are:

x y
0 0
1 1
2 1
2 0
3 0
4 0
4 0.5
5 -0.5
6 -0.5
6 0
7 0
8 0

The program I've used is excel.
0
4 years ago
#20
(Original post by Ggdf)
x
Excellent. There's just two points that you need to change and then it's complete. One is that it starts at (0,1), though I guess you could also have (0,0) below it. The other is that you've put (4,0.5) instead of (4,-0.5), but apart from that it should be fine. There's an example of what an acceleration-time graph looks like - though it doesn't have any bits where it's negative - so you can see what it'd look like if you weren't using Excel.
0
X

new posts Back
to top
Latest
My Feed

Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

University open days

• Bournemouth University
Midwifery Open Day at Portsmouth Campus Undergraduate
Wed, 16 Oct '19
• Teesside University
All faculties open Undergraduate
Wed, 16 Oct '19
• University of the Arts London
London College of Fashion – Cordwainers Footwear and Bags & Accessories Undergraduate
Wed, 16 Oct '19

Poll

Join the discussion

How has the start of this academic year been for you?

Loving it - gonna be a great year (114)
17.73%
It's just nice to be back! (177)
27.53%
Not great so far... (229)
35.61%
I want to drop out! (123)
19.13%