Physics 2 - Motion In One-Dimension (3 of 22) Graphing Position

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  • Опубліковано 3 січ 2025

КОМЕНТАРІ • 95

  • @Lostwolf16
    @Lostwolf16 5 років тому +26

    Thank you sir. So clear and easy but looks so hard in class. I would have dropped out of college if there weren't kind smart people like you uploading knowledge for free for world to learn

  • @nycityzen
    @nycityzen 8 років тому +20

    Thanks for all your work. It's greatly appreciated.

  • @ongcunon1982
    @ongcunon1982 11 років тому +4

    best physic video I have ever seen! Thank you Dr!

  • @jazminedeocares6377
    @jazminedeocares6377 4 роки тому +6

    Thank you sir, this is really helping me through physics 6A!

  • @bernsbuenaobra473
    @bernsbuenaobra473 4 роки тому +1

    Years back when I was teaching undergraduate physics one the first few meetings in the class are actually to have some coaching the are how to take notes in class, then how to study and solve physics problems sort of build some confidence, outlook, and attitude to the difficult subject matter. Many high school graduates come out of the school dropping everything they learned or had classes on and we have to assume its ground zero again. Educational research in the Philippines shows that Filipino students are graphical thinkers so graphing and plotting and visualization is really an important content in-class delivery of not so popular classroom classes among freshmen.

  • @luacaballero4762
    @luacaballero4762 4 роки тому

    Thanh you so much! In this state of quarantine i have terrible virtual lessons and you are my only help. From ARGENTINA 🇦🇷 I appreciate your work!

    • @MichelvanBiezen
      @MichelvanBiezen  4 роки тому +3

      Hi Lua, Hang it there, we'll get through this. Welcome to the channel!

  • @elmore1026
    @elmore1026 2 роки тому +2

    You are a life saver

  • @PawnUp
    @PawnUp 3 роки тому +3

    Thank you sir your are the best

  • @bernsbuenaobra473
    @bernsbuenaobra473 4 роки тому

    For consistency we could rewrite for the second segment x_sub2 to have x_sub1 in and v_sub1 while following your 3rd segment notation x_sub3 that is written to be with x_sub2 and v_sub2 components in the equation

  • @1995sanchezbaybee
    @1995sanchezbaybee 8 років тому +1

    I love the Bow Tie!
    Thank you for the video

  • @MoneyBaank
    @MoneyBaank 7 місяців тому +1

    Your explanations are so very detailed and understandable. Thank You so much for this, I really appreciate it.
    Concerning the acceleration in finding v subnot for the second segment = final v for first segment, you put acceleration as zero whereas it is 0.5m/s in the first segment. Please why is this so?

    • @MichelvanBiezen
      @MichelvanBiezen  7 місяців тому +1

      The train is only accelerating in the first section, holding the same speed in the second section, and decelerating in the third section.

  • @osmanyalvac3004
    @osmanyalvac3004 9 років тому +4

    ı am so grateful sir , thank you for all the videos:D

  • @kaiadler4916
    @kaiadler4916 Рік тому +2

    Sir, im in 9th grade and didnt understand anything explainined in the video. We are being taught the same thing in class and i dont understand anything. Please help.

    • @MichelvanBiezen
      @MichelvanBiezen  Рік тому +1

      It is difficult to respond to your request since you stated that you don't understand "anything", so that we don't have a basis to start from. It is better to ask specific questions about a specific singular concept or equation.

  • @sutieng8932
    @sutieng8932 7 років тому +2

    thank you very much sir really appreciate your work it's so helpful 💚

  • @kunheepark2093
    @kunheepark2093 7 років тому +1

    Thank you very much, professor!

  • @mitziruth_1316
    @mitziruth_1316 9 років тому

    Thank you for making these videos!! They're the best! :)

  • @robertonoz4142
    @robertonoz4142 4 роки тому +1

    Excellent kinematics video. I have one question though. Why is it when Kinematics is taught, there are no vector arrows drawn over the variables in the 3 equations? Is this because in 1D motion there is only a + and - direction? Thank you!

    • @MichelvanBiezen
      @MichelvanBiezen  4 роки тому +3

      It can be done both ways, as long as one keeps track of the directions.

  • @AR_BSME
    @AR_BSME 7 років тому +2

    Ive read the comments... and I am wondering why for X of 2 (240 seconds) was not squared ? when it is for X1 and X2 ?? I have rewinded but cannot understand ??

    • @MichelvanBiezen
      @MichelvanBiezen  7 років тому +7

      During the time period 2, there is no acceleration. Therefore distance traveled (x2) is simply = v t

  • @AliCool412
    @AliCool412 2 роки тому +3

    The Amount of Subscribers You Deserve is The Same as Your Current Amount , Just Replace "Thousand" With "Million" .

    • @MichelvanBiezen
      @MichelvanBiezen  2 роки тому +2

      Thank you. We appreciate the value you place on our videos. Much appreciated.

