Maximizing Average Velocity

1. Sep 22, 2015

Cosmophile

1. The problem statement, all variables and given/known data
This is not a homework or coursework question for myself. A friend sent it to me and I'm simply looking for assistance in mathematically proving my answer to myself. Here is the question:

"A cart rolls across a table two meters in length. Half of the length of the table is covered with felt, which slows the cart with a constant acceleration. Where should the felt be placed so that the cart crosses the table in the least amount of time?"

2. Relevant equations

Frankly, I'm having a difficult time finding the a footing here. I know that the frictional force $\vec {F_f} = \mu \vec{F_n}$. It's obvious that we are wanting to maximize the average velocity (thus reducing the total time for the trip), but I'm not sure how to express this mathematically, or even if that is the best approach. I have also considered attempting to minimize the work done by friction.

3. The attempt at a solution
I know that $W = \Delta K_E = \frac {1}{2} m{v_2}^2 - \frac {1}{2}m{v_1}^2 = \frac {1}{2}m(\Delta v)$
Because the cart eventually comes to a rest, the above equation can be simplifed to say

$$W = -\frac {1}{2}m{v_1}^2$$

I also know that, for an inclined plane, $F_a = mg\sin\theta$, $F_N = mg\cos\theta$, $F_f = \mu F_n = \mu mg\cos\theta$

I also know that the work done by friction $W_f = F_f \bullet d$, where $d$ is the displacement of the block. This gives $$W_f = \mu mg\cos\theta \bullet d = |\mu mg\cos\theta||d|\cos\phi$$, where $\phi$ is the angle between $F_f$ and $d$. Because the two are in the same direction, $\cos \phi = 1$, thus

$$W_f = \mu mgd\cos\theta =-\frac {1}{2}m{v_1}^2$$
$$W_f = \mu gd\cos\theta = -\frac{1}{2}{v_1}^2$$

And here is where I hit my wall.

Another friend has told me that approaching this via the Work-Energy principle is a dead-end and that I should instead approach the problem as a pure kinematics problem as opposed to dynamics. That being said, I'm exhausted and need to rest. Any help is greatly appreciated, as always.

2. Sep 22, 2015

Cosmophile

Just realized I was acting as if the cart's final velocity were 0 m/s, which isn't true. Sigh.

3. Sep 22, 2015

BvU

Hello worldfriend,

First of all you want to complete the problem statement: is the table horizontal ? No friction on the table itself ?
Any other initial or boundary conditions (like does the cart come to a stop?) ? Can we assume an initial speed v0 ?

Then you want to make problem and remainder consistent, so there is no confusion about cart or block. Or confusion about whether the cart (block ?) comes to rest on the table, or is allowed to fall off.

Perhaps provide a few more variable names: table from 0 to 2 m, so felt from x to x+1 with x ≤ 1 (otherwise it would be a bit corny).

The simplest approach I can think of is to look at the speed as a function of position. Assuming no friction on the table itself and a horizontal table, you have v0 from 0 to x, then something involving a square root from x to x + 1 and finally some v1 from x+1 to 2. In the section from x to x+1 the speed decreases monotonically, so the average is maximum if x is maximum. Position the felt as far away from the start as possible.

4. Sep 22, 2015

Cosmophile

So sorry! I didn't realize that I hadn't copied the entire question. The original question involves an inclined plane (hence all of the trigonometric functions). Though, I have no problem with instead dealing with a horizontal table. What I've presented is all the information I myself was presented with. I would like to come to a symbolic solution so that I may better analyze it all.

Are you saying that we have three intervals we need to be looking at? I see that from $x_0$ to $x=1m$ we have a velocity $v_0$, and I see that in the interval $1m \leq x \leq 2m$, the velocity decreases at a constant rate (assuming the felt is placed on the second half of the table). Where did you get the notion that a square root comes into play?

5. Sep 22, 2015

BvU

If you slow a cart with a constant acceleration, the speed decreases linearly with time (so at a constant rate, as you say). The distance covered has something like $x(t) = x(t_0) + v(t_0) t + {1\over 2} a t^2$ with a < 0. so v as a function of position might just have a square root in it. Not really nice, but you don't have to calculate it: it's sufficient to know that the speed decreases monotonically with distance. Obviously the average speed is maximum if the slowing down starts as late (far to the right) as possible.

Example with felt at 0.6 m , v0 = 2 m/s and devceleration on felt -1.8 m/s2 :

6. Sep 22, 2015

Cosmophile

I'm sure that this is something I should know already, but how exactly would I write velocity as a function of position? Thanks again!

7. Sep 22, 2015

BvU

You have $$v(t) = v(t_0) + at \quad {\rm and} \\ x(t) = x(t_0) + v(t_0) t + {1\over 2} a t^2$$so all you have to do is write t as a function of t using the second equation, and then insert it in the first one.
I'm basically very lazy, so I conveniently pick $x(t_0) = v(t_0) = 0 \quad$ to get $v = - \sqrt {x/a}$
(projectile trajectories are parabolas upside down -- the acceleration is in the negative y direction -- this is a parabola on its side, so to speak).

8. Sep 24, 2015

Cosmophile

What do you mean "write t as a function of t?" I'm assuming that was a typo? In my physics text (Kleppner/Kolenkow, 1st Ed), they eliminate time from an equation during an example involving motion in a uniform gravitational field:

$$x= v_0t$$
$$z=v_{0z}t - \frac {1}{2}gt^2$$
$$z = \frac {v_{0z}}{v_{0x}}x-\frac{g}{2v_{0x}^2}x^2$$

I'm not sure how you did it in your example, though.

