# At what angle must the boat travel so that it moves in a straight line

• paulimerci
In summary: If you're not sure, ask for help.Thanks for catching that.It's a good reminder to always check your work.
paulimerci
Homework Statement
A boat moves at 8m/s across a river that flows at 2m/s. At what angle must the boat travel so that it moves in a straight line that is perpendicular to the direction of the river's flow?
Relevant Equations
Trignometry
I'm assuming the boat is traveling north at an 8 m/s and the river is flowing east at a 2 m/s. For the boat to move in a straight line, it has to aim at an upstream angle given by #theta#.
Using SOH CAH TOA, ##v_r = 2m/s##, ##v_b = 8m/s##

$$\theta =\sin^{-1} \frac{v_r}{v_b}$$
$$\theta = 14 ~degrees$$

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Close : how did you calculate it ?

hmmm27 said:
Close : how did you calculate it ?
I just used trigonometry to find. According to the answer sheet, the angle is 75.5.

paulimerci said:
Homework Statement:: A boat moves at 8m/s across a river that flows at 2m/s. At what angle must the boat travel so that it moves in a straight line that is perpendicular to the direction of the river's flow?
Relevant Equations:: Trignometry

I'm assuming the boat is traveling north at an 8 m/s and the river is flowing east at a 2 m/s. For the boat to move in a straight line, it has to aim at an upstream angle given by #theta#.
Using SOH CAH TOA, ##v_r = 2m/s##, ##v_b = 8m/s##

$$\theta =\sin^{-1} \frac{v_r}{v_b}$$
$$\theta = 14 ~degrees$$

Thanks @berkeman for fixing it into Latex.

berkeman
paulimerci said:
Always show ALL of your calculations when asking such a question here.

paulimerci said:
I just used trigonometry to find.
That's NOT an answer to "how did you calculate it" Show ALL of your work.

phinds said:
Always show ALL of your calculations when asking such a question here.
I don't understand, that's what I did in post #1.

paulimerci said:
For the boat to move in a straight line, it has to aim at an upstream angle given by #theta#.
It should be noted that the boat will move in a straight line regardless how it is aimed because its acceleration is zero. Of all the possible straight lines, we are looking for the one that is perpendicular to the current.

Reason I asked is the answer's ~14.5 give or take, which is what a calculator should spew out. "14" seems odd. Also, brackets in your calculation.

paulimerci said:
I don't understand, that's what I did in post #1.
No, it is not. You went from an equation with variables straight to an answer without showing your calculation.

Anyway, you need to review basic trig. The answer of 75.5 degrees is trivial.

I followed this video, and I was quite confused with the answer I got.
I think I understand where my mistake was.
So I need to do 90-theta = 75.5, right?

paulimerci said:
So I need to do 90-theta = 75.5, right?
Now, you got it, although you STILL didn't show your work.

phinds said:
Always show ALL of your calculations when asking such a question here.
I thought he did show his reasoning and calculation that ##\theta = sin^{-1}(\frac{2}{8})##

phinds said:
Anyway, you need to review basic trig. The answer of 75.5 degrees is trivial.
Are you defining ##\theta## with respect to a different direction than directly across the river?

berkeman said:
I thought he did show his reasoning and calculation that ##\theta = sin^{-1}(\frac{2}{8})##
Hm ... either I missed the 2 and the 8 in the original post or he backfilled. It would not surpise me to find that I overlooked it.
berkeman said:
Are you defining ##\theta## with respect to a different direction than directly across the river?

View attachment 322972
I using the theta that he drew. sin theta is NOT Vr / Vb. As he correctly figured out in post #11, the answer is sin (90 MINUS theta) = Vr/Vb

phinds said:
I using the theta that he drew. sin theta is NOT Vr / Vb. As he correctly figured out in post #11, the answer is sin (90 MINUS theta) = Vr/Vb
Sorry, I'm probably being dense, but how is it possible that the boat has a speed of 8m/s and the river is moving at 2m/s and it takes an angle of greater than 45 degrees to move straight across the river...?

