Calculus / Physics: What is Acceleration given Velocity = V(x)

At any instantaneous point in time, acceleration is a derivative function. That should help you start the homework assignment.

Hurst gave you the physics of the problem.
But you still have difficulties. Are you sure that the equation you posted is correct? What is the meaning of x? Is it position? or is it time?

The units you posted with your equation are incompatible with the usual definition of x (as a distance)

x could be position as a function of t (time).
What is the domain of x? If it gets too large, then you end up with sqrt of negative numbers.

If x is position, then to get acceleration, just do what John suggested by taking dV/dt where x = x(t). Use chain rule. To get full answer, you need to know what x(t) is.
Otherwise, you can leave result with a dx(t)/dt factor.

note that when v = 0   x = 11.25
phoffric says
" To get full answer, you need to know what x(t) is."
which is very true. Ie you cannot find the acceleration from the info given. The acceteration can be anything from 0 to a VERY large number.
Suppose that initially v is a constant v = sq rt (90) so that a = 0.  Then after the object reaches 11.25 the acceleration will have to be VERY large as it comes to a screeching halt
alternatively the object undergoes a VERY large acceleration as soon as it leaves x = 0. It will then be going very slowly and will take a VERY long time to get to x = 11.25. Reality can be anything inbetween these two actions. Thus no definitive answer can be obtained to this question without knowing x (t).

the equation can also be written as:
V= (180)^1/2 -(16x)^1/2
which is easier to differentiate

I do not want to do this completely for you as I assume you are the one who needs to learn ... the constant drops off and the exponent is multiplied by the value in front of x -- be sure to keep the sign (negative). 1 is then subtracted from the exponent and 'x' will be kept.

using general terms:
if y = c+ Ax^n
then
dy/dx= (n * A) x^(n-1)

when plotted, acceleration would be the slope of the line at each point  (in time) for the velocity.  I assume this is a single point in time as, usually, 'x' means distance (offset from start) and 't' means time. However, the variable names might mean other things in this equation if it is not using standard terms.  

-------------------------------------------------------------------------------------------
Generally:
x=x0 + vt + at^2

WHERE
x = position at a point in time (t)
x0 = starting point
v = velocity (start, as velocity changes over time)
a = acceleration (assuming this is constant)
t = time

>> Generally:
>> x=x0 + vt + at^2
This equation has limited specific, not general, application.
Generally, this equation is not true. There is no reason to assume that acceleration is constant over time from the OP.

the information under
-------------------------------------------------------------------------------------------

was added as an fyi -- the question asked is answered above that line and main points are bolded.  <<EDIT> Answer deleted as we want the author to derive the answer.

The equation provided is standard physics, which I studied.  Velocity, which does change, can be calculated from the starting velocity, acceleration, and point in time.  The acceleration constant, denoted by 'a', generally IS constant although the result of the equation is different depending on where you are in time and varies with time squared (until critical acceleration is reached) .  Acceleration is the change in velocity. Velocity is the speed.

The poster's question simply asked for the acceleration*, which can be calculated by differentiating velocity, which I wrote how to do.  The other stuff was added to put things into perspective.

* and, obviously, for a single point since there was no time component.  I agree that it does not make sense for velocity to be determined based on distance (which 'x' usually refers to) otherwise ... also noted by John by his "instantaneous point in time" comment.

unit for velocity is distance/time -- ie: m/s or ft/s

The biggest difficulty with the posts is that

(180-16x)^1/2   IS NOT    = (180)^1/2 -(16x)^1/2

no matter how much easier it is to differentiate

oh, what a dumb mistake -- of course you are right! My calculus is a bit rusty -- I'll have to unpack my books and fix that

original equation:
V= (180-16x)^1/2

rearranging to put the term with x first:
(-16x + 180)^(1/2)

this is in the form:
(ax + b) ^(1/2)

WHERE
a = -16 = -4 * 4
b = 180 = 45 * 4

to differentiate (since acceleration is change in velocity), multiply by the exponent and  the coefficient of x , then subtract 1 from the exponent:
(1/2)a * (ax + b) ^(-1/2)

remember to simplify -- the square root of 4 is 2

No response from the author and we had questions about the OP that were unresolved.
Recommend deletion of the question.

Yes. Please Delete.

The question has been answered completely. It was not homework because no general physics course would have asked this question. The question may not have been exactly as the author intended but the question as asked, was answered. Comments on the comments follow (the number represents the fraction of total points to be awarded to that comment).

41783075    0.1     the basics outlined
41783090    0.6     the correct mathematics given
41783260    0.3     a word picture of the problem showing the application of the answer and illustrating why x(t) is necessary

Since this is, as John suggested, a likely homework problem, I just tried to offer some guidance.
Assuming that V = V(t), velocity; and x = x(t), position, then
Given in OP: V= (180-16x)^1/2

I must be making a mistake. Here are my first steps:
(1) V(t)= sqrt( 180-16 * x(t) ) assuming that x(t) <= 11.25 to avoid imaginary numbers.

