2D Relative Velocity regarding boats

In summary, the boat must be pointed at an angle relative to the direction the current is flowing in order to cross the river. It takes the boat about .5 minutes to cross the river.
  • #1
in the rye
83
6

Homework Statement


A boat propelled so as to travel with a speed of .5m/s in still water moves directly across a river that is 60m wide. The river flows with a speed of .3 m/s. At what angle relative to the straight across direction must the boat be pointed? How long does it take the boat to cross the river?

Homework Equations

The Attempt at a Solution

Okay, so my question really isn't on the math. It's on the question itself. I can solve it once I have a correct diagram. My question is, how do I draw my vectors? When they don't give details, I seem to frequently draw my resultant incorrectly. In this problem, I took the boats speed in still water to be directly across stream at .5 m/s considering it is moving "directly across the river." However, in the book, they take the boat in still water to be the resultant. Why? In fact, in reviewing my notes, my professor gave a similar example in which he took the "in still water" to mean the path directly from shore to shore if we consider our initial position to be the origin, it only moved one direction in his example. But the book does it differently and it screws up

It seems most of my issues are due more to reading comprehension or the way the problems are worded. I understand that directly can mean it is moving linear. However, if I told someone to walk directly across the street from our point of reference I wouldn't expect them to move diagonally across it. I would expect them to move in a horizontal straight line from the frame of reference where we are standing. Then, if there happens to be a strong wind that blows them right or left, the resultant would be from their vector walking directly from where there were plus the effects of the wind...

It's just frustrating that the most difficult part is just setting up the problems, and the way the text seems to do it just doesn't make any sense to me. When I have a diagram, or I am told angles, or direction, I can draw it perfectly and solve with no issues. But the moment there includes some sort of interpretation of the problem, I screw up. It doesn't matter how many examples I do. This keeps recurring.

I can kind of see how it may be inferred since you are getting the angle at which the boat must be pointed from the horizontal might mean that our "in still water" is the resultant vector, but I still think it's assuming that you want to go in the across stream direction

I need some help :/
 
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  • #2
Hi - try thinking of the extreme examples and see if that helps.

For example.. if the boat stood still, then it gets pushed by the current at 90 degrees to the indended motion, if it moves at infinite speed, it travels staight across; if it traveled at 0.3m/s (same as the current). then it would travel at 45 degrees down stream (and therefore at if it wanted to go straight across would need to aim 45 degrees upstream)

I think your professor is asking you to consider the addition of two vectors at right angles to each other; the boat and the current. Apologies in advance if I've missed the thrust of your question.
 
  • #3
mgkii said:
Hi - try thinking of the extreme examples and see if that helps.

For example.. if the boat stood still, then it gets pushed by the current at 90 degrees to the indended motion, if it moves at infinite speed, it travels staight across; if it traveled at 0.3m/s (same as the current). then it would travel at 45 degrees down stream (and therefore at if it wanted to go straight across would need to aim 45 degrees upstream)

I think your professor is asking you to consider the addition of two vectors at right angles to each other; the boat and the current. Apologies in advance if I've missed the thrust of your question.

This is what it turns out to be. However, I'm confused as to what the question is really after. I feel like it's implying too much based on its wording. That is, when I drew my right triangle I drew my directly across stream to be the .5m/s. I realized this would mean that the current would push it down stream, therefore my resultant should be upstream. Thus, when I read the question I read it as asking, "What should the angle of the boats velocity be if it wanted to go .5m/s across stream." Not, "What should the angle be if the boats velocity is 0.5m/s". It's these types of nuances where I keep screwing up the questions. I guess it's not always apparent to me what is being asked based on how I read it.
 
  • #4
I think maybe the confusion comes from the statement "propelled so as to travel with a speed of .5m/s in still water". The question (not just this one, but all test questions) will give you some "givens" to work with; the trick is to ignore the fluff around this - you're simply being told that the speed of the boat (when not impacted by another vector) is 0.5m/s.

