AP Physics 1: Kinematics 25: River-Crossing Problems

[@8lindamae]

I was asked to check my son's homework, and I needed to refresh my memory on how to solve these problems. I watched quite a few videos. Yours was the best. I sent the link to my son. Thank you!!!!

[@JP_145]

Amazing video, for those who didn't understand some parts (been there) please take the time to understand it, she explained it really well! Thank you!

[@onlearningcurve]

You are welcome. And thanks for the comment : )

[@roflolsh]

WOW, You are the only teacher who MADE ACTUAL SENSE TO ME, THANK YOU for making this lesson!!!

[@janeaura-hf6lc]

❤

[@onlearningcurve]

That is right, only when the boat ends up directly opposite the starting point, we have Vboatx = Vriver . A general way to analyze the boat's x-direction motion is to combine the boat's own velocity's x-component and the river current's velocity to get the boat's Vx relative to the shore.

[@onlearningcurve]

Yes, you can use the complimentary angle and say that the angle is 48 degrees from the upstream direction. Or upstream at a 48 degree angle to the river bank. .

[@alirezakarimi4977]

Dear Yau-Jong TWu, Thanks a lot for your explicit explanation.I just didn't understand what heappened when you changed the angle and more important than that the direction of the swimmer velocity.COULDN'T WE SHOW IT IN INCLINED LINE BUT TOWARDS THE RIGHT SIDE??And why we should do this. I will be immensely thankful if u answer me

[@kellex98]

Thank you so much for this! I finally feel prepared for my physics test after watching these videos!

[@armani16898]

You summarized everything I didn't understand in just 10 minutes You are amazing

[@angelbravo9317]

I was absent when my professor taught this in class and was confused. However after watching this, I feel more confident. Thank you so much and good job :)

[@justblayne4533]

Thank you! This was a very helpful guide, and covered all of the questions typically asked during river crossing questions. I'm in grade 12 now, and I've always had problems with physics, but this guide was really helpful! Thanks :)

[@onlearningcurve]

+JustBlayne Thank you for the comment. I am glad that I can help : )

[@MeherrFirdouse]

Thank you so much!!! Finally I got my doubt cleared.

[@onlearningcurve]

And thank you for the comment : )

[@shru5151]

thankyou so much !! tomorrow is my physics exam and this helped a lot !!

[@paulproofmath323]

God bless you!

[@preetidarshita3089]

Thnk u so much ma'am...this really helped me a lot. Taking numerical values to explain something is easier than using variables

[@kavitaray9047]

Ma'am you are legend Thank you so so so so much

[@rajatbachhawat5155]

Superbly explained👍👍 before this i was having so much trouble to grasp this concept

[@physicsaayatatulpednekar1856]

Very nice explanation with good presentation.

[@gamingandgeekery]

Amazing video! I can't believe that I can understand it now!

[@bryandeleon-vargas4535]

Wow! This is amazing and extremely helpful!! Thanks for the help!

[@shahbaazali1195]

Thank you so much ma'am!!

[@ariannaferretti6889]

Can you only say Vbx=Vr if the boat ends up directly opposite on the shore but there is a different Vx for the boat if the angle makes it end up somewhere else?

[@onlearningcurve]

Yes, only if Vbx = - Vr, meaning the boat's velocity's x component equals to the magnitude of river current's velocity but in the opposite direction, the boat will end up directly opposite the starting point.

[@onlearningcurve]

You are very welcome : )

[@christinal7041]

谢谢吴老师

[@sophialin2865]

this is a really good, well-explained video :D thanks!

[@abhayabhi5351]

wow are u kidding me..........all my doubts were cleared with this

[@mrsamlim]

hi if the boat aims at the stream of current to go across. Will it be faster

[@onlearningcurve]

Hi, If the boat aims at the stream of current to go across. It will move downstream faster. However, it won't be able to get to the other side because the boat has to have velocity component perpendicular to the current to move towards the other side.

[@mrsamlim]

@@onlearningcurve thank you

[@ariannaferretti6889]

Can you say the angle is 48 degrees from the upstream direction? complimentary angles

[@onlearningcurve]

Yes, at an 42 degrees upstream angle is the same as 48 degrees from the upstream direction.

[@onlearningcurve]

Thanks!

[@viveksajjan9681]

awesome teaching

[@marleneromo6617]

Thank you so much, u are such a great teacher. Yet I do have a question, how do u know where to put the angle?

[@onlearningcurve]

I placed the angle theta there because I wanted to able to specify it as an upstream angle - please see 6:55 to 7:22.

[@dannypipewrench533]

Thank you.

[@gabrielbitti2114]

The first boat is travelling at a longer distance than 600metres though? it is travelling at the distance of the hypotenuse, so why would the time taken for the boat to cross the river still be 600/3?

[@onlearningcurve]

You are right that the actual distance traveled by the first boat relative to the shore is not 600 meters - it is > 600 meters. It travels (square root of (600^2 + 400^2)) = 721m. The boats actual speed relative to the shore is not 3m/s either - it is square root of (3^2 + 2^2) = 3.6m/s. So to find the time using t = 721/(3.6) = 200 seconds. We still get the same time. It is just that it's easier and more straight forward to use the y-component only to find the time.

[@gabrielbitti2114]

thank you i understand now this helps a lot

[@onlearningcurve]

Gabriel Bitti You are welcome : )

[@chiragsaxena7251]

Thanks soo much great explanation!!!

[@anishabodapati6544]

You are the best🙏

[@Thesoccerdood]

Thanks so much!! You're the best

[@kopakanuva]

Very helpful, thank you!

[@xintc5703]

great vid Great explanation :)

[@allentmonachan]

Superb

[@gunal4353]

can u put the video for all neet physics prolem

[@devilzone756]

the angle should be cos^-1(-2/3) right!!!!??

[@onlearningcurve]

It's inverse sine because 2 is the opposite side and 3 is the hypotenuse for the angle I drew in the video..

[@anush7386]

Thnk you mam

[@THELaserApple]

THANK YOU SO MUCH M'LADY

[@rushikeshwadile7007]

Amazing

[@anush7386]

Aweosme teacher

[@onlearningcurve]

Thank you!

[@ramikbal5195]

relve.cm

[@ShreeRamJayaRamJayaJayaRam]

u made it easy for me !!!!

[@sambhav1808]

thanks mam

[@AoniumZ]

Very Helpful!!

[@EshwarChoudhary]

thanks !

[@suman9575]

thanks a lot ! it really helped!

[@005vinay3]

awsm madam...:D!t.q so much

[@KolluruSivanageswararao]

nice - thanks

[@navk7920]

Thanks:)

[@rumahin9702]

awesome