An Introductory Relative Motion Problem

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  • Опубліковано 21 жов 2024

КОМЕНТАРІ • 26

  • @FlippingPhysics
    @FlippingPhysics  10 років тому +9

    The best way to understand relative motion is to be able to visualize it. Therefore I used a large piece of paper and a slow moving toy car to create a simple, introductory vector addition problem that we can actually see. #PhysicsED #flipclass

  • @ahmed99094
    @ahmed99094 9 років тому +13

    That's lots of work you're putting in. I really appreciate that, Kudos! and nice socks!

    • @FlippingPhysics
      @FlippingPhysics  9 років тому +4

      +Ahmed Ali It is all about the socks really.

  • @Ukkaxah
    @Ukkaxah 8 років тому +4

    good style of teaching..

  • @samisiddiqi5411
    @samisiddiqi5411 3 роки тому

    These are genuinely fun to watch, and spare no detail.

  • @DisiCoco-nm2gw
    @DisiCoco-nm2gw 3 місяці тому +1

    YES I UNDESTAND finally what the velocity of the respect to the paper is 😢❤

  • @Adumbb
    @Adumbb 5 місяців тому +1

    how do you know if it’s 58 degrees west of north or 32degrees north of west? how do we know the direction is counterclockwise?

    • @FlippingPhysics
      @FlippingPhysics  5 місяців тому

      Please watch this: www.flippingphysics.com/cardinal-directions.html

  • @sjaoenvf
    @sjaoenvf 4 роки тому +2

    6:43 question is in cm but diagram is in mm. (awesome videos BTW. I sometimes forgot billy, bobby and bo are you)

    • @FlippingPhysics
      @FlippingPhysics  4 роки тому +1

      Yeah. Wish UA-cam would let us upload fixed versions of videos...

  • @arafe-zawad-sajid
    @arafe-zawad-sajid 7 років тому +2

    Hi. Can you please tell me why we used the Pythagorean theorem to solve for the velocity of the car with respect to the Earth(Vce) @3:30 instead of just doing Vce=Vcp+Vpe

    • @FlippingPhysics
      @FlippingPhysics  7 років тому +1

      Vce=Vcp+Vpe would be correct if velocity were a scalar. Because velocity is a vector, you need to do vector addition: www.flippingphysics.com/tip-to-tail-vector-addition.html

    • @arafe-zawad-sajid
      @arafe-zawad-sajid 7 років тому

      thank you :D

  • @sundaram58
    @sundaram58 2 роки тому

    Thank you so much

  • @juliojuarez2953
    @juliojuarez2953 8 років тому +2

    i didn't understand how you turned velocities into the length of x
    cos=a/h when adjacent was given. We know the angles of the velocity triangle but i thought they had nothing in common with the displacement triangle

    • @FlippingPhysics
      @FlippingPhysics  8 років тому

      The triangle with velocities on all three sides and the triangle with displacements on all three sides are _similar triangles_.

    • @juliojuarez2953
      @juliojuarez2953 8 років тому

      ok so would cos(58)= (692/h) [since the got the same angles]
      be correct?
      i get 1305.

    • @FlippingPhysics
      @FlippingPhysics  8 років тому +2

      cosθ=A/H => H=A/cosθ => H=692/cos(58.478) = 1323.57 ≈ 1300 mm
      Yes, it's the same answer.
      Please don't round in the middle of a problem.

  • @VeerSingh-fu9ln
    @VeerSingh-fu9ln 4 роки тому

    Sir can you please help out with optics.

  • @anshur3021
    @anshur3021 5 років тому +2

    I can't believe that you have only this much views

  • @helifynoe6956
    @helifynoe6956 2 роки тому

    I prefer motion relative to space-time. Looking at motion in this way, you can use simple geometry to construct some fascinating mathematical representations of that which is the outcome of this 4D motion that takes place within the 4D space-time environment. Then, once having done that, you end up being completely surprised that you came up with the same results as did Einstein, even though you knew nothing about physics at all.

  • @kimsahl8555
    @kimsahl8555 3 роки тому

    Rest and motion is't about relativity (or absolute), it is just "rest" and "motion". Simple.

  • @thedmitryguy
    @thedmitryguy 4 роки тому

    Is there anybody here from Foxford?