# Homework Help: Velocity-time graph issue

1. Feb 8, 2012

### Ste

Hi all, apologies if this is in the wrong section but I thought it would be ok as it's kind of homework related.

I'm 35 years old and I'm trying to learn a bit about physics. With that in mind I thought I'd start at the G.C.S.E level seeing as I can't remember a thing from school. I borrowed a book from the library and the information is sinking in ok, but there's one equation that I'm having a spot of bother with. I'll do my best to explain and hopefully someone will be able to help.

I'm currently learning about how the area under a velocity-time graph is equal to distance travelled. The book says that the equation for working out the area of a triangle, 1/2 base x height, is equal to 1/2at^2 However when I work out the two equations using the following information I get two different answers.

When t=4s
And V=5m/s

For the area of a triangle equation I get the answer s=10

For 1/2at^2 I get the answer s=12.5

It seems that I'm either working out the equation incorrectly, or I've somehow misunderstood the book. Any help would be great, thanks.

2. Feb 8, 2012

Hi Ste! Welcome to PF! It's tough to say without seeing the graph. My guess is that there is some detail being missed. Perhaps the initial velocity in the graph (where t=0) is not zero?

If you can post the graph it would help. The easiest way is to upload the image via a site like photobucket or imageshack (I use the photobucket one and they're free).

3. Feb 8, 2012

### Ste

I've checked the book and the initial velocity does equal zero. Am I right in thinking that the two equations are interchangeable ? Maybe that's where I'm going wrong?

I'll look into photo bucket when I get home.

Thanks again.

4. Feb 8, 2012

Doh! I think I know what it is: The equation (1/2)*at^2 is only valid when you have constant acceleration. This means that the rate at which velocity changes with time remains constant. This is also equivalent having to a velocity versus time graph that is a straight line. So if the graph is anything but a single straight line, then no, the equations are not interchangeable. Only the area equation will work.

5. Feb 8, 2012

### Ste

It is constant acceleration because the book says it will go into non-constant acceleration later on. This is supposed to be the easy bit lol.

Also, the line is straight. Like a right angled triangle.

6. Feb 8, 2012

### OJFord

Yes that's right - I'm pretty sure that's all that is taught at GCSE, mainly because of the emphasis on working out something like distance from the graph, which is more complex if the lines are not straight (and would need maths know-how also not taught in GCSE Mathematics).

The equation works as a normal triangle because velocity (one side of triangle/axis), $v = at + v_{0}$ and then obviously time, $t$ on the other side/axis makes:
$\frac{1}{2} * vt * t = \frac{1}{2} * at * t$, which is equal to $\frac{1}{2} at^{2}$.

7. Feb 9, 2012

### Ste

Well, to be honest, that last bit was a bit over my head as I'm still learning how to rearrange equations. Why didn't I pay attention at school lol.

The bit I still don't get is why I get the answer s=12.5, when common sense says the answer is s=10.

Here's how I work through the equations when v=5 and t=4

S=1/2at^2......at^2=25..........25/2= 12.5.......so s=12.5

Half base x height.......half base=2......2x5=10....so s=10

Hopefully I explained that ok.

8. Feb 9, 2012

### Sourabh N

How is at^2 equal to 25?

9. Feb 9, 2012

### Ste

Hi sourabh.

In the book it shows that at=5 (with at being the distance travelled along the y axis)

5 squared =25

Cheers.

10. Feb 9, 2012

### Sourabh N

Actually, i think at^2 here means at times t and NOT at times at

11. Feb 9, 2012

### Ste

I think I've just been looking at the equation in the wrong way because of the way the graph is labeled in the book. Let me explain how I now understand the equation and we'll see if I'm getting the hang of all this lol.

S=1/2at^2........displacement equals half the acceleration, multiplied by the time taken, then that outcome is marked with the square symbol because it's an area.

How's that? Once again thanks for your help all.

Last edited: Feb 9, 2012
The text passage is a discussion thread from a physics forum where a user named Ste is seeking help with understanding the relationship between the area under a velocity-time graph and the equation for displacement under constant acceleration. The user is confused about why they get different results when using the area of a triangle formula and the \( \frac{1}{2}at^2 \) formula.
single_question_with_answer

the passage contains a single mathematical question regarding the discrepancy between the results of the area of a triangle formula and the \( \frac{1}{2}at^2 \) formula. the answer is provided in the discussion, where other users explain that the \( \frac{1}{2}at^2 \) formula is only valid for constant acceleration and a straight-line velocity-time graph, and that the user's confusion arises from a misunderstanding of the equation.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Homework Help: Velocity-time graph issue

1. Feb 8, 2012

### Ste

Hi all, apologies if this is in the wrong section but I thought it would be ok as it's kind of homework related.

