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**CIV****Member**- Registered: 2014-11-09
- Posts: 74

http://www.nxtbook.com/nxtbooks/ngl/collegephysics11eAP/index.php?startid=72#/76

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 with out any issues, but Exercise 3.9 completely escapes me.

The illustration they provided has me a bit confused. I think the illustration is drawn to accommodate both parts of Exercise 3.9, but I'm not sure because I'm having trouble solving this.

I read the problem carefully over and over and I keep drawing a right triangle to represent the components of the vector and the vector itself, but the answer I'm getting isn't making any sense.

It says the river is flowing east at 5km/h. I made this the x-component which is named Vre(velocity of the river with respect to earth).

The boat is traveling 45deg south of east with respect to the the earth. I made this the vector with a magnitude of Vbe(velocity of boat with respect to earth). It wants me to solve for Vbe.

Im assuming Vbr(velocity of boat with respect to the river) is the y component, as it was for the other two similar problems I solved before this.

I keep ending up with a right triangle with two unknown sides, Vbe and Vbr. I know all the angles. I know that tan(45) = 1, therefore Opp = Adj, so 5km/h = 5km/h. Using Pythagorean Thereom I get Vbe = 7.07km/h, but that's not the answer the book has.

I could't help but assume that Vbr is 10km/h because that was the boats velocity for the last two problems, but if that true the right triangle with the 45deg angle doesn't make sense because both sides would have to be equal.

I'm completely lost and wasted so much time trying to solve this. Please help.

*Last edited by CIV (2019-02-20 03:28:13)*

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 122

The problem is simple, but you have made a mistake in assuming that the velocity triangle is a right triangle while in the question the instructor emphasizes that the boat is heading 60 degree south of east, so the triangle is in fact a 45-120-15 triangle and to solve the problem you can use the law of sines (I think you know what that means since you have reached this point in your course). Therefore the solution goes as follows (I will utilize the notation which you have used):

and for the velocity of the boat with respect to the river we would have:

which are exactly the answers given in the book. Good luck!

*Last edited by Grantingriver (2019-02-20 19:39:50)*

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 122

If you are interested, as an alternative solution you can draw the velocity triangle such that the river flows toward the positive x axis (the east direction) and the south direction to the negative y axis. In this case if you drop a vertical line from the head of the resultant vector (the velocity of the boat relative to Earth) on the x axis and take the reflection of that triangle about the x axis, you will have an identical triangle in which the Cartesian coordinates of the resultant is:

from which the magnitude of this velocity (the speed) is given by:

however, from the law of cosines we have:

Solving equations (1) and (2) simultaneously would give the required solutions which are identical to those found by the previous mothed.

*Last edited by Grantingriver (2019-02-20 19:46:42)*

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**CIV****Member**- Registered: 2014-11-09
- Posts: 74

I have to study what you did there. I am familiar with the law of sines, but haven't done anything with it in soooo loong. I'm actually more interested in your alternative method and I don't know why. I'm grateful that you provided me with the alternative. I have to comb through that to figure out what's going on there. Thank you so much. Might have a question about it later. I have to step away from this a bit. I have been up all night burning myself out trying to figure this out.

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 122

If you would like, the following is also another solution. In the velocity triangle described in the second solution if you drop a vertical line on the resultant (the velocity of the boat with respect to Earth) and labeled the obtained two sections of the resultant and the perpendicular line segment

and respectively. you would get:but you would get also

so from equation (2) it follows that

and hence substituting into equation (1) you would have:

which are also the required results.

*Last edited by Grantingriver (2019-02-20 19:49:48)*

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