(2) a = dV/dt = d/dt((180-16x)^1/2 ) = .5 * (-16) dx/dt / sqrt(180-16x)

Could you correct my above mistake. If you do not think there is an error, then I must be making a mistake in my next two steps. How about if you post the next two steps and I will compare results. Better yet, since this is homework, just let me know if you get a surprising result. I did, but it must be wrong.

generally, it would be:
x =x0 + v0(t) + a(t)^2

WHERE
v0 is the original velocity
a = constant acceleration

the original equation did not have an acceleration component -- nor did it have time.  However, X might be time or it might be distance (as X is generally used to represent distance -- but not necessarily -- in which case, it would be for a point in time).

differentiating yields:
(1/2)(-16) * (-16x + 180)^(-1/2) = -8 * (180 - 16x)^(-1/2)

simplifying, 2 can be taken out of everything:
= -4 / (45 - 4x)^(1/2)

The question as asked is not homework (see 41804105)

'I think you work shown is correct.

"How about if you post the next two steps and I will compare results. Better yet, since this is homework, just let me know if you get a surprising result.. "

But what are the next two steps? and what is your surprising result?

The question as asked is not homework (see 41804105)  That is your own assertion that it is not homework. No one else's.

in my immediate comment, "your work" refers to comment 41805438, proffric

"The question as asked is not homework (see 41804105)  That is your own assertion that it is not homework. No one else's. :

true, but reasons were given for that conclusion and no reasons have been given for any conclusion that it is homework.
In my mind the question does not represent exactly what the author thought he was asking. What he thought he was asking might very well be homework but the question as asked was interesting in itself and represent an opportunity to learn about calculus, physics, and the visulization of problems.

I agree ~ and if it was homework, it is probably late ;)

I didn't think of this as a physics problem, but as a parameterization calculus problem with an application in physics. The title, "Calculus", really makes me think this is a homework or self-study problem, rather than a professional issue.

@Crystal,
Too late or not for the author, not sure we are supposed to provide solutions. After all, if it is a homework problem, then someone could look up the answers in the future.

@Crystal,
>> generally, it would be:
>> x =x0 + v0(t) + a(t)^2
I didn't think your formula is true when a(t) varies over time?

Recap: Lecture #2 Constant Acceleration
(1) v = vo + a t
(2) x = xo + vo t + ½ a t^2
https://www.physics.ohio-state.edu/~dws/class/131/lecture_recap.pdf

@aburr,
You agreed with my first two steps. Since I think I am making a mistake. Here are my first 3 steps:
 (1) V(t)= sqrt( 180 - 16 * x(t) ) assuming that x(t) <= 11.25 to avoid imaginary numbers.

 (2) a = dV/dt = d/dt((180-16x)^1/2 ) = .5 * (-16) dx/dt / sqrt(180-16x)

Since, V(t) == dx/dt, can't I substitute dx/dt with V(t) to get the following?
(3) a(t) = dV/dt = .5 * (-16) V(t) / sqrt( 180 - 16 * x(t) )

If you agree with this Step3, could you try reducing this further, as I am pretty sure that I am reaching a ridiculous conclusion (that contradicts other numerical approaches).

One reason I argued for deleting this question is that we have said various things about the OP - like units don't make sense; like maybe x is actually t (and I'll add now just to be ornery, maybe V is t, and x is V). Depending on how we choose the problem to be, it may be interesting or not. For this reason, I think the PAQ will not benefit at all from this question.

But if aburr can figure out what my 4th step is, I would like his best guess, so keep it open for a bit longer. (Or give out points when we are done. I made my case - you can make yours.)

phoffric,  I should have looked it up -- obviously forgot the 1/2 in front of acceleration

I did come up with the right answer in 41805500 though but it took a lot of mistakes first! this thread was good to bring things back to my mind before they are really gone ;)

@crystal,
Of course you can always check your work when taking derivatives:
http://www.wolframalpha.com/input/?i=derivative+(180-16x)%5E1%2F2

In your post, #a41805500, I guess you are in the camp that x is t. For some reason, call it habit, I assumed that x is x. But who knows. I don't think the author knows given his lack of responses to our questions.

BTW, I didn't even notice that you left out the 1/2 in your equation. I was trying to emphasize that your equation only applies for constant acceleration, which is often not the case.

it appears this was his one and only question -- probably right before the assignment was due. I am guessing that because it is not a real-world example and only schools do that, lol

> "I was trying to emphasize that your equation only applies for constant acceleration, which is often not the case"

"a" is a force -- not the value of an object's speed

if you drop an object, it will pick up speed until it cannot go any faster (when the force of gravity and air resistance are equal). The rate at which acceleration happens, at first, is mostly due to the force of gravity (which is constant) but there are other factors -- resistance /friction, shape, ....  At first, gravity is big compared to other things but sometimes shape makes a huge difference  -- consider parachutes.

If you roll a puck on a smooth, flat surface with no incline (and, therefore, gravity is not considered), there is friction and it eventually stops.  The rate at which it slows can be determined by the force of friction as other factors are generally negligible.

It is years since I took physics ... and calculus! So my memory for these things isn't what it used to be. I have enjoyed thinking about them again.

aburr,
Are you inclined now to delete this question? If not, I will not argue. I honestly don't know what the question is. I made some assumptions about V and x, but obviously, others have made different assumptions.

