# How does position uncertainty change in time?

I have an online homework for my Modern Physics class, that requires me to find the uncertainty in velocity and position of a duck. The question is as below:

Suppose a duck lives in a universe in which h = 2π J · s. The duck has a mass of 2.55 kg and is initially known to be within a pond 1.70 m wide. (a) What is the minimum uncertainty in the component of the duck's velocity parallel to the pond's width? (b) Assuming this uncertainty in speed prevails for 4.10 s, determine the uncertainty in the duck's position after this time interval.

Now, I got the solution to the first part using the equation

$$\Delta(x).\Delta(p) = h / 4\pi$$

This gives me $$\Delta(v) > 0.1153$$

I assume, the second part of the question would be calculated this way:

$$\Delta(x) = \Delta(v) . t$$ or, $$\Delta(x) = 0.1153 * 4.10$$

But this shows up as the wrong answer. What am I doing wrong? Wouldn't the uncertainty in the duck's position after time t equal the product of velocity uncertainty and time?

(Disclaimer: Yes, this is a homework question as I mentioned above, but I have tried to solve it and seem to be hitting a wall. I suppose I have some concept wrong. A nudge in the right direction would be appreciated.)

-
Is the duck confined to the pond, or can it get out and start walking over land? (And I suppose we should ask, does its wavefunction continue to evolve in the same way over land as it does over water? But I have a hunch that's more detail than you are meant to think about for this... one of the reasons I'm starting to think I don't like this problem.) On another note, here's a hint: if $\Delta x = \Delta v t$, what would be the uncertainty at $t=0$? Does that agree with what you know to be the uncertainty at $t=0$? –  David Z May 21 '12 at 1:55
@DavidZaslavsky: Thanks for the hint. That helped me find the solution. I was doing $$\Delta(x) = \Delta(v)t$$, but as you said, that does not agree with our initial information. I got the answer by doing $$\Delta(x) = \Delta(v)t. \Delta(x_{initial})$$ –  xbonez May 21 '12 at 5:24
OK, good. If you were able to find the solution, could you write it up and post it as an answer? –  David Z May 21 '12 at 6:01

Following David Zaslavsky hint (see comments under question), I realized what I was doing wrong.

I was solving using

$$\Delta(x) = \Delta(v).t$$

This would give me $$\Delta(x) = 0.47273$$

However, I wasn't taking into consideration the initial uncertainty in position.

Thus the correct was of doing it was,

$$\Delta(x) = \Delta(v).t + \Delta(x_{initial})$$ $$\Delta(x) = 0.47273 + 1.70 = 2.17273$$

-
Although you did the problem right, this problem is terrible is many ways. –  Ron Maimon May 21 '12 at 8:57