Points of inflection on acceleration graph

In summary, the conversation discusses points of inflection and the correct method for finding displacement from a velocity-time graph. The mistake in the book's solution is due to finding the average speed over a longer time interval instead of the correct time interval. The conversation also touches on the difference between average speed and average velocity, and how the average velocity formula is only valid for constant acceleration.
  • #1
ChiralSuperfields
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Homework Statement
Please see below
Relevant Equations
Please see below
EDIT: For this part(b) of this problem,
1676497354870.png

The solution is
1676508669551.png

1676508731673.png

However, isn't there more points of inflection than just ##t = 3,5 s ##? Points of inflection is when ##x'' = a = 0## so it should be ## 3 ≤ t ≤ 5 s##

I also have a question about part(d):

The solution is
1676511464795.png

However, could I tried solving for the position by finding the area under the velocity curve from t = 0 to t = 6 sec.

## 0.5 \times 3 \times 8 + 2 \times 8 + 1\times 4 + 1 \times 1 \times 4 = 34 m ##.

I don't understand what I have done wrong since I should be able to get the correct answer from areas.

Can someone please help?

Many thanks!
 

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  • #2
Callumnc1 said:
## 0.5 \times 2.5 \times 8 = 10 m ##
2.5? Look again.
 
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  • #3
haruspex said:
2.5? Look again.
Thank you for your reply @haruspex! Do you please know the answer to my new confusion?

Many thanks!
 
  • #4
Callumnc1 said:
isn't there more points of inflection than just t=3,5s?
It should not refer to inflection points here. The first and second derivatives of v would need to exist and be continuous.
Callumnc1 said:
## 0.5 \times 3 \times 8 + 2 \times 8 + 1\times 4 + 1 \times 1 \times 4 = 34 m ##.
For the purpose of finding e.g. displacement from a v-t graph, you have to count areas below the x axis as negative.
 
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  • #5
haruspex said:
It should not refer to inflection points here. The first and second derivatives of v would need to exist and be continuous.

For the purpose of finding e.g. displacement from a v-t graph, you have to count areas below the x axis as negative.
Thank you for your reply @haruspex !

Why would the second derivative of v need to exist and be continuous?

Also I still don't understand how I calculated the displacement wrong since there is no negative velocities from t = 0 to t = 6 sec

1676515381951.png

Many thanks!
 

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  • #6
Callumnc1 said:
Why would the second derivative of v need to exist and be continuous?
https://en.wikipedia.org/wiki/Inflection_point
Callumnc1 said:
there is no negative velocities from t = 0 to t = 6 sec
How true, but you fooled me with the last two terms in
## 0.5 \times 3 \times 8 + 2 \times 8 + 1\times 4 + 1 \times 1 \times 4##.
If stopping at 6sec, where do those come from?
 
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  • #7
haruspex said:
https://en.wikipedia.org/wiki/Inflection_point

How true, but you fooled me with the last two terms in

If stopping at 6sec, where do those come from?
Thank you for your reply @haruspex!

Sorry I made a slight mistake with the last term. It should be

## 0.5 \times 3 \times 8 + 2 \times 8 + 1\times 4 + 0.5 \times 1 \times 4 = 34 m ##.

The last two terms ## 1\times 4 + 0.5 \times 1 \times 4 = 34 m ## are accounted for here on the graph in red and orange.
1676516831971.png

The solution seems to account for only the orange triangle (##1 \times 4##) giving ##32 m##. However, I though displacement should the total area under the velocity curve over the interval ##[0, 6] s## so I included the red triangle to get ##34 m##

Let me know what you think!

Many thanks!
 

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  • #8
Callumnc1 said:
Thank you for your reply @haruspex!

Sorry I made a slight mistake with the last term. It should be

## 0.5 \times 3 \times 8 + 2 \times 8 + 1\times 4 + 0.5 \times 1 \times 4 = 34 m ##.

The last two terms ## 1\times 4 + 0.5 \times 1 \times 4 = 34 m ## are accounted for here on the graph in red and orange.
View attachment 322355
The solution seems to account for only the orange triangle (##1 \times 4##) giving ##32 m##. However, I though displacement should the total area under the velocity curve over the interval ##[0, 6] s## so I included the red triangle to get ##34 m##

Let me know what you think!

Many thanks!
Well, when you put it like that, I agree.
The book mistake is that they found the average speed between 5s and 7s, instead of between 5s and 6s.
 
