Vectors - Boat & Stream problem

In summary: SinθSubstituting this value in (1)(vbCosθ-vr)(3/vbSinθ)=23(vbCosθ-vr)=2vbSinθ3vbCosθ-3vr=2vbSinθ3vbCosθ-2vbSinθ=3vrvb(3Cosθ-2Sinθ)=3vrvb=3vr/(3Cosθ-2Sinθ)vb=3(3.5)/(3(3.5)-2(13)Sinθ)vb=10.5/(10.5-26Sinθ)It is not clear what the question is asking for, but using the information given,
  • #1
smize
78
1
Homework Statement

A boatman wants to cross a canal that is 3 km wide and wants to land at a point 2 km upstream from his starting point. The current in the canal flows at 3.5 km/h and the speed of his boat is 13 km/h.
(a) In what direction should he steer?
(b) How long will the trip take?

This is not for an actual course...this is from a textbook, but I am self teaching. I am having an issue finding out how they solved for this. I need a solution to this so I can understand it in the future.

The answers:

(a) At an angle of 43.4 deg from the bank, toward upstream
(b) 20.2min
 
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  • #2
Your destination is across and upstream, 3km north(y=3) and 2km east(x=2).
Your resultant velocity in direction tan-1(3/2).
This resultant velocity is made up of your velocity plus river current velocity.
You can resolve your boatman velocity into vertical(y) and horizontal(x) to obtain the magnitude and direction required.
 
  • #3
There may be a quicker way to do this, and I hope there is since you posted it in the calculus and beyond section, but off the top of my head the obvious one is to simply draw a right triangle with sides (3) and (3.5t+2) with hypotenuse (13t). Just use your standard pythagorean theorem to get a quadratic...

I got something like 156.75t^{2}-14t-13 = 0

Use the quadratic formula, which is a little messy here, but no big deal if you are allowed to use even a basic calculator on the problem. Solving this for the positive t value gives you something like t =.336 hrs = 20.165 min

Just plug in this t value to get the values of the sides of the original triangle, and now you can use any of your basic trig functions to find the angle you need.

To explain the set up: hopefully you see why I set up the triangle this way, but if you don't, it's very simple. Rate x Time = Distance, so we can use the given current and boat speed rates, multiply them by the unknown t to get distances. 13t is the distance traveled by the boat relative to the surface of the water. 3.5t is the distance the surface of the water "shifts" the boat downstream during the trip. And then of course 2 and 3 are distances between/on the stationary shores.
 
  • #4
Thank-you bossman, and you solved it the way they probably wanted (or in a similar fashion). This is from a chapter about vectors etc. This shall help me whenever I come across anything similar to this...I kinda feel stupid that I didn't figure it out though.
 
  • #5
smize said:
Thank-you bossman, and you solved it the way they probably wanted (or in a similar fashion). This is from a chapter about vectors etc. This shall help me whenever I come across anything similar to this...I kinda feel stupid that I didn't figure it out though.

No worries, I'm learning much more advanced math now and it still took me a minute to see which way to set it up. The takeaway here is just to remember that you can often set up problems easily by interconverting rates, times and distances, using the unknown one as a variable. In this case we can just convert the rates into distances with a variable for time, etc. etc.

That said, do be mindful of where you're posting these questions. This question belongs in the precalculus or introductory physics section.
 
  • #6
smize said:
Homework Statement

A boatman wants to cross a canal that is 3 km wide and wants to land at a point 2 km upstream from his starting point. The current in the canal flows at 3.5 km/h and the speed of his boat is 13 km/h.
(a) In what direction should he steer?
(b) How long will the trip take?

This is not for an actual course...this is from a textbook, but I am self teaching. I am having an issue finding out how they solved for this. I need a solution to this so I can understand it in the future.

The answers:

(a) At an angle of 43.4 deg from the bank, toward upstream
(b) 20.2min

Like to know is there reasons for sequence of the questions.
Do the first is easier to solve than the next.
I'm doing self teaching too.
 
  • #7
azizlwl said:
Like to know is there reasons for sequence of the questions.
Do the first is easier to solve than the next.

Although the questions in textbooks are usually ordered with intent, as may be the case here since there is a way you can find the angle first, it's good practice to look at problems like this as being one situation with two (or more) unknowns and then figuring out for yourself the best, or most efficient way to find solutions. If a textbook constantly asks you to use inefficient methods, it's probably not a very good book.
 
  • #8
bossman27 said:
That said, do be mindful of where you're posting these questions. This question belongs in the precalculus or introductory physics section.

This is from Stewart 7E Calculus textbook in Chapter 12 section 2 problems. It is over vectors; this is the first problem I came across in the practice questions I had issues with, mainly because I'm a bit rusty on my solution from a Physics standpoint.
 
  • #9
Boatman upstream component velocity = vbCosθ
Boatman across stream component velocity = vbSinθ

River downstream velocity=-vr

(vbCosθ-vr)t=2 ...(1)

(vbSinθ)t=3 ...(2)
 
Last edited:

Related to Vectors - Boat & Stream problem

1. What is a vector in the context of boat and stream problems?

A vector is a mathematical representation of a quantity that has both magnitude and direction. In the context of boat and stream problems, a vector represents the velocity of the boat and the velocity of the stream, which are both necessary to determine the boat's resultant velocity.

2. How do you find the resultant velocity of a boat in a stream using vectors?

To find the resultant velocity of a boat in a stream, you must first determine the velocity of the boat and the velocity of the stream. These velocities can be represented as vectors, with the boat's velocity pointing in the direction of the boat's movement and the stream's velocity pointing in the direction of the stream's current. Then, using vector addition, you can find the resultant velocity by adding the two vectors together.

3. What is the difference between a boat moving with the stream and against the stream?

A boat moving with the stream means that the boat's velocity and the stream's velocity are in the same direction, resulting in a higher resultant velocity. On the other hand, a boat moving against the stream means that the boat's velocity and the stream's velocity are in opposite directions, resulting in a lower resultant velocity.

4. Can the velocity of the stream affect the boat's direction of movement?

Yes, the velocity of the stream can affect the boat's direction of movement. If the stream's velocity is strong enough and in a different direction than the boat's velocity, it can cause the boat to change its direction of movement.

5. How can you use vectors to solve boat and stream problems?

Vectors are essential in solving boat and stream problems because they allow us to accurately represent and calculate the different velocities involved. By using vector addition and considering the direction and magnitude of each velocity, we can determine the resultant velocity of the boat and solve these types of problems.

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