Do you think my Step3 is correct? If so, I will have to look again tomorrow at my analytical Step4. And also, I will have to re-evaluate my numerical methods.

The analytical method and the numerical methods (using Matlab) are completely at odds with each other. Not even close. I am inclined to think that the numerical methods I use are more correct where I simply pick an arbitrary x(t) subject to the constraints to avoid imaginary numbers.

I am hoping that someone will independently tell me that something in my 3 steps is incorrect because Step4 seems ridiculously simple and yet ridiculously wrong.

I didn't want to bias anyone and lead them down a false path. Maybe tomorrow.

I may have to call in the cavalry.

Phoffric
My picture of the problem is all wrong. Your analysis may be correct. As a check on my work (which is often wrong) what is your unusual result?

Proffric
This is not quite right
V(t)= sqrt( 180-16 * x(t) )            Your equation  is not quite right. it is V(x)= sqrt( 180-16 * x(t) ) which is not quite V(t)

"Since, V(t) == dx/dt, can't I substitute dx/dt with V(t) to get the following?" You can, but V(x) not=V(t).
I thought I had it all worked out but I fell asleep so will have to continue tomorrow.

By the way, I have no idea how to solve this analytically.

hang on, I may have an easy solution but I have a couple of other things for next hour

If we defined a different x(t) where we know that A varies, then Step4 above still gives A = -8.0; a paradox.
Not sure what happens if we do that in d-glitch's spreadsheet.
But, if I have a concrete x(t), then for all values of t, then, we better have V(t) from the OP matching the dx/dt.
So, the x(t) cannot be arbitrary. I think the non-linear differential equation needs to be solved if we are interested in what family of x(t) are legitimate for the OP.

x(t) is not arbitrary.

V(x)  =  dx/dt  =  f(x)  =  sqrt(180-16x)   is a differential equation in x and t.  

Integrate it with respect to time to find x(t).  This is what I did in the spreadsheet, and can not do analytically.   The solution is defined except for an arbitrary constant of integration  x(t=0) = C

No matter how the projectile gets to x=0, its subsequent behavior is defined.

I can redo my calculations starting from a different point, for example  x=-5  at  t=0.

The projectile takes a while to reach x=0, but it still stops at x=11.25
The acceleration is still the same constant value.

Things are much simpler than we realized. We were put off by the title of the problem. No calculus need be applied although as d-glitch showed it can be used. The problem, correctly understood, is of a type often presented in general physics so could very well have been a homework problem. Part of the problem is that the numbers in the problem have units with 16 having the units of acceleration and 180 being related to the initial position of the particle. Proffric, calculations are not as uncertain as I indicated although I have not really cleared that point up. (and probably will not). The next post restates the solution (and problem)

I now think that our last few posts should be the accepted answers representing a (1) math analytic solution, (2) a math numerical solution, and (3) a physics modeling solution.

but first..

@aburr,
>> I considered this problem to be that of a particle in a uniform force field
Not sure how you made this uniform force field assumption w/o knowing from the analytic and numerical approaches that we get a constant acceleration of -8 m/s^2. Please help me understand this point.

(BTW, d-glitch, you have m/s^3 in your posts.)

>> A good physics student should recognize this as a particle subjected to a constant force where
>>  v = (c -2ax)^1/2    where c is related to the initial starting point.
Good point. I went back to that link I posted earlier:
https://www.physics.ohio-state.edu/~dws/class/131/lecture_recap.pdf
Letting x1 = 0, and v1 = sqrt(180) (see below), we have the form in the OP.
V= (180-16x)^1/2
So immediately, we get a = -8 m/s^2.

v2^2 = v1^2 + 2a(x2 - x1)
v2 = sqrt( v1^2 + 2a(x2 - x1) )

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Gee, that was easy.

@crystal,
Re: #a41805500 which you said is the right answer. We now believe the right answer is -8 m/s^2.
I didn't see that answer in your post. If you believe your answer is the right one and ours is wrong, now would be a good time to clarify.
Thanks,
phoffric

@d-glitch,
Thanks for joining us at this late time. I thought my -8 that I was getting in Step4 was a mistake. Your spreadsheet certainly helped.

I modified the question a little to make it seem more sensible.
Given how long this thread took, I marked it unavailable for the search engines.

">> I considered this problem to be that of a particle in a uniform force field
Not sure how you made this uniform force field assumption w/o knowing from the analytic and numerical approaches that we get a constant acceleration of -8 m/s^2. Please help me understand this point."
The graph of V(x) is a straight line. If the field were not uniform the line would have bends in it.

@aburr,
>> 180 being related to the initial position of the particle
Couldn't 180 be related to the initial velocity of the particle?

@aburr,
>> The graph of V(x) is a straight line.
>> Plot V vs x              v(0) = 180^1/2
>>                                   V is 0 when x = 11.25
>>  One can connect these two points by a straight line

I am a bit confused, since V(x) = sqrt(180-16x).
Can you help me out of my pickle, since V(x) doesn't look like a straight line.

in the real world, 180 would not be initial velocity or it would be adjusted by acceleration forces acting upon it . Rather, I think the example was dreamed up by some math professor who found out that differentiating velocity is acceleration so he just threw that in. The original question states, "since we are assuming one dimension" and indicates why time is not part of the equation.