  • @ahmedal-ebrashy3691
    @ahmedal-ebrashy3691 6 років тому +2

    Excuse me SIr, But why is the velocity 30m/s in the final segment of the graph?

    • @MichelvanBiezen
      @MichelvanBiezen  6 років тому +3

      That is the velocity at the beginning of the last segment, before it slows down to 0 m/sec

    • @ahmedal-ebrashy3691
      @ahmedal-ebrashy3691 6 років тому

      Marvelous work thank you very much.

  • @RebellionOG
    @RebellionOG 7 років тому +1

    thanks for the efforts!

  • @ZRShay
    @ZRShay 10 років тому

    Clear and very useful

  • @frankcarey250
    @frankcarey250 5 років тому +1

    Sir please clear my doubt. To get distance traveled we calculate the area under Velocity Time curve, which we usually do considering curve above X axis. But we could have compute the area under that curve with respect to Y axis. But why don't we do so? Please reply.

    • @MichelvanBiezen
      @MichelvanBiezen  5 років тому +2

      If we have a negative velocity, the velocity curve will be below the horizontal axis and there will be negative area representing distance traveled in the opposite direction.

  • @MariahBaillie1
    @MariahBaillie1 8 років тому

    why is the initial X 900m instead of 300m? @7:07

    • @MichelvanBiezen
      @MichelvanBiezen  8 років тому +1

      The problem is divided into 3 parts. 1) The acceleration part 2) the constant velocity part 3) The deceleration part. In the second part, the initial distance for the second part is the distance covered during the first part which is 900 m.

  • @ahmadhusayn539
    @ahmadhusayn539 8 років тому +1

    hello sir i have a question, how do you when to use which kinematic equation?

    • @MichelvanBiezen
      @MichelvanBiezen  8 років тому +2

      They can always be used as long as the acceleration is constant. (They don't work if the acceleration is not constant).

    • @ahmadhusayn539
      @ahmadhusayn539 8 років тому

      Michel van Biezen ok and how do we know which problems to use the kinematic equations on?

    • @MichelvanBiezen
      @MichelvanBiezen  8 років тому +1

      Every motion problem uses the equations of kinematics. (Or are you asking when to use the graphical method and when just to use the equations?) If so, it is a personal preference. You can use both methods for every problem.

    • @usery_user
      @usery_user 3 роки тому +1

      You see what is missing from your equation. If for ex you have time missing you use the time independent equation

  • @shadowkxm
    @shadowkxm Рік тому +1

    I noticed we couldnt use the formula: s = ((v+v0)/2)t - why is that?

    • @MichelvanBiezen
      @MichelvanBiezen  Рік тому +1

      That equation is not one of the three equations of kinematics. If there is acceleration then s = (1/2) a t^2

  • @advikdutta
    @advikdutta 3 роки тому +1

    So X2 = X1 + V1 t + 0.5 at squared

    • @MichelvanBiezen
      @MichelvanBiezen  3 роки тому +1

      X2 = X1 + v t X3 = X2 + (1/2) at^2 (for the last part)

  • @moezzatehseen5831
    @moezzatehseen5831 6 років тому +2

    @7.07,,why can't we find the velocity by finding the slope from the graph,instead of using 1st equation?

    • @MichelvanBiezen
      @MichelvanBiezen  6 років тому +2

      You can. There are often multiple ways to find the same solution.

    • @cihangonen9746
      @cihangonen9746 6 років тому +1

      there are many ways to find velocity

  • @jeodemp
    @jeodemp 9 років тому

    At X3 8:47, before subway moved 900(x1) + 8100 (x2).However, you just added 8100 why ? x0=9000 at there Isn't it ?

    • @MichelvanBiezen
      @MichelvanBiezen  9 років тому

      +CloneComCody
      The 8100 m for x2 already includes the 900 m from x1.
      x1, x2, and x3 are all relative to the origin.

  • @ajinkyamaurya6481
    @ajinkyamaurya6481 8 років тому

    a body dropped from a tower travels half of the total distance in the last second of its motion. The total time of fall will be?

    • @MichelvanBiezen
      @MichelvanBiezen  8 років тому +1

      There are a number of ways to do this, but this may be the simplest: the distance an object falls is proportional to t^2 Thus if you want half the distance, when is t^2 = 1/2? When t = sqrt(2)/2 = 0.707 Thus a falling object spends 70.7% of the time to travel the first half and 29.3% of the time to travel the second half. Total time = 1/0.293 = 3.41 sec.

  • @kemiomodele6708
    @kemiomodele6708 9 років тому +1

    hi , why wasn't the time in x2 squared ?

    • @MichelvanBiezen
      @MichelvanBiezen  9 років тому

      +Kemi Omodele
      x2 represents the distance traveled by the train at constant speed. Therefore there is no acceleration and thus the third term goes to zero (a = 0)

  • @physics67
    @physics67 2 роки тому +1

    Thank you so much.