Question practice is key here - knowing your math is one thing, but interpreting the questions correctly is the different between a good pass and a bad fail!

Best of luck in your practice.
 
  • #5
mgkii said:
I think maybe the confusion comes from the statement "propelled so as to travel with a speed of .5m/s in still water". The question (not just this one, but all test questions) will give you some "givens" to work with; the trick is to ignore the fluff around this - you're simply being told that the speed of the boat (when not impacted by another vector) is 0.5m/s.

Question practice is key here - knowing your math is one thing, but interpreting the questions correctly is the different between a good pass and a bad fail!

Best of luck in your practice.

Right, this is actually why I took the 0.5 to be the component to be perpendicular to my stream current. I guess I don't see why this is actually my resultant velocity, so to speak, or my hypotenuse. I guess I'll just look at more examples. We only had 2 assigned for this on mastering physics.
 
  • #6
Hi again. I'm not sure I understood exactly what you meant in that last post, but if you meant 0.5m/s in respect to the statement "I guess I don't see why this is actually my resultant velocity", you may not be on the right track.

Go back to the extreme examples - of your boat was "propelled so as to travel with a speed of 0.3m/s in still water" and you headed straight up stream, you'd end up at a stand-still. Decrease the angle a little and you'll start moving across stream, but very slowly.

Back to the real question of your boat "propelled so as to travel with a speed of .5m/s in still water" - If you're heading upstream to counter the 0.3m/s flow, then your resulting velocity will be less that .5m/s
 
  • #7
mgkii said:
Hi again. I'm not sure I understood exactly what you meant in that last post, but if you meant 0.5m/s in respect to the statement "I guess I don't see why this is actually my resultant velocity", you may not be on the right track.

Go back to the extreme examples - of your boat was "propelled so as to travel with a speed of 0.3m/s in still water" and you headed straight up stream, you'd end up at a stand-still. Decrease the angle a little and you'll start moving across stream, but very slowly.

Back to the real question of your boat "propelled so as to travel with a speed of .5m/s in still water" - If you're heading upstream to counter the 0.3m/s flow, then your resulting velocity will be less that .5m/s

Yeah, I see that. I guess when I read the in still water I thought the driver would be directing it across the stream directly from the point he was going. So that if we broke it into components the .5m/s would be headed in a 1Dimensional direction. Then the stream would shift the .5m/s down stream at .3m/s. So when I added my components I got my y direction to be .5 and my x direction to be -.3 (opposite of what it is actually flowing) to give a resultant in order that they can travel .5m/s in the y direction. But thinking of it in terms of what it is designed to do helps. Because it clears up my confusion in that he resultant vector in the way I drew it would be greater than .5m/s, which the boat can't do.
 

Related to 2D Relative Velocity regarding boats

1. What is 2D relative velocity and how does it apply to boats?

2D relative velocity is the measurement of an object's speed and direction in relation to another object. In the case of boats, it refers to the speed and direction of one boat compared to another. This is important in understanding how boats move and navigate in relation to each other.

2. How is 2D relative velocity calculated for boats?

The 2D relative velocity of boats is calculated by taking into account the speed and direction of each boat, as well as their positions in relation to each other. This can be done using vector addition or by using the Pythagorean theorem and trigonometric functions.

3. What factors affect 2D relative velocity between boats?

Several factors can affect the 2D relative velocity between boats, including the speed and direction of each boat, the shape and size of the boats, the water conditions, and external forces such as wind and currents.

4. How does 2D relative velocity impact boat collisions?

Understanding 2D relative velocity is crucial in avoiding collisions between boats. If two boats have similar or opposing 2D relative velocities, they are more likely to collide. Boats with significantly different 2D relative velocities can pass by each other without incident.

5. Can 2D relative velocity be used for other types of vehicles besides boats?

Yes, 2D relative velocity can be applied to any moving object, including airplanes, cars, and trains. It is used in many fields, such as navigation, engineering, and physics, to understand the movement of objects in relation to each other.

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