We are in deep trouble.  The acceleration is NOT a constant -8.  The velocity is NOT a straight line.
regard Proffric comments
1.   @aburr,
>> 180 being related to the initial position of the particle
>> Couldn't 180 be related to the initial velocity of the particle?
It IS related to the initial velocity. I mis-wrote myself

2.  @aburr,
>> The graph of V(x) is a straight line.
>> Plot V vs x              v(0) = 180^1/2
>>                                   V is 0 when x = 11.25
>>  One can connect these two points by a straight line

I am a bit confused, since V(x) = sqrt(180-16x)
Of course you are confused. I am wrong.

I think we are going wrong when we ask as did Proffric

"Since, V(t) == dx/dt, can't I substitute dx/dt with V(t) to get the following?
(3) a(t) = dV/dt = .5 * (-16) V(t) / sqrt( 180 - 16 * x(t) )"

The answer is Yes you can

BUT

V(t) does NOT = V(x).
We do not know what V(t) is.
Most of my comment 41807810 is wrong. It will take me a while to rework it.

V(x) is not a straight line. Therefore the acceleration is NOT constant

d-glitch
Your spreadsheet is very interesting. Did you carry x as far as the region around x = 10 to 11 ?
Your general idea is guarenteed to give us insight. I will be interested to see the results with x near 11.

I got tied up with other things this evening and possibly for two more days. Certainly by Friday or Saturday I will provide a better analytic solution.
Feel free to post it if you have one on your mind.

BTW
>> Proffric
I am neither a professor, nor a prophet, and not much profit either.
I am just a phoffric.

I fail to understand how a simple question can result in 51 postings!

The given function for the velocity is dependent on "x". Since velocity is by definition a time related event the equation should be written V(x,t) = .....

That said the acceleration is then dependent on "x", but one differentiates the velocity with respect to time NOT "x", so the differentiation becomes one half of the term in brackets raised to the power minus one half times the differential of the term in brackets. The 180 being a constant contributes nothing and the -16x term becomes -16*dx/dt. So the acceleration becomes minus 8 divided by the square root of the term in brackets multiplied by dx/dt.

Now here we have a problem of terminology because we were never told what x was - there is a topic adviser comment that it is the position, but it could have been the amount of resistance experienced during travel or anything else. But if one assumes that it IS the position then we are forced to conclude  that dx/dt is the rate of change of position, and that is by definition the velocity. That is that dx/dt is V(x,t) and substituting that for dx/dt in the equation for the acceleration we get -8, which is of course a deacceleration.

We have fallen into a sticky calculus trap.  
Acceleration is the time derivative of velocity, but we don't have an explicit function of time.

Go to a different planet and drop a unit mass projectile from height x=h
From energy considerations: [aburr]
   mv²/2 =  mg(h-x)
    v²   =  2gh - 2gx
    v    =  ±sqrt(2gh - 2gx)     

This has the form of the original equation with  
    g= 8 and h= 11.25

Positive acceleration  a= -g = -8 m/s²  

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Now do the drop at t=0 and apply some calculus:
     a(t)= -8   x(0)= 11.25   v(0)= 0    
     v(t) =        -8*t  ==>   t = -v/8
     x(t) = 11.25 - 4*t² ==>   t = ±sqrt(11.25 - x)/2

Eliminate t for the parametric equation   
     v = 4*sqrt(11.25 - x)

and we are back to where we started.

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We have fallen into a sticky calculus trap.  [d-glitch]
Yes, by failing to simply apply the rules.     [BigRat]

I agree, but if I didn't dig myself into a hole and dig myself out once in a while I probably wouldn't recognize me.  
I remember my multivariable calculus course.  It was in 1971.  Actually, that is the only thing I remember about it.
As a result of this thread, I just realized that I don't remember anything about the chain rule either.  And that is why I am here.

I had this yesterday, but I didn't complete my verification. Maybe this weekend..

http://www.wolframalpha.com/input/?i=x'(t)+%3D+(180-16x)%5E1%2F2
If wolframalpha solved the DE correctly, then:

OP: V= (180-16x)^1/2 m/s
Since V = dx/dt:
x'(t) = (180-16x)^1/2 

x(t) = ( - 16 t^2 -16 c1 t  - 4 c1^2 + 45 ) / 4
If this satisfies the DE, then
V(t) = x'(t) = -8t - 4*c1;   where c1 = -V(0)/4
x''(t) = -8;

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I started checking results, but I got a sign wrong in my haste, so I will review again when I have more time. The solution does allow for a perfect square expression within the sqrt, so that part is good since that allows removal of the sqrt in the check.

Another round-about solution, phoffic. Make a differential equation, then solve it and then find the first and then the second differential. And why does one have to "allow a perfect square expression within the sqrt"? This is all too complicated and error prone - as you right point out when you wrote "but I got the sign wrong".

@Crystal,
What is the meaning of your result?
What are the units of your result?
What is the solution to the OP?

acceleration is the change in velocity, which is the slope of the line/curve by graphing the equation at each point. The equation for slope can be determined by differentiating the equation for velocity.  

