Incorrect
These formulae are valid for Linear Motion with constant acceleration only. For example, you cannot use those to analyze the motion of a race-car on an elliptical race-track.
i need to find the speed of acceleration
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Dr.Peterson
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Because acceleration is not "change in velocity"; it is "rate of change in velocity". If your speed increases from 30 mph to 50 mph over one hour, your average acceleration would be 50 mph divided by 1 hour. In your problem, you have to divide by 4 seconds, converted to hours.
These formulas relate to motion in a straight line, so that velocity is just signed speed: + means forward, and - means backward. If you are told that the motion is always in one direction, then the speeds are the same as velocities.
They also assume a constant acceleration.
How is this relevant to the problem you are solving?
Your work here suggest that you want to find the total distance traveled, in feet, given that a vehicle went 46 fps for 3.1 seconds, then 49 fps for the next 3.3 seconds, and so on. Where did that come from?
The Highlander
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Just to clarify a couple of things, would you please post a picture of the original question you are trying to answer here, thank you.
NB: It doesn't matter if the question is not in English, it would be helpful if we could see it (including any figures or diagrams that are part of it) in its original form.
It would also be helpful if you would tell us a bit more about why you want to solve it and what you already know.
You say you are an artist. What kind of artist are you? A painter, a sculptor, a poet, a writer, an actor or what?
What is you field of expertise? What do you already know a lot about?
Why do you now wish to learn about Maths? What level of Maths have you already been taught?
And why do you now wish to learn about Physics? What have you already been taught or know about that subject?
We can help you better if we know more about what you already know and what your objectives are.
(PS: I believe it has already been discussed and agreed(?) that when you write "tap", a better word to use might be "button". Please write button instead of tap to avoid confusion. Thank you.)
The car traveled for 4 seconds from 30 mph, steadily increasing to 50 mph,
so the upper bound on the number of feet traveled (it won't reach this) is
about 4*(73 feet) = 292 feet.
The problem posted by the OP is correct. Look at here:
Suppose that a car can accelerate from 30 mph to 50 mph in 4 seconds. Assuming a constant acceleration, find the acceleration (in miles per second squared) of the car and find the distance traveled by the car during the 4 seconds. | Numerade
VIDEO ANSWER: Suppose that a car can accelerate from 30 mph to 50 mph in 4 seconds. Assuming a constant acceleration, find the acceleration (in miles per secon…
No need for a picture or anything.
Why are you making the life of the OP difficult by asking him these non-sense questions?
Instead of writing a novel that will probably increase the OP confusion, try to help him numerically.
OP, you don't need to waste a lot of time in a simple problem. The problem is asking you to find the acceleration with the assumption it is constant. Therefore, you can directly use the formulas you provided.
For example, you can use this formula to find the acceleration:
[imath]\displaystyle v = v_0 + at[/imath]
Just make sure that you convert the units so that they are consistent with each other. Then, you can use the second or third formula to find the distance.
[imath]\displaystyle x = x_0 + v_0t + 0.5at^2\\v^2 = v_0^2 + 2ax - 2ax_0[/imath]
And you are done.
Or you can go the other way. Give Highlander a picture of the problem. Then, he will probably ask you about the name of the book. Then, the page, then the author, then the edition. And after wasting two months, he will probably tell you to follow what Dr.Peterson suggested.
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The Highlander
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Oh Dear! I see the old Mario is back!
(Leopards & spots come to mind.)
(Leopards & spots come to mind.
)Please refrain from personal attacks in this forum.
It was clear from the ambiguous subject line of this problem - "
It was clear from the ambiguous subject line of this problem - "
i need to find the speed of acceleration
that there could be many misinterpretation in the problem. In that case, a picture of the original problem will be helpful.Professor Khan,
I don't have anything personal against Highlander. I know that he is smart and his guidance in other threads was perfect, but in this thread, he went completely wrong.
The original problem was well written, despite the confusion of the OP between speed and velocity. Even Highlander himself does not understand that we can define velocity as the derivative of position when the initial position is zero. So this is not a beginner stuff. If we cannot solve the original problem posted by the OP in a way that he understands, this does not mean that there is something wrong in that problem. Be a gentleman, and please, don't blame the problem.
And lastly, the title of the OP problem is not ambiguous, it is wrong. I advise Highlander, instead of asking the OP personal stuff, he better starts by guiding him with the post title.
Small note:
You are not seeing the old Mario. If the old Mario comes back, Chaos will happen!
Only Dan can see him.
The Highlander
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Dear @mario99,
Regardless of whether it was intended to be a Maths question or a Physics question, in thirty years of teaching both, it was one of the worst I have ever seen!
