Help with a problem involving motion in two dimensions

In summary, the conversation is about a physics problem involving a boat traveling on a river and the difficulties the person is having in solving it. They have tried using the law of cosines and drawing vector diagrams but have not been successful. They are seeking help from others and are feeling frustrated with their progress.
  • #1
CVRIV
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5
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I attached an image with problem and the associated illustration.

I don't know what the relevant equations are for this problem.

Each and every time I read the problem I keep drawing a right triangle to represent the velocity components. The problem says the river is flowing east at 5km/h. I am assuming this is the x-component named Vre. It then says the boat is traveling south of eat at 45deg. I am assuming this is the vector with an unknown magnitude, named Vbe. The boat's velocity, with respect to the river(Vbr), is also unknown. In the previous similar problems Vbr was 10km/h and I can't help but assume it's the same for this problem. Also, vBR was perpendicular to the river, but somehow I feel like it's not, for this problem.

I don't know and I am literally ready to just give up and move on because I just don't have time for this.
 

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  • #2
Welcome to the PF.
CVRIV said:
I was very fortunate to find the very pages on the internet. I'm doing Exercise #39 on page 73. I did Example 3.9 and 3.8(Page 71) without any issues, but Exercise 3.9 completely escapes me.
Do you mean lines 3.9 on page 73? Can you upload a PDF or JPEG snapshot of the part of the problem that is confusing you? That will work a lot better than us all opening your full text and trying to zoom in and decode which part you are asking about. Thanks.

(Click the Upload button at the lower right of the Edit window, or just use the Windows Snapshot tool to grab a copy of the problem, and Cntrl-V Paste it into the Edit window).
 
  • #3
I edited my original post to be more easily understood.
 
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  • #4
Can you post your work so far (as vector drawings and equations, preferably)?

As the problem says, you will end up with two equations and two unknowns to solve for.
 
  • #5
I don't have any work for this problem saved because the work I've done so far was by no means worth saving. I think I figured out what I'm doing wrong though.

I was assuming that the boat is traveling across the river at an angle of 90deg; this is why I kept drawing a right triangle to represent my work. I realized that this isn't the case because the last two problems I solved. Each problem involved the very same boat traveling at the very same speed, 10km/h with respect to the river. In one problem, the boat is traveling at 90deg with respect to the river and i was to solve for the velocity of the boat and angle with respect to earth. I got 11.2km/h @ 26.6deg. The other problem stated that the boat WANTED to travel at 90 with respect to the river. It turned out the boat had to travel at an angle of 30deg with respect to earth, to travel at 90deg with respect to the river.

If the two known magnitudes are the same, 5km/h for the river and 10km/h for the boat with respect to the river, and the angle of the boat traveling with respect to Earth is 45deg, then the angle of the boat traveling with respect to the river can't be 90deg. It has to be less than that.

I think I know what I have to do now.
 
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  • #6
Sounds like good progress. For me, I have to draw vector diagrams to help me with problems like this. It's easy for me to make mistakes unless I'm careful keeping track of which reference frame my vectors are drawn with respect to (the Earth reference frame, the boat's frame, or one moving with respect to the river, etc.). Keep us posted on what you find. :smile:
 
  • #7
I give up. I tried the law of cosines and still got it wrong. I didn't even want to use the law of cosines. I wanted to solve it without using LOC. I'm super frustrated. I need to walk away from this. I really need to move on to new stuff. I'm already behind.
 

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  • #8
CVRIV said:
I give up. I tried the law of cosines and still got it wrong. I didn't even want to use the law of cosines. I wanted to solve it without using LOC. I'm super frustrated. I need to walk away from this. I really need to move on to new stuff. I'm already behind.
In your diagram, the dashed line perimeter is a square. Let its sides be h.
Can you get an expression for the known length AB in terms of h?
 
  • #9
CVRIV said:
Each and every time I read the problem I keep drawing a right triangle to represent the velocity components. The problem says the river is flowing east at 5km/h. I am assuming this is the x-component named Vre. It then says the boat is traveling south of eat at 45deg. I am assuming this is the vector with an unknown magnitude, named Vbe. The boat's velocity, with respect to the river(Vbr), is also unknown.
It is a good start. Let be x the East direction and y the South one. Write up the x and y components of all velocity vectors. How are the velocity vectors ##\vec v_{re}##, ##\vec v_{be}## and ##\vec v_{br}## related?
Do not mind the previous problems, this is independent of them.
 
  • #10
This is my best effort. I had a little more help from another forum but don't fully understand how that person got what they got. I just done get it.
 

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  • #11
CVRIV said:
This is my best effort. I had a little more help from another forum but don't fully understand how that person got what they got. I just done get it.
In the 17a image, look at the two x= and two y= equations. Sin(45)=cos(45), right? What does that tell you about x and y? What do you get if you use that in the other two?
 
  • #12
Finally! Oh my god! I'm suppose to know how to do this! Sad. Firstly, I remember solving for two unknowns in Algebra. It came back to me when I read what haruspex said.

The next mistake I made was that I didnt make each expression equal to each other. I was substituting one inside the other.

I can't tell you how happy I am to have figured this out. I have trouble not finishing stuff. If I gave up that woupd of really bothered me.
 

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FAQ: Help with a problem involving motion in two dimensions

What is motion in two dimensions?

Motion in two dimensions refers to the movement of an object in both the horizontal and vertical directions, as opposed to just one direction. This can also be referred to as two-dimensional motion or projectile motion.

What is the difference between distance and displacement?

Distance is the total length traveled by an object, while displacement is the straight line distance between the starting and ending points of an object's motion. Displacement takes into account the direction of motion, while distance does not.

How do you calculate velocity in two dimensions?

Velocity in two dimensions is calculated by dividing the displacement in each direction by the time taken. This can be written as v = (Δx/Δt)i + (Δy/Δt)j, where i and j represent the horizontal and vertical directions, respectively.

What is the difference between speed and velocity?

Speed is a measure of how fast an object is moving, while velocity is a measure of both the speed and direction of an object's motion. While speed is a scalar quantity, velocity is a vector quantity.

How do you incorporate the effects of gravity in two-dimensional motion?

In two-dimensional motion, gravity can be incorporated by using the acceleration due to gravity, which is typically represented as -9.8 m/s² in the downward direction. This value can be used in equations to calculate the effects of gravity on the object's motion.

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