> What are the units of your result?
m/s^2

@crystal,
>> acceleration is the change in velocity
Ok, but we still need change in velocity "with respect to <some unit>".

I think earlier, someone wrote dV/dx, and later someone wrote dV/dt.
In order to be clear, you need to spell out which one you are doing.

derivitive of the the given equation would be called dv/dx
the rate of change is the difference in V over the difference in X -- and this is the acceleration

@crytstal,
You might be interested in this article:
https://www.khanacademy.org/science/physics/one-dimensional-motion/acceleration-tutorial/a/acceleration-article

thanks, phoffric, I did find that interesting.

The given equation had no time component unless it was x.  All we can know is that x is a variable. No  assumptions can be made about what x really is since no more information was given.

@crystal,
>> All we can know is that x is a variable. No  assumptions can be made about what x really is since no more information was given.
You are right. I guess all you had to do to be clearer in your last math post was to say that you are assuming that x is t, which in earlier posts had been stated. And given all the possible interpretations, I suggested that this question be deleted since the author was not clarifying. In the future, I would be not take for granted that we all understand what may be obvious to you, and to explicitly state all your assumptions and be complete in your description of your differentiation.

But to make the discussion single-tracked after seeing that we were understandably going in multiple directions, I decided to arbitrarily make some assumptions about x and V, and that is why I edited the OP so that we could assume that x = x(t) m and that V = V(x(t)) m/s.

Hope we are all getting something out of this. I am looking forward in reviewing d-glitch's spreadsheet this weekend.

it doesn't matter if x is time or not.  If x is position, then the equation is probably for a point in time (hence the "one-dimension" directive). If x is time, then it still doesn't matter for differentiating. x is the only variable in the equation.

> "explicitly state all your assumptions "

No assumptions were made to derive the answer. Only what was given by the original poster was considered.

differentiation is finding derivitives

>> it doesn't matter if x is time or not.
If you are trying to get acceleration, then if x is not time, then dV/dx is not acceleration, as its units is 1/sec.
For acceleration, we need dV/dt, whose units are m/s^2.

Acceleration is probably m/s^2 since velocity is expressed in m/s.  Yes, that is an assumption -- but not an assumption required to derive the equation.

>> Acceleration is probably m/s^2
Actually, Acceleration is m/s^2 by definition. No assumptions needed for this.

not necessarily -- it is any unit of distance over a unit of time squared. And while possible but not likely, the equation could be converting units.

A Google search for   velocity. 180 - 16x   indicates that this problem is all over the web.  
The definition of the variables is usually clarified, but not always.  One early instance actually indicates uniform acceleration in a straight line, which would have been nice to know.  
Some of the on-line solutions are incorrect.  I have not seen much in the way of discussion.

The earliest version I have found so far is in a 1999 issue of Competition Science Vision, which is a practice prep magazine for the Indian version of the MCAT.
https://books.google.com/books?id=7ecDAAAAMBAJ&pg=PA814&lpg=PA814&dq=velocity.+180+-+16x&source=bl&ots=Ap25wKHddu&sig=0Uw-WlQ52aHcjjFTsgy-6Y_PL94&hl=en&sa=X&ved=0ahUKEwiHv5WvuaTPAhXGMSYKHT10Do8Q6AEIITAI#v=onepage&q=velocity.%20180%20-%2016x&f=false

This link also provides a solution with some unfortunate typography on the penultimate step.  [A product term is broken over two lines.]  
I am willing to forgive our substantial confusion here due to the ambiguity of the original post.  
I note that aburr pointed this out early in the process.

I am happy to have worked on this question.  I really think there is a subtlety here that is not captured in the terse statement of the problem.  I hope most of us have learned (or relearned) something.

>> >> Acceleration is probably m/s^2
>> Actually, Acceleration is m/s^2 by definition. No assumptions needed for this.

>> not necessarily -- it is any unit of distance over a unit of time squared.

I bet we are not talking about the same thing. Maybe you are talking about interpretations of the OP. I am just talking about the units of acceleration which is m/s^2, and if V is velocity, then dV/dt is the acceleration and has those same units, m/s^2.

If you still disagree with this point, then let me know by showing an online reference supporting your position.

Sometime this weekend, I would like to either delete this question, or if you do not desire that, then we can remove most of the posts so that the results are more focused. Given d-glitch's research just posted about this problem being common, I am inclined to say that this has been an interesting reverse engineering project on the question itself, and just let it go into the delete basket.

t is a conventional term but it was not a variable used in the equation. The equation used x, which is what was used to get the result.  If the equation had been rewritten to substitute t for x,  or z for x, or bananas for x, then it would have been called something else.

I was happy to work with you all.

If you didn't recognize that a is constant from the OP, then does anyone know how to solve the ODE in #a41809821 whose solution is provided by wolframalpha. If we guessed that A is constant, then we would guess that x(t) is a parabola and just work out the coefficients using boundary conditions.

x'(t) = (180-16x)^1/2

Answer:
x(t) = ( - 16 t^2 -16 c1 t  - 4 c1^2 + 45 ) / 4
 

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Try the substitution 16x/180 = sin²u

>> Try the substitution 16x/180 = sin²u
Thanks for the tip.