9. Sep 24, 2015

haruspex

I believe BvU meant t as a function of x. But I would do it the other way around, writing t as a function of v and substituting for t in the 2nd equation. It keeps the square root out of the picture for a bit longer.

10. Sep 24, 2015

Cosmophile

Ah, I see. So, I've fiddled with what you suggested and encounter a problem. First, here's what I've come up with:

$$Eq. 1: v(t) = v(t_0)+at$$
$$Eq. 2: x(t) = x(t_0) + v(t_0)t + \frac {1}{2}at^2$$

After solving $Eq. 1$ for $t$, I get

$$t = \frac {v(t)-v_0}{a}$$

Before I plug this into $Eq. 2$, I already have an issue with what I've come up with, and that is that by solving $Eq. 1$ for $t$, I still have the $v(t)$ term, which is itself a function of $t$ (isn't it?). Am I wrong to toss up a red flag here? Thanks for the help thus far!

11. Sep 24, 2015

haruspex

Since there is a one-to-one relationship between position and time, you can equally consider v to be a function of x.

However, my responses so far have been purely in relation to the last few posts. Going back to the original problem, it seems wrong to me to be trying to eliminate time when that is precisely the variable you are trying to minimise.
I gather that speed is constant until the mat is reached. What does that tell you about the time taken to cross the mat, in regard to how it depends on where the mat is placed? Next, what can you say about the change in speed from crossing the mat? How does that depend on where the mat is placed?

12. Sep 24, 2015

Cosmophile

Would you mind elaborating on the first part of your response? Regarding it being equally possible to consider v to be a function of x due to there being a 1:1 ratio between x and t. I apologize if this is trivial.

The time it takes to cross the covered half is longer than the time to cross the non-covered section because the velocity for the latter is constant, while the velocity during the former is constantly decelerating. If we place the mat on the first half, then the velocity of the cart will constantly be what it was as the cart came off of the mat (because there is no other source of acceleration). So, placing the mat on the second half results in a faster overall trip, because the cart at first has a velocity which has a higher magnitude than the cart will have at the end of the mat. I'm not sure if I'm wording this adequately...

The average velocity will always be higher when the mat is placed on the last half than when it is placed on the first half because placing it on the last half allows for the cart to move for some time at a constant velocity $v_c$, which is higher than the velocity at any point after touching the mat, due to the constant acceleration.

Sorry for the rambling and redundancy; I hope I'm at least making some sense.

13. Sep 24, 2015

BvU

You certainly are making sense ! If we look back to the picture in post #5: there is a section with $v_0$, then a section where $v$ drops, and finally a section where $v$ is constant again, but less than $v_0$. It is evident that the average speed is maximum if the section where the speed drops off is shifted tothe right as much as possible.
All my bla was just to point out that on the mat the speed drops off linearly with time , but not linearly when considered as a function of x on the v(x) plot.

And I considered the exercise as a maximization problem for v(x) in the range 0 < x < 2 m. However, the exact shape of the v(x) on the mat is not relevant for the outcome; but the position on the x-axis is the deciding quantity for achieving the maximum $$\ \ v_{\rm average} \equiv { \int_0^2 v(x) dx \over \int_0^2 dx} \\ \mathstrut$$

14. Sep 24, 2015

Cosmophile

Glad to know my rambling was sufficiently coherent. Now, as I said in my OP, I'm really looking for a way to mathematically validate what it is I am saying in my last post.

15. Sep 24, 2015

BvU

Well, if I work out the vaverage expression (let the mat start at $x_1$ and the speed at the end of the mat be $v_1$) :$$\ \ v_{\rm average} \equiv { \int_0^2 v(x) dx \over \int_0^2 dx} \\ \mathstrut \quad = {1\over 2} \left ( \int_0^{x_1} v_0 dx + \int_{x_1}^{x_1 + 1 } v(x) dx + \int_{x_1+1}^{2} v_1 dx \ \right ) \\ \quad = {1\over 2} \left (v_0 \int_0^{x_1} dx + {\rm some \ constant} + v_1 \int_{x_1+1}^{2} dx \ \right )\\ \quad = {1\over 2} \left (v_0 x_1 + {\rm some \ constant} + v_1 (2-x_1) \ \right )\\ \quad = {1\over 2} \left ( (v_0 - v_1) x_1 + {\rm some \ constant} + 2 v_1 \ \right ) \\ \quad = a x_1 + b$$ with a and b constants. No need to differentiate wrt $x_1$ because this is increasing linearly with x, so the biggest x possible gives the maximum.

Last edited: Sep 24, 2015
16. Sep 24, 2015

haruspex

Not a 1:1 ratio, a 1:1 relationship. That is, there is only one time at which it is at any given position, and only one position at any given time (obviously). That means it is possible to express the velocity as a function of position, which is exactly what you did.
Maybe your confusion comes from the y=y(x) notation. It is rather misleading and can lead to confusion. If instead you were to write v=f(t) and v=g(x) perhaps you would understand the shift better. For more on the subject, see https://www.physicsforums.com/insights/conceptual-difficulties-roles-variables/.

Your argument works, as BvU says, but I would add the observation that the time taken to traverse the mat does not depend on where it is placed. Neither does the change in speed as the mat is crossed. Thus, you can replace the mat by an instantaneous drop in speed of a given amount, and simply ask where that should be placed.

17. Sep 24, 2015

Cosmophile

Yes, my confusion certainly involved the notation $v(t)$. I had thought to say $v = v(t) = v_0 + at$, and simply said $t = \frac {v-v_0}{a}$, but I'm not sure that really helps my case at all, haha. Truthfully, I'm not sure why this is giving me such a hard time.