Simple Vector Boat Problem, Conceptual Misunderstanding

  • Thread starter WRS
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  • #1
WRS
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Homework Statement:
Yes
Relevant Equations:
Pythagoras, Trigonometry
Hi there,

I have attached the problem I'm working with.

I believe I must have the wrong idea of how to approach this question.

My issue is with the stated width and calculating how long the boat will take to cross the river.

It's using width; 110m and the boats velocity to determine how long the boat will take to cross the river. Later in the section it states that the boat will have drifted 71m downstream.

My assumption is that if the river was 110m (shore to shore) and the boat drifted 71m down stream the actual distance traveled by the boat would be found using c^2 = a^2 + b^2

Distance traveled = 110^2 + 71^2
Distance traveled = 130.9m (1.dp)

And therefore the time taken to travel would be founding using width 130.9m, rather than 110m?

Thank you very much
 

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Answers and Replies

  • #2
hutchphd
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But the speed of the boat along the diagonal path is increased by the current. That increase is exactly enough to not affect the crossing time. If you try to fight the current and cross directly by angling upstream, your crossing time will increase. This of course can all be shown most easily using vectors.
 
  • #3
Like @hutchphd said, working it through with vectors might help to better your understanding. All we are really saying is that, in a frame fixed to the shore,

##\Delta \mathbf{r} = \mathbf{v} \Delta t##

Where ##\mathbf{v} = \mathbf{v}_{BS}##. From this you can get two other component equations,

##\Delta y = v_y \Delta t##
##\Delta x = v_x \Delta t##

Luckily in this case it's easy to see from a diagram that ##v_y## is just that of the boat with respect to the water, and ##v_x## is the velocity of the river (make sure to insert appropriate negative signs depending on how you setup your coordinate axes!).

So the easiest way to get the time is just to use ##\Delta t = \frac{\Delta y}{v_y}##, as your textbook did. But you can do your method also, just so long as you use the right speed! For instance, take the magnitude of both sides of the first equation I wrote:

##||\Delta \mathbf{r}|| = ||\mathbf{v}|| \Delta t##

So now ##||\Delta \mathbf{r}||## is the full (##a^2 + b^2##) distance, but ##||\mathbf{v}||## is also the magnitude of the total velocity with respect to the shore.
 

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