Now, all I have to do is check my work and see whether this matches Wolfram and d-glitch's spreadsheet.

x'(t) = (180-16x)^1/2
x'(t) = 180(1-16x/180)^1/2

Let 16x/180 = sin²u
--> x = 180/16 sin²u
Then 16/180 dx = 2 sin(u) cos(u) du
--> dx = 2 (180/16) sin(u) cos(u) du
--> dx/dt = 2 (180/16) sin(u) cos(u) du/dt

dx/dt = 180(1-16x/180)^1/2 = 180(1-sin²u)^1/2 = 180 cos(u)
2 (180/16) sin(u) cos(u) du/dt = 180 cos(u)
2/16 sin(u) du/dt = 1
sin(u) du/dt = 8
sin(u) du = 8 dt

sin(u) du = 8 dt
-cos(u) = 8t - C1
cos(u) = -8t + C1
cos²u = (-8t + C1)²
1 - sin²u = (-8t + C1)²
1 - 16x/180 = (-8t + C1)²
16x/180 = (-8t + C1)² - 1
x = (180/16) [ (-8t + C1)² - 1 ]
x(t) = (180/16) [ 64t² + C1² - 16 C1 t - 1]

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The powers that be believe that, thanks to our lengthy discussion, that this question should not be deleted.
IMO, I think that points should be awarded not based upon the original ambiguous OP, but rather on my <<EDIT>> assumptions to the OP.

Let me know if you think we need more discussion and/or checking of work.

with all due respect, I do not agree with the edit to the original post ... t  was not in the equation and it got added it in.  Additional comments should have been placed below what was already there.  If anything, the extra stuff dug up from the web to define the problem should be added.  Other comments have been edited also.

I sill do not understand why it is necessary to solve a differential equation to then take a second differential to produce a result. Nor do I see why it is necessary to calculate spreadsheets. It may be all very interesting but a straightforward differentiation and substitution is all that is necessary.

As far as the <EDIT> is concerned I have already asked when this took place and by whom. It seems now from post 41813323 that phoffic did this. Good, But when?

By the way, phoffic, line 2 is wrong. When the 180 comes out of the root it gets reduced to 6 root 5.

Thanks for identifying the 180 slip up.

I don't see how knowing exactly when I did the <<EDIT>> of the OP is going to help us functionally. It was some time after I did an RA requesting that this question be deleted since the contributing experts would not agree on what the question actually represented and since there was no feedback from the author. I am guessing it was some time after #a41804105 .

Even now, there is objection to my <<EDIT>> from crystal.

I said that I had come up with a solution in #a41805438 , but did not show my steps 3 and 4 (which produced a -8 m/s^2 acceleration), hoping that others would confirm or deny my first two steps. I did show my 4 steps after d-glitch came up with a numerical solution that agreed with my analytic solution in #a41807393 , which shows a simple analytic solution.

As early as #a41804059 and #a41804060 two of us recommended deletion of this question.
Now that d-glitch researched the problem on the web ( #a41812525 ) and found it is usually better defined there, I am even more inclined to recommend a deletion.

I think we were just having fun trying to reverse engineer the question with different assumptions and working out what we thought the answer might be.

Had the question been as clear as elsewhere on the web, there might have been 2-3 posts instead of 400 posts. (If we are not at 400 yet, we may get there in the next week.)

I'm sorry phoffic, but there seems to be a lack of transparency here. Contributing to a thread, which you were the TA who added important information, which was basically guesswork, then deleting members comments, including your own, all seems to me not very fair. OK, having fun I can understand, but the playing fields ought to be level. This is or ought to be a technical thread - it is not the Lounge and I for one posted a solution in the hope of helpful. I find the overall lack of clarity and confusion not up to E-E standards. Sorry but that is how I feel.

We should have deleted this question as proposed early and just made it a blog.

Both time and space are implicit when you talk about velocity and acceleration.  The second post by aburr clearly flagged the lack of an explicit definition for x before anyone had shown any work.  

There are two obvious choices:
You can choose x as time because you need to differentiate with respect to time to find acceleration.  The calculus required is fairly straight forward.  The physical interpretation of the solution is:

  t=  0.00 s     v= +13.42 m/s     a= -0.60 m/s² 
  t= 11.25 s     v=   0.00 m/s     a= -INF  m/s²
  Total distance traveled is 100.6 meters.  

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Or you can choose x as position because that is its usual meaning in physics problems.  You need slightly trickier calculus, energy methods, or numerical techniques.  The physical interpretation in this case is:

    
  x=  0.00 s     v= +13.42 m/s     a= -8.00 m/s² 
  x= 11.25 s     v=   0.00 m/s     a= -8.00 m/s²
  Total time elapsed is 1.68 seconds.

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I am sure phoffic's edit to the original post was a good faith effort to clarify an ambiguous question from a first day user, as was my digging stuff up on the web.

I think it might be appropriate to modify the edit on the question:
    Identifying the time/postiton ambiguity for x
    Indicating that we will look at both interpretations
    Noting that x = position is probably the intended one.

I recommend the following four-way split:
     crystal for the correct solution to the x=time interpretation
     phoffic for the classic calculus solution
     aburr  for the energy method contribution
     d-glitch for the numerical solution

>> crystal
    I do appreciate you contributions to this question.  I have found other websites that take
    your view of the problem and duplicate your solution.

