What is wanted here?

[allegansveritatem]

Here is the problem:


I am stuck on part a. Here is all I could come up with in the way of a formula for the stopping distance in terms of Vo and a:


Something tells me I am falling short here but somehow I can't come up with anything better. Is this what is being asked for in part a?

 

[Dr.Peterson]

I don't think what you've written makes any sense. How can you replace t with v_0 on the LHS and v_0 with x on the right, and call this a function of v_0, with no v_0 in the expression at all? I wonder if you really know what function notation means.

What they are asking is, what is the farthest the particle can go -- the greatest possible value of s? What does the graph of s(t) look like?

 

[Steven G]

As already mentioned, you were given that s(t) = (v₀/a)(1-e^-at) so why would you change it to what you did?

UNITS matter. UNITS are important. UNITS should be used whenever possible. This way you will never say that 10 + x = 315 when 10 represents 10 minutes, x represents x minutes and 315 represents 315 gallons. minutes + minutes NEVER equal gallons.

It was given that t is in seconds so when it comes to s(t) you can only replace t with a number of seconds. You replaced t with v₀ which is in meters/seconds NOT seconds. You need to catch these mistakes when you make them. Just check to see if the units match what they should be.

Since I was so hard on you above I will show you how to solve this problem.

1st of all, s(some number of seconds) tell you how many meters the particle traveled in that length of time. v(some number of seconds) tells you the velocity at that exact time.

We know that the particle has stopped moving when v(t) = 0.

So we need to solve for v(t)=0. Ie solve v₀e^-at= 0. Or v₀/e^at= 0
The problem is that this never equals 0, but it does approach 0 as t gets larger and larger and larger.
So the total distance the particle traveled will be s( a very very very large number)= (v₀/a)(1-e^{-a(very very large number)}).
Now e^{-a(very very large number)})is close to 0. So s= (v₀/a)(1-0) = v₀/a

To other helpers: Understand my language above is due to the fact that this problem is posted in the algebra section

 

[allegansveritatem]

I don't think what you've written makes any sense. How can you replace t with v_0 on the LHS and v_0 with x on the right, and call this a function of v_0, with no v_0 in the expression at all? I wonder if you really know what function notation means.

What they are asking is, what is the farthest the particle can go -- the greatest possible value of s? What does the graph of s(t) look like?

well, I realized later that I had not written exactly what I meant. What I should have put on the right of the equals sign is f(x) with the understanding that the x stood for Vo. I know they wanted a formula but what is a function if not a formula? Anyway, I have not yet studied your post thoroughly so I will come back to it later tonight. But I wanted to explain what I meant with my sloppy solution.

 

[allegansveritatem]

As already mentioned, you were given that s(t) = (v₀/a)(1-e^-at) so why would you change it to what you did?

UNITS matter. UNITS are important. UNITS should be used whenever possible. This way you will never say that 10 + x = 315 when 10 represents 10 minutes, x represents x minutes and 315 represents 315 gallons. minutes + minutes NEVER equal gallons.

It was given that t is in seconds so when it comes to s(t) you can only replace t with a number of seconds. You replaced t with v₀ which is in meters/seconds NOT seconds. You need to catch these mistakes when you make them. Just check to see if the units match what they should be.

Since I was so hard on you above I will show you how to solve this problem.

1st of all, s(some number of seconds) tell you how many meters the particle traveled in that length of time. v(some number of seconds) tells you the velocity at that exact time.

We know that the particle has stopped moving when v(t) = 0.

So we need to solve for v(t)=0. Ie solve v₀e^-at= 0. Or v₀/e^at= 0
The problem is that this never equals 0, but it does approach 0 as t gets larger and larger and larger.
So the total distance the particle traveled will be s( a very very very large number)= (v₀/a)(1-e^{-a(very very large number)}).
Now e^{-a(very very large number)})is close to 0. So s= (v₀/a)(1-0) = v₀/a

To other helpers: Understand my language above is due to the fact that this problem is posted in the algebra section

Thanks for your reply. I have not yet read it for lack of time right now but later this evening I will come back to it and go over it with attention. I will reply.

 

[allegansveritatem]

As already mentioned, you were given that s(t) = (v₀/a)(1-e^-at) so why would you change it to what you did?

UNITS matter. UNITS are important. UNITS should be used whenever possible. This way you will never say that 10 + x = 315 when 10 represents 10 minutes, x represents x minutes and 315 represents 315 gallons. minutes + minutes NEVER equal gallons.

It was given that t is in seconds so when it comes to s(t) you can only replace t with a number of seconds. You replaced t with v₀ which is in meters/seconds NOT seconds. You need to catch these mistakes when you make them. Just check to see if the units match what they should be.

Since I was so hard on you above I will show you how to solve this problem.

1st of all, s(some number of seconds) tell you how many meters the particle traveled in that length of time. v(some number of seconds) tells you the velocity at that exact time.

We know that the particle has stopped moving when v(t) = 0.

