Am I making a fatal error with this simplification? I'm teaching a grade 9 math course that deals with volumes of pyramids, cones, prisms, and cylinders and I wanted to come up with an interesting situation that uses the 1/3 volume factor. I know that fluids are tricky but I was wondering if the logic I've used here is a valid simplification or not? It's fine if they learn more details should they take physics some other time, but I don't want to tell them anything wrong. Here's the example:

"Water flowing through a cylindrical hose flows through a cone-shaped spout at the end. If water is flowing through the hose at 3 cm/s, how fast is it flowing through the cone tip? You can assume that water cannot be compressed to take up less volume.


Answer: The volume of the spout is one-third of the volume of the hose. As the water gets to the end of the hose, the water is squished into a smaller space. So, the water must flow faster to compensate. Logically, the water must flow through the cone 3 times as fast. So, water should be passing through the cone at a rate of 9 cm/s.
This 9cm/s should be considered an average rate. In reality, the water coming out of the tiny tip would be moving faster, but water at the base of the cone where it is the widest would be traveling more slowly."

EDIT: Thanks so much for helping me think through this problem, everyone. I enjoyed reading all of your answers and am grateful you took the time to write at such length. I am going to shelve this problem as is for the grade 9s, but if I do get a chance to pose it with another age group then I'll be able to do it properly then.
 A: *

*The tip cannot come to a point - otherwise the water cannot escape from its end -  so actually its shape is a truncated cone also known as a conical frustrum.

*You say


If water is flowing through the hose at 3 cm/s, how fast is it flowing through the cone tip?

This wording is ambiguous. I have taken this to mean "how fast is the water leaving the end of the tip". I see other answers have used different interpretations.


*The ratio of the speed of the water in the hose pipe and the speed of the water leaving the end of the tip depends on their respective cross-sectional areas. The water leaving the end of tip only travels three times as fast as in the hose if the cross-sectional area at the end of the tip is one third that of the hose. The volume (and, indeed, the exact shape) of the tip are irrelevant.