>> BigRat
    I appreciate your on-point criticism here as well, but I don't want you to get any points.

>> phoffic
     If this question is deleted, I will miss it, but that's okay too.  I am saving a copy for my
     personal archive though.

Thank you, d-glitch - but I did not say x is time, just that it could be since not more was given -- x could also be distance at a point in time in which case x would have a particular value.  BigRat brought up sin, which gives meaning to 180 since a circle is 360° and this puts a whole new twist on things -- maybe the object is a particle. His posts were insightful and on-target as to what was going on in the discussion.  This question was muddled by those who don't understand differentiating and while it may have been interesting to them, some of the original comments before editing were insulting and hurtful to those actually trying to solve differentiating the equation as written. Apologies are in order.

why don't you just put the question back as it was and keep your guesswork below? Speed is a scalar quantity with no direction. Velocity is a vector quantity.  I only drew an analogy to speed since others obviously did not know the difference between velocity and acceleration.

Assuming that x meant position, when I first got an answer of A = -8 m/s^2, I thought I had made a mistake. Then d-glitch came in with a numerical solution #a41807270 , which I recommend be the accepted post.

After d-glitch got the same answer that I had, I posted my 4-step analytic solution #a41807393 (assist).

Please identify other posts that should be assists (or accepted if you do not agree with my proposal).

> "I thought I had made a mistake"
and you did -- an exact value cannot be determined because it depends on X, which is not defined. The spreadsheet is not correct either.  Its whole basis is off since there is no time component in the equation.  The answer is an equation, not a spreadsheet of guessed-at values.

>"Please identify other posts"
asked and answered

I think aburr's insight in #a41807810 was very helpful:

I am not sure which of crystal's posts should be accepted. We can discuss that after you make your suggestions.
Crystal did say that her post, #a41805500 is the right answer. Could one of you confirm. I was a little confused by the wording.

>> why don't you just put the question back as it was and keep your guesswork below?

I just saw your comment. I thought that d-glitch's suggestion allowed for the two prevalent views in this thread. I could remove all the <<EDIT>> comments if you and d-glitch prefer that. Then we could even allow for x to be "the amount of resistance experienced during travel or anything else".

What X is does not matter to solve the equation.  Invalid assumptions are being made and then attributed to others to gain acceptance.  

Best would be to put the question back the way it originally was and keep your theories out of there -- put them in one of your posts, where they belong, not in the original question.

I'll do whatever all of you want. :)

For the purposes of discussing Maths questions without filling up technical threads, I've opened a Group "Maths Forum" where one can post such things.

phoffric, your posts keep changing ... just now seeing this, "I am not sure which ... should be accepted"

The answer is best explained in #41811100
___
thanks, BigRat

Crystal:

The answer is best explained in #41811100

you have to be kidding!

I very much appreciate your Microsoft MVP -  I myself am a VSOP (Very Superior Old Pale) - but 41811100 is not an answer.

Thanks. I guess I am confused. That post (ID: 41811100) shows the original equation, explains that acceleration is the differential of velocity, and shows how to arrive at the answer,  -4 / (45-4x)^1/2

the answer has to be in terms of x because a was not defined.

if you think another one is better, that is fine with me.

I'm afraid that the answer is not -4/sqrt(45-4x) but -8. Try the question I've just posted.

-4 / sqrt(45-4x) is correct

> Try the question I've just posted.
can you give me a link? I don't see it.  sorry I haven't been on EE long enough to figure out how to do much on the site

https://www.experts-exchange.com/questions/28972216/Identify-Crisis.html

I assumed what you said was right -- but I simplified everything and divided by 2 so what I wrote is correct

the link you gave me is something different.  Just hopping on today for light messages but actually working on a project I can't be distracted (too much) from. I am going to be somewhat tied up the next couple days but will look at your link when I have some time ~ thanks

>> That post (ID: 41811100) shows the original equation, explains that acceleration is the differential of velocity

When I responded to your original statement like this, I wasn't trying to hurt you; rather I was trying to help you state things in a little more precise manner. I was trying to get you to clarify what the differential was with respect to <what??>.

I wrote earlier that acceleration is defined to be dV/dt (where t is time).

It does matter in the OP whether we think x is position vs time. If you believe that x is meant to be time, then dV/dx is the acceleration. But, we cannot just say that acceleration is the differential of velocity. We can say that acceleration is the derivative of velocity with respect to time.

-------------

>> I did not say x is time, just that it could be since not more was given -- x could also be distance at a point in time in which case x would have a particular value.
If we accept that x represents a distance (with respect to some origin, or in other words, a position along the x-axis), then I don't see how we would have the OP equation where x is a point at a point in time and therefore, x(t) is just a particular value. With this interpretation, then the OP equation has nothing but constants in it which implies that V is a constant. And, as you know, if that is the interpretation, then the acceleration = dV/dt = 0.

If x represents position, then, if not by science/math, then by democracy, we believe that acceleration = -8 m/s^2.

http://www.wolframalpha.com/input/?i=differentiate++(180-16x)%5E1%2F2

The answer is not necessarily just a number. Since x was not known (or a range defined), it cannot be substituted in to arrive at a specific number(/s).  Therefore, the answer must be in terms of x.