DaveC426913
HA, he mixed up the 8 and the 2 and I just went along with him. Good thing SOMEBODY here is paying attention. I was also fooled by the fact that he said the answer is 75. Had he got his numbers right he would have gotten 15 (well, 14.47) and would have questioned the book's answer. The book's answer is right ONLY if you take theta from the perpendicular to the bank, not to the bank, which it normally is taken as in such problems.

Good catch, Mike. That's why Greg pays you the big bucks, huh?

berkeman
Gads. He pays you with animals? I'm going to call the SPCA.

berkeman
berkeman said:
Sorry, I'm probably being dense, but how is it possible that the boat has a speed of 8m/s and the river is moving at 2m/s and it takes an angle of greater than 45 degrees to move straight across the river...?
Sanity check. The final critical but disappearing step to problem-solving.

Tom.G, BvU, phinds and 1 other person
DaveC426913 said:
Sanity check. The final critical but disappearing step to problem-solving.
Exactly. I USED to do that all the time, before I went insane.

Tom.G and SammyS
phinds said:
Hm ... either I missed the 2 and the 8 in the original post or he backfilled. It would not surpise me to find that I overlooked it.

I using the theta that he drew. sin theta is NOT Vr / Vb. As he correctly figured out in post #11, the answer is sin (90 MINUS theta) = Vr/Vb
I truly appreciate your time and explaining! What do you mean I backfilled?
I’ve learning disabilities and I really want to learn something. Guide me what mistakes I did. As a non-native English speaker, it also takes me some time to comprehend and communicate.

phinds said:
Hm ... either I missed the 2 and the 8 in the original post or he backfilled. It would not surpise me to find that I overlooked it.
It was in his OP in two places. It was in his LaTeX work, but initially the LaTeX had an extra "{" character in it that messed up the rendering, so likely you saw his OP before that was fixed. It was also in his hand-drawn sketch that was attached to the OP, but I usually don't click into those unless I have to.

paulimerci said:
What do you mean I backfilled?
He just means that maybe you fixed it later, after others had read the OP (your Original Post to start the thread). But as I say above, it was likely just due to the typo in the LaTeX making it not readable, which I fixed a little while after you posted your OP.

berkeman said:
It was in his OP in two places. It was in his LaTeX work, but initially the LaTeX had an extra "{" character in it that messed up the rendering, so likely you saw his OP before that was fixed. It was also in his hand-drawn sketch that was attached to the OP, but I usually don't click into those unless I have to.He just means that maybe you fixed it later, after others had read the OP (your Original Post to start the thread). But as I say above, it was likely just due to the typo in the LaTeX making it not readable, which I fixed a little while after you posted your OP.
Thank you for clarifying!

berkeman
phinds said:
Hm ... either I missed the 2 and the 8 in the original post or he backfilled. It would not surpise me to find that I overlooked it.

I using the theta that he drew. sin theta is NOT Vr / Vb. As he correctly figured out in post #11, the answer is sin (90 MINUS theta) = Vr/Vb
I truly appreciate your time and explaining! What do you mean I backfilled?
I’ve learning disabilities and I really want to learn something and I’m not a native English speaker either as you all.
paulimerci said:
Thank you for clarifying!
actually I didn’t edit anything in post #1 so far.

paulimerci said:
What do you mean I backfilled?
See berkeman's post #22. The second paragraph addresses this.

phinds said:
HA, he mixed up the 8 and the 2 and I just went along with him. Good thing SOMEBODY here is paying attention. I was also fooled by the fact that he said the answer is 75. Had he got his numbers right he would have gotten 15 (well, 14.47) and would have questioned the book's answer. The book's answer is right ONLY if you take theta from the perpendicular to the bank, not to the bank, which it normally is taken as in such problems.

Good catch, Mike. That's why Greg pays you the big bucks, huh?
What do you mean I mixed up 8 and 2? Can you point out where the mistake is?

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