However, it has become clear, over a number of threads, that English is not this OP's first language and my principal concern was that there may well have been something lost in translation. That was precisely why I asked for a picture of the original problem (in whatever language it was originally posed).
I'm afraid your reference to the Numerade website and accompanying assertion that what the OP posted was "correct" prove absolutely nothing!
For all we (you or I) know it could have been this OP who posted that question on that site (and a couple of others) only to discover that, in order to see the solution(s), signing up to those sites was required!
Unlike on this forum, where help and advice are provided freely and for free.
You can re-define velocity any way you like (in your own little world) but (as I pointed out), in Physics, velocity is defined as the rate of change of displacement with respect to time...
[math]\left(ie: \frac{\mathrm{ds}}{\mathrm{dt}}\text{, where s is defined as displacement.}\right)[/math]
In my opinion it would have been a waste of time addressing the inconsistency in the thread title when the OP had already displayed a much deeper lack of understanding in his second (and, subsequently, third) posts in the thread.
As it happens, I did try to help him understand how to approach the problem and prepared a lengthy explanation of how to go about it that would have been the 4th post in the thread.
However, the site stopped working for me (it does that occasionally) for several days and the only way I could get into the site was by using a VPN and even then I had to keep bouncing around servers in different countries (from the Netherlands to Romania to the USA and thence Poland). And when I was able to log in, I was only able to post one or two paragraphs; anything longer than that and the site simply refused to Preview or Post it!
That was why I took the opportunity to place a short post (#17) illustrating how to enter expressions like x2 and v0 (without the use of LaTeX) and provided a link to another thread where I had shown how to use the Equations of Motion (but I suspect you didn't bother to read that?).
When, finally, I was able to get back into the forum properly, there had been far too much 'water under the bridge' for me to make, what I considered to be, any more useful contributions but I did look back at all 14 posts that this OP has made and reading through them did give me cause to wonder exactly where s/he was 'coming from'.
The OP has claimed the ability to deal with quadratics and Eigen values and has also learned all about LaTeX in two days (that took me several weeks just to grapple with the basics) and yet s/he doesn't understand the difference between scalar & vector quantities or what the Domain of a function is!
It seems clear to me that (even if you don't agree) it would better enable us to address this OP's particular problems if we knew more about exactly what levels of knowledge s/he already possesses and what specific objectives s/he is working towards (as well as a clear picture of what this original problem looked like
).
You are, of course, entitled to disagree if you wish but, perhaps, in future you might do so a bit more politely (or even amicably)?
My Regards.
TH.
PS: Is this (below) intended to be some kind of threat???
Well what you wrote certainly sounded very "personal" indeed but I accept your apology.
You mean in your (not very humble) opinion!
The original problem was not at all "well written"!
Regardless of whether it was intended to be a Maths question or a Physics question, in thirty years of teaching both, it was one of the worst I have ever seen!
However, it has become clear, over a number of threads, that English is not this OP's first language and my principal concern was that there may well have been something lost in translation. That was precisely why I asked for a picture of the original problem (in whatever language it was originally posed).
I'm afraid your reference to the Numerade website and accompanying assertion that what the OP posted was "correct" prove absolutely nothing!
For all we (you or I) know it could have been this OP who posted that question on that site (and a couple of others) only to discover that, in order to see the solution(s), signing up to those sites was required!
Unlike on this forum, where help and advice are provided freely and for free.
Please don't presume to have any idea of what I "understand"; you know nothing about me or what I know.
You can re-define velocity any way you like (in your own little world) but (as I pointed out), in Physics, velocity is defined as the rate of change of displacement with respect to time...
[math]\left(ie: \frac{\mathrm{ds}}{\mathrm{dt}}\text{, where s is defined as displacement.}\right)[/math]
I am no "beginner" and, having taught Physics for thirty odd years (quite possibly since before you were even a twinkle in your daddy's eye), I strongly resent your suggestion in that respect; another apology wouldn't go amiss there!
I did not "blame" anything or anyone. I simply sought to obtain further information that might shed some light on exactly what difficulties the OP was actually having (and how best to address them).
Thank you for your suggestion but I think I will take a rain check on that.
In my opinion it would have been a waste of time addressing the inconsistency in the thread title when the OP had already displayed a much deeper lack of understanding in his second (and, subsequently, third) posts in the thread.
As it happens, I did try to help him understand how to approach the problem and prepared a lengthy explanation of how to go about it that would have been the 4th post in the thread.