A: This does work out to be correct, but there's a pitfall hiding in the question in the idea of what the "average speed" is, and (IMO) the subtleties involved make it a bad fit for a ninth-grade math class.
To be a bit more realistic, let's assume that the nozzle is a frustum with height $h$, a radius $r_0$ at the base and radius $\epsilon r_0$ (with $\epsilon \ll 1$) at the tip.  Assuming the frustum tapers uniformly, the radius of the cone at a distance $x$ from the base would be $r = r_0 + xr_0(\epsilon - 1)/h $, and the speed of the water a distance $x$ from the base would be
$$
v = v_0 \frac{r_0^2}{(r_0 + x(\epsilon r_0 - r_0)/h)^2} = \frac{v_0}{(1 + x (\epsilon-1)/h)^2}.
$$
This means that we can calculate the distance-averaged speed of the water over the length of the nozzle.  This would be what we would get if we stuck a large number of evenly-spaced flowmeters along the length of the nozzle and then averaged their readings.   It works out to be
$$
\langle v \rangle_x = \frac{v_0}{h} \int_0^h \frac{dx}{(1 + x (\epsilon-1)/h)^2} = \frac{v_0}{h} \left( \frac{h}{\epsilon} \right) = \frac{v_0}{\epsilon}.
$$
We can see that this is not equal to $3 v_0$ as your argument would predict.
But wait!  Maybe if we take the average over the volume of the nozzle, things will work better.  We can accomplish this by "weighting" the above integral by the cross-sectional area of the nozzle.  This does work out;  we have
$$
\langle v \rangle_V = \frac{v_0}{V} \int_0^h \frac{\pi r_0^2 (1 + x (\epsilon-1)/h)^2 dx}{(1 + x (\epsilon-1)/h)^2} = \frac{\pi h v_0 r_0^2}{\frac{1}{3} \pi h r_0^2(1 + \epsilon + \epsilon^2)} = \frac{3 v_0}{1 + \epsilon + \epsilon^2}.
$$
which does in fact approach $3 v_0$ as $\epsilon \to 0$.
Alternately, we can also try to calculate the average speed of a water molecule as it passes through the nozzle.  The time taken for the water molecule to pass through this nozzle will be
$$
\Delta t = \int_{x=0}^{x=h} dt = \int_0^h \frac{dx}{v} = \frac{h}{3 v_0} (1 + \epsilon + \epsilon^2)
$$
and the time-averaged speed of the water molecule will then be the distance traveled $h$ divided by $\delta t$, or
$$
\langle v \rangle_t = \frac{3 v_0}{1 + \epsilon + \epsilon^2},
$$
as we found above by the volume average.
So if you define "average speed" to be distance traveled over time taken, then your answer is correct.  To be fair, this is how "average speed" is defined in physics, so it's technically correct.   But it's also true that the speed of the water gets very very very large for $\epsilon \ll 1$;  for a proper cone ($\epsilon = 0$), it is moving infinitely fast as it exits the nozzle.  You might get a student asking you "if the water is moving infinitely fast as it leaves the cone, how can the average speed be finite?"  I'm not sure that there's a good answer to this question without breaking out the calculus like I did above, and presumably ninth-graders aren't ready for that.
(This idea that an infinite speed at one point in a trip is compatible with a finite average speed over that trip is more clearly (?) illustrated in the old chestnut of a brain-teaser.  A person needs to drive a 10-mile journey.  They run into traffic over the first 5 miles and only drive at 30 miles per hour over that distance.  How fast must they drive for the remaining 5 miles to average 60 miles per hour over the whole trip?  The intuitive answer of 90 mph is, of course, wrong, but very smart people have reportedly fallen into that trap.)
A: That's not right, unfortunately. The principle governing this situation is the continuity equation, which says that the total flow rate past any given point is constant.  Since the flow rate is given by the flow speed $v$ times the cross-sectional area of the hose $A$, one has that
$$v_1 A_1 = v_2 A_2$$
for any two points along the flow.  In particular, the velocity of the water coming out through the nozzle is
$$v_{spout} = \frac{\big(3 \mathrm{cm/s} \big)A_{hose}}{A_{spout}}$$
where $A_{hose}$ is the cross-sectional area of the hose and $A_{spout}$ is the cross-sectional area of the aperature at the end of the nozzle.

The term average might be used too loosely. But, wouldn't it still be true that after 1 second of flow, the equivalent amount of water in the cylindrical portion of the hose would need to escape the cone shape at the end. And that the cone would have to expel its volume 3 times to match the volume in the cylinder portion?

Yes.  There is a sense in which the 9 cm/s is a relevant figure. If you divide the water within the cone into many tiny little parcels, compute the speed of each parcel, and then average them all together, the average of all of those speeds approaches 9 cm/s as the area of the aperture at the end of the cone goes to zero. In particular it is not the velocity at the half-way point of the cone (in fact, it is the velocity at the point $1-1/\sqrt{3}\approx 42.26\%$ of the way from the base to the tip).
If that's what you're asking your students to compute, the volume of a cylinder vs. the volume of a cone with the same length and base area is a reasonable shortcut to the solution. However, for grade 9 students I fear that the question would be completely opaque.
Your intent appears to be to get them to divide the volume of a cylinder by the volume of a cone. It is in principle possible to devise a question to which that is the shortest path to a solution, but my personal opinion is that this route would simply confuse them, and that any attempt at explanation would ultimately lead you to "look, just divide these two volumes ... " which defeats the point of having a creative question.
But then again, you're their instructor, so that's a call you'd have to make.
A: There is a different tack that you can take to ask a similar question and arrive at the correct final concept (see the other excellent answers for directly answering the original question).

Question: Water flowing through a cylindrical hose flows through a cone-shaped spout of height $h$ at the end. If water is flowing through the hose at a constant rate (water volume per unit time) and it takes three seconds for the final length $h$ of the hose to be filled, how much more time will it take to fill the cone tip? You can assume that water cannot be compressed to take up less volume.


Answer: The volume of the spout is one third of the volume of the hose. Since the water is flowing at a constant rate, it will take one third of the time: one second.

