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 add them all together, the arithmetic 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.

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.

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.

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 add them all together, the arithmetic 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.