  • @ajinkyamaurya6481
    @ajinkyamaurya6481 8 років тому

    sir i have a question .
    from top of tower of height 20m, a boy drops a stone, after one second he throws another stone vertically downward so that both hit the ground together. The initial velocity of the second stone is?

    • @MichelvanBiezen
      @MichelvanBiezen  8 років тому +1

      You first figure out how long it takes the first stone to drop to the ground. (2.02 sec). Then you use that information to find the initial velocity of the second stone: Y = Yo + V0y t + (1/2) g t^2

    • @ajinkyamaurya6481
      @ajinkyamaurya6481 8 років тому

      thank u sir
      can i ask u more questions

    • @MichelvanBiezen
      @MichelvanBiezen  8 років тому

      When I have time (I am very busy).

    • @Borz091
      @Borz091 7 років тому

      Michel van Biezen check my calculations, but wouldn't the initial velocity then be 14.6 m/sec?

  • @marals1061
    @marals1061 7 років тому +1

    Thank you

  • @snethembamsomi9390
    @snethembamsomi9390 8 років тому

    Do you have like videos which go into the harder aspects of this topic because I'm kind of looking for that before my test. The simple parts I get, it's the non- constant acceleration that troubles me

    • @MichelvanBiezen
      @MichelvanBiezen  8 років тому

      Did you check out the playlists on physics? (There are over 70 of them with close to 2000 videos). They are all in order.

  • @ernestaskriauciunas3790
    @ernestaskriauciunas3790 9 років тому

    Hi Michael, I wanted to ask you why we have sec, sec^2, sec^3 etc.. same with many other physics quantities where something is squared or cubed what does it mean ? Thanks

    • @bernsbuenaobra473
      @bernsbuenaobra473 4 роки тому

      If a variable is squared mathematically the order of that variable is raised to 2 also an exponent of zero on a variable is to mean it raised to 1. A second-order variable say in a function if you plot them will result to a curve and not a line. Thus the component 1/2at^2 will result in a curve as our dear teacher has shown.

  • @jeroincababat565
    @jeroincababat565 3 місяці тому

    I understand your lecture than my professor

  • @darksin2509
    @darksin2509 5 років тому

    i have a doubt in my head. in the second segment, to find x2 , a=0 so why time is included?? If a=0 , then the whole 1/2 at^2 should be 0 right ? Anyone clear this up.

    • @MichelvanBiezen
      @MichelvanBiezen  5 років тому

      There are 3 terms in the equation (not just the (1/2) a t^2 term)

  • @yis.grabie1424
    @yis.grabie1424 9 років тому

    in finding X2, why did you use a different formula than x1

    • @MichelvanBiezen
      @MichelvanBiezen  9 років тому

      Yisroel Yaakov Grabie
      I used the same equation for both

  • @ahmadbelhaj1756
    @ahmadbelhaj1756 8 років тому

    for X2 = X1 + Vt + (the time at which the speed was costant starts from 60s to 240s. which is 180s. if not, explain why?

    • @MichelvanBiezen
      @MichelvanBiezen  8 років тому

      Based on the description of the problem, the DURATION of the period during which the velocity remains constant is 240 sec (from 60 sec to 300 sec).

  • @curtpiazza1688
    @curtpiazza1688 8 місяців тому +1

    This is GREAT! Thanx! 😂

  • @amanullahzahid9303
    @amanullahzahid9303 4 роки тому

    This Gentleman is very kind, because nowadays people even don't fart without fucking money

  • @surendrakverma555
    @surendrakverma555 3 роки тому

    Very good 🙏🙏🙏🙏

  • @snethembamsomi9390
    @snethembamsomi9390 8 років тому

    The acceleration is not constant at X1 and X3 so how come you using that equation

    • @MichelvanBiezen
      @MichelvanBiezen  8 років тому +1

      That is a good observation. The acceleration is constant over the entire period the train is accelerating, except at the very "end" points of the acceleration. We ignore those for the sake of simplicity as is done in all of lower division physics. If we don't these problems would become much harder as you will see in upper division physics or graduate level physics. Thus in this context it is fine to use that equation. Remain inquisitive, that is great.

  • @damien4836
    @damien4836 4 роки тому

    What is x?

    • @MichelvanBiezen
      @MichelvanBiezen  4 роки тому +1

      x represents displacement (distance traveled)

    • @damien4836
      @damien4836 4 роки тому

      @@MichelvanBiezen thank you so much!

  • @HoomanxD
    @HoomanxD 10 років тому

    great one

  • @buildingphase9712
    @buildingphase9712 10 років тому +1

    Love the Outfits

  • @michaelmoleele1754
    @michaelmoleele1754 8 років тому

    very helpful
    thanks hey

  • @austinchang1529
    @austinchang1529 10 років тому

    great video