I'm 35 years old and I'm trying to learn a bit about physics. With that in mind I thought I'd start at the G.C.S.E level seeing as I can't remember a thing from school. I borrowed a book from the library and the information is sinking in ok, but there's one equation that I'm having a spot of bother with. I'll do my best to explain and hopefully someone will be able to help.

I'm currently learning about how the area under a velocity-time graph is equal to distance travelled. The book says that the equation for working out the area of a triangle, 1/2 base x height, is equal to 1/2at^2 However when I work out the two equations using the following information I get two different answers.

When t=4s
And V=5m/s

For the area of a triangle equation I get the answer s=10

For 1/2at^2 I get the answer s=12.5

It seems that I'm either working out the equation incorrectly, or I've somehow misunderstood the book. Any help would be great, thanks.

2. Feb 8, 2012

Hi Ste! Welcome to PF! It's tough to say without seeing the graph. My guess is that there is some detail being missed. Perhaps the initial velocity in the graph (where t=0) is not zero?

If you can post the graph it would help. The easiest way is to upload the image via a site like photobucket or imageshack (I use the photobucket one and they're free).

3. Feb 8, 2012

### Ste

I've checked the book and the initial velocity does equal zero. Am I right in thinking that the two equations are interchangeable ? Maybe that's where I'm going wrong?

I'll look into photo bucket when I get home.

Thanks again.

4. Feb 8, 2012

Doh! I think I know what it is: The equation (1/2)*at^2 is only valid when you have constant acceleration. This means that the rate at which velocity changes with time remains constant. This is also equivalent having to a velocity versus time graph that is a straight line. So if the graph is anything but a single straight line, then no, the equations are not interchangeable. Only the area equation will work.

5. Feb 8, 2012

### Ste

It is constant acceleration because the book says it will go into non-constant acceleration later on. This is supposed to be the easy bit lol.

Also, the line is straight. Like a right angled triangle.

6. Feb 8, 2012

### OJFord

Yes that's right - I'm pretty sure that's all that is taught at GCSE, mainly because of the emphasis on working out something like distance from the graph, which is more complex if the lines are not straight (and would need maths know-how also not taught in GCSE Mathematics).

The equation works as a normal triangle because velocity (one side of triangle/axis), $v = at + v_{0}$ and then obviously time, $t$ on the other side/axis makes:
$\frac{1}{2} * vt * t = \frac{1}{2} * at * t$, which is equal to $\frac{1}{2} at^{2}$.

7. Feb 9, 2012

### Ste

Well, to be honest, that last bit was a bit over my head as I'm still learning how to rearrange equations. Why didn't I pay attention at school lol.

The bit I still don't get is why I get the answer s=12.5, when common sense says the answer is s=10.

Here's how I work through the equations when v=5 and t=4

S=1/2at^2......at^2=25..........25/2= 12.5.......so s=12.5

Half base x height.......half base=2......2x5=10....so s=10

Hopefully I explained that ok.

8. Feb 9, 2012

### Sourabh N

How is at^2 equal to 25?

9. Feb 9, 2012

### Ste

Hi sourabh.

In the book it shows that at=5 (with at being the distance travelled along the y axis)

5 squared =25

Cheers.

10. Feb 9, 2012

### Sourabh N

Actually, i think at^2 here means at times t and NOT at times at

11. Feb 9, 2012

### Ste

I think I've just been looking at the equation in the wrong way because of the way the graph is labeled in the book. Let me explain how I now understand the equation and we'll see if I'm getting the hang of all this lol.

S=1/2at^2........displacement equals half the acceleration, multiplied by the time taken, then that outcome is marked with the square symbol because it's an area.

How's that? Once again thanks for your help all.

Last edited: Feb 9, 2012
user
The text passage is a discussion thread from a physics forum where a user named Ste is seeking help with understanding the relationship between the area under a velocity-time graph and the equation for displacement under constant acceleration. The user is confused about why they get different results when using the area of a triangle formula and the \( \frac{1}{2}at^2 \) formula.

TYPE: SINGLE_QUESTION_WITH_ANSWER

The passage contains a single mathematical question regarding the discrepancy between the results of the area of a triangle formula and the \( \frac{1}{2}at^2 \) formula. The answer is provided in the discussion, where other users explain that the \( \frac{1}{2}at^2 \) formula is only valid for constant acceleration and a straight-line velocity-time graph, and that the user's confusion arises from a misunderstanding of the equation.