However, the site stopped working for me (it does that occasionally) for several days and the only way I could get into the site was by using a VPN and even then I had to keep bouncing around servers in different countries (from the Netherlands to Romania to the USA and thence Poland). And when I was able to log in, I was only able to post one or two paragraphs; anything longer than that and the site simply refused to Preview or Post it!
That was why I took the opportunity to place a short post (#17) illustrating how to enter expressions like x2 and v0 (without the use of LaTeX) and provided a link to another thread where I had shown how to use the Equations of Motion (but I suspect you didn't bother to read that?).
When, finally, I was able to get back into the forum properly, there had been far too much 'water under the bridge' for me to make, what I considered to be, any more useful contributions but I did look back at all 14 posts that this OP has made and reading through them did give me cause to wonder exactly where s/he was 'coming from'.
The OP has claimed the ability to deal with quadratics and Eigen values and has also learned all about LaTeX in two days (that took me several weeks just to grapple with the basics) and yet s/he doesn't understand the difference between scalar & vector quantities or what the Domain of a function is!
It seems clear to me that (even if you don't agree) it would better enable us to address this OP's particular problems if we knew more about exactly what levels of knowledge s/he already possesses and what specific objectives s/he is working towards (as well as a clear picture of what this original problem looked like
).You are, of course, entitled to disagree if you wish but, perhaps, in future you might do so a bit more politely (or even amicably)?
My Regards.
TH.
PS: Is this (below) intended to be some kind of threat???
Small note:
You are not seeing the old Mario. If the old Mario comes back, Chaos will happen!
Only Dan can see him.
Dr.Peterson
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I'm confused. I wonder if you are confusing the problem with the OP's misunderstandings.
The problem was this:
I see nothing wrong with that. Can you show specifically what you think is "poorly written" about it?
The poster obviously has major misunderstandings, starting with the title; but we need to keep that separate from the problem itself.
The problem is, indeed, straightforward, and I see nothing unclear about it. The car's speed increases by 20 mph, which is 20/3600 = 1/180 mi/sec, in 4 seconds, so the acceleration is (1/180)/4 = 1/720 mi/s^2. The initial speed is 30 mph, which is 30/3600 = 1/120 mi/sec. The distance traveled in that time is [math]\frac{1}{2}at^2+v_0t=\frac{1}{2}\frac{1}{720}(4)^2+\frac{1}{120}(4)=\frac{2}{45}\text{mi}[/math]
This is 0.0444... miles, or 234 2/3 feet.
(This is also the average of the distances that would have been traveled at 30 and 50 mph in 4 seconds.)
The units are awkward, but I see no reason to think the problem was mistranslated from another language, or was written by someone who doesn't understand the subject.
I will start with physics and anything I say that you don't like has nothing to do with hating.picture of what this original problem looked like
You are, of course, entitled to disagree if you wish but, perhaps, in future you might do so a bit more politely (or even amicably)?
My Regards.
TH.
PS: Is this (below) intended to be some kind of threat???
If [imath]\bold{a} = \bold{p} - \bold{p_0}[/imath] And [imath]\bold{p_0 = 0}[/imath]
Then
[imath]\displaystyle \frac{d\bold{v}}{dt} = \frac{d\bold{a}}{dt} = \frac{d\bold{p}}{dt}[/imath]
Look above and convince me that it is wrong to say velocity is the derivative of position! I am not against the definition of velocity. I am against you when you said velocity is not the derivative of position.
If you cannot convince me while I am an amateur, then you are whether a not very good physics teacher, or you are stuck and don't want to embarrass yourself (an amateur was right), so keep saying velocity is the derivative of displacement which everyone knows and agrees.
Teaching physics and mathematics for 30 years does not make you better than any senior student who solved every single problem.
I have myself seen a lot of this kind who knows a lot of nothing else. When I was studying physics, I loved to solve all the problems of the chapter, especially, those with level III because they require Calculus. One time I was stuck in a simple flux problem which can be solved with or without calculus.
I took this problem to the teacher and I was shocked that he did not even understand the problem which is basically level I, not even III. Later, I solved it by myself by both ways. In the same time, I was studying Electric Circuits course.
Physics has also a chapter about electric circuits. When we arrived to this chapter in physics, I realized that the physics teacher method in solving problems in circuits was different than the teacher of the electric circuit course and when I and some of my colleagues applied the latter technique on physics, the physics teacher marked wrong on our answers, however, the answers were correct, only because we used a different approach than his. His excuse was that our steps to solve the problem were wrong and by coincidence we got the correct answer.
Then I realized that this Physics' teacher understands nothing else (than what he was teaching) and he was just a machine which was used to solve the same problem 1000 times in the last 30 or 40 years. He could give us the numerical approximation solution to any problem he solved rounded to even the 10th decimal place before we even started solving. (Proof he was a machine.)