So we need to solve for v(t)=0. Ie solve v₀e^-at= 0. Or v₀/e^at= 0
The problem is that this never equals 0, but it does approach 0 as t gets larger and larger and larger.
So the total distance the particle traveled will be s( a very very very large number)= (v₀/a)(1-e^{-a(very very large number)}).
Now e^{-a(very very large number)})is close to 0. So s= (v₀/a)(1-0) = v₀/a

To other helpers: Understand my language above is due to the fact that this problem is posted in the algebra section

You say that t is a very large number, but so is a--8x10^5 to be exact....but I see what you are saying, and I see the way to go now but I still need to explain it to myself in action.I will work on this again tomorrow and post my results. I will try to keep the units in the picture. I guess part of my problem was not understanding the injunction to "find a formula in terms of v and a" I didn't get that this meant ONLY v and a should figure on the business side of the equal sign.

 

[allegansveritatem]

As already mentioned, you were given that s(t) = (v₀/a)(1-e^-at) so why would you change it to what you did?

UNITS matter. UNITS are important. UNITS should be used whenever possible. This way you will never say that 10 + x = 315 when 10 represents 10 minutes, x represents x minutes and 315 represents 315 gallons. minutes + minutes NEVER equal gallons.

It was given that t is in seconds so when it comes to s(t) you can only replace t with a number of seconds. You replaced t with v₀ which is in meters/seconds NOT seconds. You need to catch these mistakes when you make them. Just check to see if the units match what they should be.

Since I was so hard on you above I will show you how to solve this problem.

1st of all, s(some number of seconds) tell you how many meters the particle traveled in that length of time. v(some number of seconds) tells you the velocity at that exact time.

We know that the particle has stopped moving when v(t) = 0.

So we need to solve for v(t)=0. Ie solve v₀e^-at= 0. Or v₀/e^at= 0
The problem is that this never equals 0, but it does approach 0 as t gets larger and larger and larger.
So the total distance the particle traveled will be s( a very very very large number)= (v₀/a)(1-e^{-a(very very large number)}).
Now e^{-a(very very large number)})is close to 0. So s= (v₀/a)(1-0) = v₀/a

To other helpers: Understand my language above is due to the fact that this problem is posted in the algebra section

In my first reply I said, to the effect, that a function was a kind of formula. I thought about this later and realized that is not so if for no other reason than: a formula can be full of variable but a function--at least the ones I know about--only has one variable. No?

 

[Steven G]

You say that t is a very large number, but so is a--8x10^5 to be exact....but I see what you are saying, and I see the way to go now but I still need to explain it to myself in action.I will work on this again tomorrow and post my results. I will try to keep the units in the picture. I guess part of my problem was not understanding the injunction to "find a formula in terms of v and a" I didn't get that this meant ONLY v and a should figure on the business side of the equal sign.

I won't argue whether or not a=8x10^5 is a very large number. But I will say that t is many many many times 8x10^5. In calculus and above we say that t approaches infinity instead of extremely large.
Just to be clear, which course are you in and did you learn about limits?

 

[Steven G]

In my first reply I said, to the effect, that a function was a kind of formula. I thought about this later and realized that is not so if for no other reason than: a formula can be full of variable but a function--at least the ones I know about--only has one variable. No?

You can have a function of 2 (or more) variables. For example f( x, y) = 2x + y^2. Then f(3, 5) = 2(3) +(5)^2 = 6+25 =31
Some functions are special formulas. Like A(b,h) =b*h (the area of a rectangle is base*height).
How about the formula for the area or a circle which is pi*r^2. The function is A(r) = pi*r^2.
Consider the quadratic formula x= ....... Now x is in terms of a,b and c. So you can write x(a,b,c) =....
However in the end I would not tend to think of a formula as a function especially since it seems to be confusing you. I even suspect that there are formulas that are not functions anyways but at 1:30am I can't think of any.

 

[Dr.Peterson]

Have you tried sketching a graph of s(t) yet? I think seeing what kind of function it is will go a long way toward helping you understand the problem. (If necessary, pick specific values of v₀ and a, and use a graphing program.)

The answer will be an expression containing only v₀ and a, so it can be called a function of v₀ and a.

 

[allegansveritatem]

In my first reply I said, to the effect, that a function was a kind of formula. I thought about this later and realized that is not so if for no other reason because a formula can be full of variable but a function--at least the ones I know about--only has one variable. No?

You can have a function of 2 (or more) variables. For example f( x, y) = 2x + y^2. Then f(3, 5) = 2(3) +(5)^2 = 6+25 =31
Some functions are special formulas. Like A(b,h) =b*h (the area of a rectangle is base*height).
How about the formula for the area or a circle which is pi*r^2. The function is A(r) = pi*r^2.
Consider the quadratic formula x= ....... Now x is in terms of a,b and c. So you can write x(a,b,c) =....
However in the end I would not tend to think of a formula as a function especially since it seems to be confusing you. I even suspect that there are formulas that are not functions anyways but at 1:30am I can't think of any.

At !:30am I can't think of anything.

 

[allegansveritatem]

I won't argue whether or not a=8x10^5 is a very large number. But I will say that t is many many many times 8x10^5. In calculus and above we say that t approaches infinity instead of extremely large.
Just to be clear, which course are you in and did you learn about limits?

