# Brain teaser involving very basic concepts?

1. Feb 25, 2009

### csprof2000

You need to deliver a package of mass M from your current location (x = 0) to a room at the end of the hall (x = D). You have an amount of time T in which to do it.

Neglecting resistive forces, what is the minimum amount of energy required to accomplish this task? NOTE: You expend energy to speed up, but you don't get it back when you slow down.

This may be very simple, but I want to see if you guys get the same thing as me.

Let us also make the following constraints explicit:

x(0) = 0, x(T) = X, v(0) = 0, v(T) = 0.

Account only for the energy used in moving the package, not yourself.

Last edited: Feb 25, 2009
2. Feb 25, 2009

### DaleSwanson

I could be wrong but wouldn't it just be the smallest amount of energy you can produce? Since you said we are ignoring resistive forces once you give the package the slightest push it will start moving, and eventually reach the destination. I suppose if you want it stopped when it reaches the destination you will need to give it an equal push in the opposite direction.

3. Feb 25, 2009

### snoopies622

But that wouldn't get it there in time T. My guess would be to use uniform acceleration all the way, but that's just my intuition; I haven't done the math. If I'm right, the total energy invested would be the final kinetic energy of the object: $$\frac {2MD^2}{T^2}$$

Edit: oh wait, v(t)=0? Then $$\frac {4MD^2}{T^2}$$, half the energy is expended in the acceleration, and half in the deceleration.

Last edited: Feb 25, 2009
4. Feb 25, 2009

### DaleSwanson

Woops, I don't know how I missed that.

5. Feb 25, 2009

### snoopies622

Wait, does this mean that you expend energy to slow it down, or do you simply not get any back?

6. Feb 25, 2009

### csprof2000

Well, I was thinking of it as not needing to expend energy to slow down (perhaps you just "coast" to a stop?) but if you'd rather include that in the solution and add the speeding up energy and the slowing down energy, that's fine too.

It seems like it could go one of several ways... maybe the best idea is to get up to the average speed of X/T as quickly as possible (energy expenditure of 1/2 M(X/T)^2) and then coast and then stop right at the end (possibly expending an additional 1/2 M(X/T)^2...).

Maybe it's constant acceleration / constant deceleration... which would be like a unit-step acceleration. Perhaps a cubic or quintic? Perhaps a sinusoidal?

Hmmm...

7. Feb 25, 2009

### confinement

According to the bang-bang optimal control theorem, the solution will correspond to immediatly at t = 0 accelerating to a constant velocity v = D/T and decelerating instantaneously to v = 0 at x = D. Mathematically we can express the accelration as a delta function, and the velocity will be a step function.

http://en.wikipedia.org/wiki/Pontryagin's_minimum_principle

The point of the theorem is that optimal solutions correspond to minimizing or maximizing the control parameters, which in this case can be acceleration.

8. Feb 25, 2009

### csprof2000

Very interesting...

So the answer would be 1/2 M (X/T)^2 (or possibly M (X/T)^2 depending on whether energy is needed to stop the system at time t = T).

That's what I thought, and here's how I arrived at the answer:

If the package goes from 0 to X in time T, then its average velocity was (X/T). Then, by the mean value theorem, the instantaneous velocity must reach (X/T) at some 0 < t < T. But this already requires (1/2)(M)(X/T)^2 work, so this is the smallest possible energy.

Also, jumping at the start and stop points (via the delta function) does not require going above (X/T), so I figured this would be optimal. Cool.

9. Feb 25, 2009

### snoopies622

I agree.

10. Mar 2, 2009

### Bob S

You imply that the only energy used is equal to the kinetic energy and no energy is recovered, so the simple solution would be to accelerate instantly to the final velocity which would be D/T. Then the total energy expended is (1/2) M (D/T)^2