Another example is of another physics teacher who has a doctorate in physics and was teaching Physics for more than 30 years, then he started MBA, and changed his career. After one year only he could not answer any physics problem I gave him, saying he forgot everything.
What I am trying to say above, teachers of this kind can focus only on the material that they teach and forget everything else with the years passing. I myself gave many problems involving differential equations to experienced math teachers who could not solve them. Why? because they were used to teach something else in these last years and were stuck on it like a machine.
On the other hand, teachers and professors like Dr.Peterson, Khan, Steven, topsquark, lookagain, and many others threw themselves in online forums to face all kinds of problems in a daily basis. Therefore, they will not be stuck like a machine on the material they are teaching in school, university, or wherever. Those deserve the word "Real" instructors. (Are you one of them? Then convince me.)
Everything said above meant that your 30 years of teaching might be not as powerful as you are expecting.
Back to the OP's main boring subject. After reading what you have written, I think that I know what was your real intention. You just want to prove to the world that the OP was wrong when he said the original problem was a physics problem. You want to be a hero by saying it was a Calculus problem. Let us assume it was a Calculus problem. What has been changed? Did you win the Olympiad Gold Medal or World Cup? No.
Your excuses are not strong enough. After that you said the OP learnt Latex quickly. What I understand is that you meant that it was impossible for a beginner to learn Latex faster than you (the Genius Highlander). Even if that was true, what that has to do with the original problem posted by the OP? Nothing!
After that you jumped to another problem posted by the OP regarding system of differential equations. You are trying to say it is impossible that the OP knows how to find the eigenvalue of a differential equation, but does not know how to find the domain of a function. Even if that was true, what that has to do with the original problem posted by the OP. Nothing!
Finally, the chaos Mario or the old Mario that will cause chaos has nothing to do with threats or anything. This is an old subject, Dan and a few others know about.
I am not going to comment on anything regarding our fight after this post, so I will advise you to do two things which I am sure that you will not do.
1. [imath]99[/imath]% of your problems in this forum because of your big font size. why not reduce it to normal?
2. don't complicate the easy things.
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What are p, p0 & a ?
Correction.
[imath]\displaystyle \bold{v} = \frac{d\bold{a}}{dt} = \frac{d\bold{p}}{dt}[/imath]
[imath]\displaystyle \bold{v} = \frac{d\bold{a}}{dt} = \frac{d\bold{p}}{dt}[/imath]
So obvious!
Dr.Peterson
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An equation is meaningless unless the variables have been defined. Clearly you don't mean that a is acceleration and p is momentum. What do they mean?
topsquark
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Mario, I've generally got you on "ignore" but this is a long (Physics) thread and I thought I'd keep up with it from time to time.
The OP cannot handle your "simple" equations and needs some work on where they came from. Yes, someone familiar with the equations can solve this in two lines. The OP clearly cannot and needs more help.
Understanding the material is far different from being able to teach it properly. Teaching Intro Physics is hard: it is far more than just showing someone how to get the right answer. Your comments and your own mistakes are taking over the thread. This isn't your question. I would appreciate it if you would back down before you confuse the OP more and let the other members do the good job they've already been doing.
-Dan
The OP cannot handle your "simple" equations and needs some work on where they came from. Yes, someone familiar with the equations can solve this in two lines. The OP clearly cannot and needs more help.
Understanding the material is far different from being able to teach it properly. Teaching Intro Physics is hard: it is far more than just showing someone how to get the right answer. Your comments and your own mistakes are taking over the thread. This isn't your question. I would appreciate it if you would back down before you confuse the OP more and let the other members do the good job they've already been doing.
-Dan
Really - please define those terms..... what do those mean......
logistic_guy
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post 15 told me divide 10 intervals to find the distance. why my calculation is wrong? how to find the speeds of 10 intervals given in post 16?
in post 24, he say the distance is 292 feet? how? i can't find this distance
in the question, he don't say the car moving in a straight line. do i assume straight line to use the 3 formulas? if i can do this i think i can find the acceleration
there are lots of posts, i'm sorry i can't read all of them, but thanks for helping me
in post 24, he say the distance is 292 feet? how? i can't find this distance
in the question, he don't say the car moving in a straight line. do i assume straight line to use the 3 formulas? if i can do this i think i can find the acceleration
there are lots of posts, i'm sorry i can't read all of them, but thanks for helping me
That sad face does not answer the questions posted in response #35 and #37.
Please read the post authored by @topsquark - and digest it!
Yes ..... in my opinion ...... and state it explicitly and add "with constant acceleration"
before starting to solve the problem.