>dV/d? -- see what Wolfram calls it

> because a was not defined.
typo -- --> because x was not defined.

>> >dV/d? -- see what Wolfram calls it
I looked and saw that Wolfram took its best guess and interpreted your remark as d/dx.
Sometimes, when I ask Wolfram for answers, they do not always interpret what I want, so I have to rephrase my query until they understand me. Here they understood me in a different derivative:
http://www.wolframalpha.com/input/?i=d%2Fdt+(180-16x(t)+)%5E1%2F2

>> -8 is only correct when x = -11.1875
I interpret this to mean that x is position (m?) and that you are saying that d-glitch's spreadsheet showing -8 m/s^2 for different values of x is incorrect. (Also, that my 4 step analytical solution is also incorrect.) If this is what you are saying, how about telling us, including Aburr and BigRat, where we went wrong, as the four of us seem to agree that if x is position = x(t), then a(t) = -8 m/s^2.

I am pretty sure that I posted this link before showing x(t) solution, and then when you take 2nd derivative, you get -8 m/s^2.
http://www.wolframalpha.com/input/?i=x'(t)+%3D+(180-16x(t)+)%5E1%2F2

got the sign wrong since it is being multiplied by -4 not 4 and that is why I removed my previous calculation.  The differential equation  MUST be written in terms of x because x was not defined.

-8 is only correct when x = 11.1875
_________________________________________ proof:
since dv/dx = -4 / (45-4x)^1/2

to get -8:
 -8 = -4/ (1/2)  so sqrt(45-4x) must be 1/2

so:
1/2 = sqrt(45-4x)

square both sides
1/4 = 45-4x

multiply by -1
-1/4 = 4x - 45

add 45 to both sides:
45-1/4= 4x
44.75 = 4x

divide both sides by 4
x = 11.1875

plugging back in:

dv/dx = -4 / (45-4x)^1/2 where x = 11.1875
=  -4 / sqrt(45-4* (11.1875))
= -4/ / sqrt(45-44.75)
= -4/ sqrt( 1/4)
= -4 / (1/2)
= -8

> "when you take 2nd derivative, you get -8 m/s^2"
 the 2nd derivative is not correct -- acceleration is the first derivative of velocity

Is x position (m) or time(t) in your proof?
What do your calculations show for acceleration (m/s^2) when x = 10.00 m ?
If your acceleration at x = 10 m differs from d-glitch's spreadsheet, then why is the spreadsheet wrong?

x doesn't matter! it is not necessarily m since we do not know the units of x ... but to humor you ...  when x = 10:
acceleration = -4 / (45-4(10))^1/2
= -4/(45-40)^1/2
= -4/sqrt(5) ~= -8.94

sorry but I can't spend any more time on this right now -- shouldn't have taken the time I just did

Well, how bizarre a question this is! The OP indicates that V is m/s, but we have to be open to the possibility that x, as a position, could be, say, in light years.

>> -8.94 for x = 10 something.
Ok, if you are right, would you be so kind to point out why the spreadsheet is wrong. This question isn't going to close today, so if this is instructive, then we should proceed to try to help each other.

Did you take a look at these two links?
http://www.wolframalpha.com/input/?i=d%2Fdt+(180-16x(t)+)%5E1%2F2
http://www.wolframalpha.com/input/?i=x'(t)+%3D+(180-16x(t)+)%5E1%2F2 

In this 2nd link, there is a solution to x(t), and when you take 2nd derivative, you get -8 m/s^2.

time is too short right now to look at the spreadsheet and point out the problems but the whole premise is wrong

on the links,  you substituted x for x(t), which is not correct since t was not in the equation and if it was, it wouldn't be x(t) it would be something like some-other-letter(x,t) like z(x,t) or, if you want to get rid of x: y(t) but the equation for y would have to be defined, which it wasn't.

repeating:
> "when you take 2nd derivative, you get -8 m/s^2"
 the 2nd derivative is not correct -- acceleration is the first derivative of velocity

>> acceleration is the first derivative of velocity
acceleration is the first derivative with respect to time of velocity.
acceleration is the 2nd derivative with respect to time of position, x(t).

the equation was for velocity, not position

Hey d-glitch,

I looked at your spreadsheet and saw that you incremented X (position).

When I took a different approach, I got numerical errors. Maybe I'll post it in a blog or a question.

The idea was that x0 = 0, and then  v0 = sqrt( 180-16 * x0 )
Set dt to, say 1/2^14

Then x(i+1) = x(I) + v(I) * dt
Then v(I+1) =  sqrt( 180-16 * x(I+1)  )

A(I) = [ v(I+1) - v(I) ] / dt

As velocity got very close to 0, then A started dropping to produce what looks like a rotated L-shape. Do you happen to know how to better compute the A array to reduce the numerical errors as V --> 0 ?

I stand by my earlier assessment:

I recommend the following four-way split:
     crystal for the correct solution to the x=time interpretation
     phoffic for the classic calculus solution
     aburr  for the energy method contribution
     d-glitch for the numerical solution