I worked on this again and lthis is what I got for part a, (with a little explanation):


Now, for part b I worked it out but I am having trouble believing the answer....it doesn't seem to make sense. But I have worked it out more than three times and keep getting the same results. Here:


Huh? How can this be right? What is wrong here?

 

[allegansveritatem]

Have you tried sketching a graph of s(t) yet? I think seeing what kind of function it is will go a long way toward helping you understand the problem. (If necessary, pick specific values of v₀ and a, and use a graphing program.)

The answer will be an expression containing only v₀ and a, so it can be called a function of v₀ and a.

I think I have done this already because I have turned this problem every way but loose yesterday and today too. I know it is a simple thing but....

 

[Dr.Peterson]

Okay, you haven't shown a graph, but you have shown that for large [MATH]t[/MATH], [MATH]s(t)[/MATH] approaches [MATH]\frac{v_0}{a}[/MATH]; that is, the graph has a horizontal asymptote there. That is the answer to the first part.

For the second part, you just plug in the numbers, as you did. But what are the appropriate units for [MATH]s[/MATH]? What are the units of [MATH]a[/MATH]?

 

[allegansveritatem]

Okay, you haven't shown a graph, but you have shown that for large [MATH]t[/MATH], [MATH]s(t)[/MATH] approaches [MATH]\frac{v_0}{a}[/MATH]; that is, the graph has a horizontal asymptote there. That is the answer to the first part.

For the second part, you just plug in the numbers, as you did. But what are the appropriate units for [MATH]s[/MATH]? What are the units of [MATH]a[/MATH]?

well, that is part of the problem with this problem, namely, we are not told what (a )stands for nor are we given any units for it. Vo is meters/sec. As for the asymptote...I knew there was somekind of asymptotic action figuring into this because of the presence of that (1-e^-at). I did not make a graph yet nor did I input a function into my calculator to generate one, mainly because that is how the problem is presented,ie. there is no calculator icon attached. I will sketch a graph tomorrow and see what I can see.

 

[allegansveritatem]

I won't argue whether or not a=8x10^5 is a very large number. But I will say that t is many many many times 8x10^5. In calculus and above we say that t approaches infinity instead of extremely large.
Just to be clear, which course are you in and did you learn about limits?

I have a vague idea what limits are but as far as I can recall have never been formally introduced to them. I am not in any course but I am learning this subject from a text--this is a precalculus text. I preceded this text with a college algebra text. Many decades ago I studied--if you want to call what I did study--algebra both in high school and a bit in college. I have time now and find myself interested in going as far in math as the integrity, such as it is, of my faculties allows me to. Old age, as my granny used to say, is catching up with me but I still have a little lead on it.

 

[Dr.Peterson]

well, that is part of the problem with this problem, namely, we are not told what (a )stands for nor are we given any units for it. Vo is meters/sec. As for the asymptote...I knew there was somekind of asymptotic action figuring into this because of the presence of that (1-e^-at). I did not make a graph yet nor did I input a function into my calculator to generate one, mainly because that is how the problem is presented,ie. there is no calculator icon attached. I will sketch a graph tomorrow and see what I can see.

Yes, the problem could be stated more clearly. But you can determine the units for a, by observing that an exponent must always be dimensionless, so a must be in units of 1/seconds. Or, you can just know that s is said to be a distance, so it is presumably in meters.

Are you saying you never sketch graphs yourself, without a calculator? Such a sketch can be an important part of understanding a problem. And since you have been taught about asymptotes but not limits (which are to some extent the same thing), the former is the way you have to think here.

 

[allegansveritatem]

Yes, the problem could be stated more clearly. But you can determine the units for a, by observing that an exponent must always be dimensionless, so a must be in units of 1/seconds. Or, you can just know that s is said to be a distance, so it is presumably in meters.

Are you saying you never sketch graphs yourself, without a calculator? Such a sketch can be an important part of understanding a problem. And since you have been taught about asymptotes but not limits (which are to some extent the same thing), the former is the way you have to think here.

yes, that makes sense that the asymptote could be looked on as a limit. I often sketch graphs but this one looked a little too mean for simple sketching so I used the calculator and indeed, there; is a limit (asymptote) at .00013.


I have seen before that sketching a graph with pen and paper has something over using a calculator--you get a sense of the nuts and bolts of the situation, so to speak. Just looking at the equations tells me that I am going to have a really small number coming forth and it also tells me that after a very very short time it doesn't matter to the result how long the particle travels.

 

[Dr.Peterson]

I don't know how much you have learned about curve sketching with regard to exponential functions, but I would just recognize that [MATH]e^{-at}[/MATH] approaches zero from above (has a horizontal asymptote at y=0) to the right, so [MATH]1-e^{-at}[/MATH] will be approaching 1 from below. You could also specifically graph it using transformations (reflection and translation) from the basic exponential.

But you see my point -- the graph gives you an overview of what this function is all about.

I will add that I figured from the start that a very small particle might not get very far at high velocity through the air; the calculation confirms that, to the extreme. It didn't get very far at all!