Relative Velocity Against the Water Current

[unoonu]

Homework Statement

A rower is paddling his canoe towards his goal 36 m away against the current with a speed of 3 m/s with respect to the water. If the speed of the water with respect to the ground is 6 m/s, how long will it take him to reach his goal?

Relevant Equations

Based on my own understanding:
1. v[SUB]canoe_wrt_ground[/SUB] = v[SUB]canoe_wrt_water[/SUB] + v[SUB]water_wrt_ground[/SUB]
2. v[SUB]canoe_wrt_water[/SUB] = 3 m/s
3. v[SUB]water_wrt_ground[/SUB] = -6 m/s

v_{canoe_wrt_ground} = 3 m/s + (-6) m/s = 3 m/s - 6 m/s = -3 m/s

Thus, if I understand this correctly, the rower will never reach his goal 36 m away as his canoe's resultant velocity is negative (i.e., his canoe is effectively going downstream even though it is trying to go upstream). My only problem with this is that this seems to be an unexpected answer to the question, which makes me think that I might be missing something important here.

I would appreciate your kind feedback!

 

[PeroK]

Yes, the rower cannot make progress upstream. You need a diagram of the scenario to see what is intended.

 

[Steve4Physics]

Homework Statement:: A rower is paddling his canoe towards his goal 36 m away against the current with a speed of 3 m/s with respect to the water. If the speed of the water with respect to the ground is 6 m/s, how long will it take him to reach his goal?
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Thus, if I understand this correctly, the rower will never reach his goal 36 m away a

Have you supplied the question completely and accurately? If so, I agree with you.
Could this simply be a 'trick' question to check your basic understanding?

 

[jbriggs444]

Alternately, if the "goal" is a hapless swimmer thrashing in the water hoping for rescue then an answer can be obtained.

Or one can work the original question, obtain an answer, note that it is negative and conclude that the rower was previously at the goal as of the indicated delta time in the past.

 

[kuruman]

There is another interpretation. The problem states that the goal is "36 m away". If the direction of the displacement vector is downstream, the solution is trivial.

For this interpretation to work, "Towards his goal" should be interpreted as "in such a way as to enable him to reach his goal", not as a vector direction.

 

[jbriggs444]

If the direction of the displacement vector is downstream, the solution is trivial.

Within $\arcsin \frac{1}{2}$ (30 degrees) of straight downstream it is still possible, albeit a bit less trivial.

 

[unoonu]

You need a diagram of the scenario to see what is intended.

Have you supplied the question completely and accurately?

Unfortunately, this is all that was provided. My instructor won't respond to any queries, too, sadly.

There is another interpretation. The problem states that the goal is "36 m away". If the direction of the displacement vector is downstream, the solution is trivial.

I had thought so, too, but the phrasing of the problem made me think otherwise. I wonder if it would be acceptable to have two answers in this case, haha.

 

[jbriggs444]

General rule is to explain why you think a problem is impossible to solve, state some reasonable assumptions to make the problem possible to solve (but not trivial) and solve it on that basis.

Several of my instructors over the years made statements of this sort. Usually when they were also telling us to "always show your work".

Naturally, whether this works depends on whether your instructor is a reasonable person.

 

[kuruman]

It is also possilbe that the instructor inadvertently swapped the speeds of the canoe and the water current.

 

[unoonu]

Thank you, everyone!