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  • #9
haruspex said:
Well, when you put it like that, I agree.
The book mistake is that they found the average speed between 5s and 7s, instead of between 5s and 6s.
Interesting that makes a difference to their calculation! Thank you for your help @haruspex!

You mention average speed which makes me wonder why they could not have just used the average velocity over the entire interval from t = 0 to t = 6 sec :

##v_{avg} = \frac {v_i + v_f}{2} = \frac {x_f - x_i}{t} ##
##\frac {0+ 4}{2} = \frac {x_f - 0}{6} ##
## 12m = x_f ##

I now realize this because the average velocity formula ##v_{avg} = \frac {v_i + v_f}{2}## is only valid for constant acceleration. But the acceleration varies over the interval.

I'm not sure how to prove that the average velocity formula is only valid for constant acceleration thought.

Many thanks!
 
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  • #10
Callumnc1 said:
how to prove that the average velocity formula is only valid for constant acceleration
You cannot prove it wrong in general because it might just happen to give the right answer sometimes. But to prove it is not a valid method you only have to construct one example where it gives the wrong answer.
 
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  • #11
haruspex said:
You cannot prove it wrong in general because it might just happen to give the right answer sometimes. But to prove it is not a valid method you only have to construct one example where it gives the wrong answer.
Thank very much for your help @haruspex !
 
  • #12
Callumnc1 said:
I'm not sure how to prove that the average velocity formula is only valid for constant acceleration thought.
You can prove that it is only right for constant acceleration.
For constant acceleration we have two main equations: (If we have ##t_0=0##)
##x=x_0+v_{0}t+\frac 1 2 at^2##
##v=v_0+at##
Using ##v=v_0+at## we know that ##a=\frac {v-v_0} {t}##. We put it in first equation then we have:
##x=x_0+v_{0}t+\frac 1 2 \frac {v-v_0} {t}t^2##
##x=x_0+v_{0}t+\frac 1 2 (v-v_0)t##
##x=x_0+v_{0}t+\frac 1 2 vt -\frac 1 2 v_{0}t##
##x-x_0=\frac 1 2 vt +\frac 1 2 v_{0}t##
##x-x_0=\frac 1 2 (v+v_0)t##
Compare it with ##v_{avg}=\frac {x-x_0} {t}##→ ##x-x_0=v_{avg}t##

You can see that ##v_{avg}=\frac 1 2 (v+v_0)##
 
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  • #13
MatinSAR said:
You can prove that it is only right for constant acceleration.
For constant acceleration we have two main equations: (If we have ##t_0=0##)
##x=x_0+v_{0}t+\frac 1 2 at^2##
##v=v_0+at##
Using ##v=v_0+at## we know that ##a=\frac {v-v_0} {t}##. We put it in first equation then we have:
##x=x_0+v_{0}t+\frac 1 2 \frac {v-v_0} {t}t^2##
##x=x_0+v_{0}t+\frac 1 2 (v-v_0)t##
##x=x_0+v_{0}t+\frac 1 2 vt -\frac 1 2 v_{0}t##
##x-x_0=\frac 1 2 vt +\frac 1 2 v_{0}t##
##x-x_0=\frac 1 2 (v+v_0)t##
Compare it with ##v_{avg}=\frac {x-x_0} {t}##→ ##x-x_0=v_{avg}t##

You can see that ##v_{avg}=\frac 1 2 (v+v_0)##
Thank you for your help @MatinSAR !
 
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1. What is a point of inflection on an acceleration graph?

A point of inflection on an acceleration graph is a point where the acceleration changes from increasing to decreasing, or vice versa. It is the point where the acceleration changes direction.

2. How can I identify a point of inflection on an acceleration graph?

To identify a point of inflection on an acceleration graph, look for a change in the concavity of the curve. At a point of inflection, the curve changes from concave up to concave down, or vice versa.

3. What does a point of inflection represent in terms of motion?

A point of inflection represents a change in the rate of change of velocity. This means that the object is either speeding up or slowing down at a different rate at that point.

4. Can there be more than one point of inflection on an acceleration graph?

Yes, there can be more than one point of inflection on an acceleration graph. This occurs when the acceleration changes direction multiple times.

5. How does the shape of an acceleration graph affect the number of points of inflection?

The shape of an acceleration graph affects the number of points of inflection. A more complex and curved graph is likely to have multiple points of inflection, while a simpler and more linear graph may only have one point of inflection.

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