AP Physics 1: Kinematics 25: River-Crossing Problems

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  • เผยแพร่เมื่อ 14 ต.ค. 2024
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ความคิดเห็น • 68

  • @8lindamae
    @8lindamae 10 หลายเดือนก่อน +3

    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
    @JP_145 ปีที่แล้ว +3

    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!

  • @kellex98
    @kellex98 10 ปีที่แล้ว +6

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

  • @onlearningcurve
    @onlearningcurve  9 ปีที่แล้ว +1

    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.

  • @roflolsh
    @roflolsh 12 ปีที่แล้ว +2

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

  • @onlearningcurve
    @onlearningcurve  9 ปีที่แล้ว +3

    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
      @alirezakarimi4977 5 ปีที่แล้ว

      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

  • @onlearningcurve
    @onlearningcurve  11 ปีที่แล้ว +1

    You are welcome. And thanks for the comment : )

  • @justblayne4533
    @justblayne4533 9 ปีที่แล้ว +3

    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
      @onlearningcurve  9 ปีที่แล้ว +1

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

  • @armani16898
    @armani16898 8 ปีที่แล้ว +1

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

  • @paulproofmath323
    @paulproofmath323 4 ปีที่แล้ว +1

    God bless you!

  • @christinal7041
    @christinal7041 5 ปีที่แล้ว +1

    谢谢吴老师

  • @MeherrFirdouse
    @MeherrFirdouse 9 ปีที่แล้ว +4

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

  • @angelbravo9317
    @angelbravo9317 7 ปีที่แล้ว

    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 :)

  • @dannypipewrench533
    @dannypipewrench533 ปีที่แล้ว +1

    Thank you.

  • @preetidarshita3089
    @preetidarshita3089 7 ปีที่แล้ว

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

  • @marleneromo6617
    @marleneromo6617 5 ปีที่แล้ว +1

    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
      @onlearningcurve  5 ปีที่แล้ว +1

      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.

  • @rajatbachhawat5155
    @rajatbachhawat5155 7 ปีที่แล้ว +4

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

  • @physicsaayatatulpednekar1856
    @physicsaayatatulpednekar1856 9 ปีที่แล้ว +1

    Very nice explanation with good presentation.

  • @kavitaray9047
    @kavitaray9047 6 ปีที่แล้ว +1

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

  • @allentmonachan
    @allentmonachan 5 ปีที่แล้ว +1

    Superb

  • @rushikeshwadile7007
    @rushikeshwadile7007 3 ปีที่แล้ว +1

    Amazing

  • @onlearningcurve
    @onlearningcurve  12 ปีที่แล้ว +2

    Thanks!

  • @gamingandgeekery
    @gamingandgeekery 6 ปีที่แล้ว

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

  • @sophialin2865
    @sophialin2865 10 ปีที่แล้ว +2

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

  • @shru5151
    @shru5151 10 ปีที่แล้ว +2

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

  • @onlearningcurve
    @onlearningcurve  12 ปีที่แล้ว +1

    And thank you for the comment : )

  • @onlearningcurve
    @onlearningcurve  12 ปีที่แล้ว +1

    You are very welcome : )

  • @shahbaazali1195
    @shahbaazali1195 9 ปีที่แล้ว +2

    Thank you so much ma'am!!

  • @viveksajjan9681
    @viveksajjan9681 7 ปีที่แล้ว +1

    awesome teaching

  • @chiragsaxena7251
    @chiragsaxena7251 7 ปีที่แล้ว

    Thanks soo much great explanation!!!

  • @gabrielbitti2114
    @gabrielbitti2114 9 ปีที่แล้ว +1

    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
      @onlearningcurve  9 ปีที่แล้ว +3

      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
      @gabrielbitti2114 9 ปีที่แล้ว +1

      thank you i understand now this helps a lot

    • @onlearningcurve
      @onlearningcurve  9 ปีที่แล้ว

      Gabriel Bitti You are welcome : )

  • @bryandeleon-vargas4535
    @bryandeleon-vargas4535 12 ปีที่แล้ว

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

  • @TheLimwuxuan2
    @TheLimwuxuan2 4 ปีที่แล้ว +1

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

    • @onlearningcurve
      @onlearningcurve  4 ปีที่แล้ว +2

      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.

    • @TheLimwuxuan2
      @TheLimwuxuan2 4 ปีที่แล้ว +1

      @@onlearningcurve thank you

  • @abhayabhi5351
    @abhayabhi5351 7 ปีที่แล้ว +6

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

  • @kopakanuva
    @kopakanuva 8 ปีที่แล้ว +1

    Very helpful, thank you!

  • @ariannaferretti6889
    @ariannaferretti6889 9 ปีที่แล้ว

    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
      @onlearningcurve  6 ปีที่แล้ว

      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.

  • @sambhav1808
    @sambhav1808 6 ปีที่แล้ว +1

    thanks mam

  • @Thesoccerdood
    @Thesoccerdood 10 ปีที่แล้ว +2

    Thanks so much!! You're the best

  • @ariannaferretti6889
    @ariannaferretti6889 9 ปีที่แล้ว

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

    • @onlearningcurve
      @onlearningcurve  6 ปีที่แล้ว

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

  • @anishabodapati6544
    @anishabodapati6544 7 ปีที่แล้ว

    You are the best🙏

  • @xintc5703
    @xintc5703 7 ปีที่แล้ว

    great vid Great explanation :)

  • @anush7386
    @anush7386 8 ปีที่แล้ว +2

    Thnk you mam

  • @AoniumZ
    @AoniumZ 11 ปีที่แล้ว

    Very Helpful!!

  • @THELaserApple
    @THELaserApple 6 ปีที่แล้ว

    THANK YOU SO MUCH M'LADY

  • @gunal4353
    @gunal4353 5 ปีที่แล้ว

    can u put the video for all neet physics prolem

  • @ShreeRamJayaRamJayaJayaRam
    @ShreeRamJayaRamJayaJayaRam 11 ปีที่แล้ว

    u made it easy for me !!!!

  • @EshwarChoudhary
    @EshwarChoudhary 9 ปีที่แล้ว +1

    thanks !

  • @KolluruSivanageswararao
    @KolluruSivanageswararao 11 ปีที่แล้ว

    nice - thanks

  • @devilzone756
    @devilzone756 8 ปีที่แล้ว

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

    • @onlearningcurve
      @onlearningcurve  8 ปีที่แล้ว

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

  • @anush7386
    @anush7386 8 ปีที่แล้ว +6

    Aweosme teacher

  • @navk7920
    @navk7920 8 ปีที่แล้ว +1

    Thanks:)

  • @rumahin9702
    @rumahin9702 7 ปีที่แล้ว

    awesome

  • @suman9575
    @suman9575 8 ปีที่แล้ว

    thanks a lot ! it really helped!

  • @005vinay3
    @005vinay3 8 ปีที่แล้ว +1

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