River-boat vector math calculation (Introductory physics- Vector)

Thread Starter

studyhard24

Joined Jun 20, 2022
35
A streamy river of 1000m wide is flowing in front of Robin's house. His college is just across the river from the house. One morning, just
10 minutes before the start of the class, he set sail for the college in a 10 kmh^-1 speed boat at 120 angles with the speed of the current and went
straight to the other side. [The distance of college from the bank of the river is negligible.]
Q:
c) What is the value of the velocity of the current of the river according to the stem?
d) Could Robin reach in class in due time? Give your opinion with mathematical analysis.

For question no c, as we know the relation of tan(theta) and do calculation accordingly I've calculated the river current.
For question no d, For a minimum time we know the value of theta in t= d/(v*cos(theta)) will be 90 degree. So minimum time is calculated by this way.

306213984_434532215196419_1729278961013638120_n.jpg

It'll be helpful to know if my approach is correct. Thanks in advanced
 

Attachments

WBahn

Joined Mar 31, 2012
27,389
In looking over your work, it is very hard to follow what you are doing and what the reasoning is. There are also several points where what you are doing, assuming I'm interpreting it correctly, make no sense.

You seem to be defining 'v' as the speed of the boat (10 km/hr) and 'u' as being the velocity of the river, with positive values going to the right.

You then ended up with a value for the river current of -12.28 km/h, which means that it is flowing in to the left (the same direction that the boat is angled) and it is flowing faster than the total speed of the boat?

Does this make any logical sense to you?

Always ask yourself if the answer makes sense.

A couple of lines above your result, you have 1/0 = (something). Since division by zero is undefined, this equation is indeterminate. So your next step is completely meaningless.

Consider this: You state that

tan(90°) = ∞ = 1/0

But why not say

tan(90°) = ∞ = 1000/0

or

tan(90°) = ∞ = 0.001/0

All three of those are just as valid, yet would yield completely different results for the rest of your work.

Whenever you see something divided by zero (or at any point where an expression in a denominator could become zero), you need to tread very carefully and make sure that what you are doing truly is valid.

In part d, you are completely ignoring the fact that the boat is angled upstream.

Finally, you REALLY need to track your units properly. You are just tacking whatever units you would like the answer to have onto the final result. Get out of that habit (regardless of the fact that your teacher and your textbook probably also have the same horrible habit).
 

Thread Starter

studyhard24

Joined Jun 20, 2022
35
Does this make any logical sense to you?
I guess it's wrong, that's why I came for help

you are completely ignoring the fact that the boat is angled upstream.
Well as we were told to find the minimum time, to perpendicular direction gives the minimum time, for this I did that way.

you REALLY need to track your units properly.
For your kind information Sir, I did add units to the answers. As for no. c) we did the calculation in kmh^-1 so the answer will have the same.
For no. d ) the question has a minimum reach time of 10 minutes, for this, I have converted 10km/h to meter/min (multiplying 1000/60 to get it).

Pardon my mistakes.

And I'll be more than glad to get some suggestions on solving the problem. Thank you.

Your help is valuable to me
 

WBahn

Joined Mar 31, 2012
27,389
I guess it's wrong, that's why I came for help
But it would better if you sanity check your results so that you can say something like, "I know the answer is wrong because the calculated river current is flowing too fast and in the wrong direction, but I need helping understanding what I did wrong."

Remember, your education is aimed at preparing you to solve problems for which you won't be able to ask others if your results are correct or not. So it is extremely important to develop the skills to validate your work as effectively as you can.

Well as we were told to find the minimum time, to perpendicular direction gives the minimum time, for this I did that way.
But give this some thought. If the boat heads directly across, it may make it to the other shore in six minutes, but the current will take it well down stream. At the very least, he would then have to head directly upstream against the current until he gets even with the school.

For your kind information Sir, I did add units to the answers. As for no. c) we did the calculation in kmh^-1 so the answer will have the same.
For no. d ) the question has a minimum reach time of 10 minutes, for this, I have converted 10km/h to meter/min (multiplying 1000/60 to get it).
As I just tacking units onto the answer is NOT properly tracking your units. Tracking your units properly means using the proper units at EVERY step of the computations and verifying that they work out correctly.

Proper units tracking is perhaps THE most powerful error detection tool available. Most (not all) mistakes that are made will mess up the units, letting you know exactly when the mistake happened and letting you catch it and fix it before wasting a bunch of time on a lot of work that is guaranteed to be wrong from that point on. But this only works if the units are there to get messed up by the mistake.
 

ericgibbs

Joined Jan 29, 2010
16,361
Hi S24,
Consider the problem in another way.

Assume the river is NOT flowing, your boat speed is 10km/hr, and the river is 1000mtrs wide.
How long will it take you to cross the river.??

Now consider the river is now flowing, you start to cross the river, again at 10km/hr, BUT always steer at right angles towards the far river bank.

You will be swept a distance downstream by the river current, at an angle of 120deg – 90deg ie: 30deg, you are still moving 10km/hr relative towards the far bank.

Knowing your speed is always 10km/hr , how many minutes will it take to cover this longer ‘hypotenuse’ distance.??

Knowing how many minutes it took to cross and the distance you were swept downstream, you should be able to work out the river current velocity.


E
 

MrChips

Joined Oct 2, 2009
27,112
This is an exercise in vector analysis. There is no need to take a simple problem and make it complicated.

The boat has a velocity vector which can be decomposed into Vx and Vy, where x is the direction along the river and y is the orthogonal direction across the river.

The resultant vector in x is zero.
The resultant vector in y remains unchanged.
 

ericgibbs

Joined Jan 29, 2010
16,361
hi C,
OK,
What I try to do in general is to get the TS's to visualize their technical problems before they haphazardly start throwing maths equations at the problem, hoping something may hit the mark.

This thread is a typical example of the TS's lack of visual understanding of this simple problem.

E
 
With an angle of 120 degrees to the flow, which is 30 degrees to the direction directly across the river you really don't need sines and cosines - it's half an equilateral triangle. If you remember the square root of three (which crops up reasonably often) you don't need a calculator or tables at all...
 
Top