Kinematics Part 1: Horizontal Motion
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- čas přidán 25. 01. 2017
- Alright, it's time to learn how mathematical equations govern the motion of all objects! Kinematics, that's the name of the game! Ready? Yes you are. Come on, it'll be fun. We will start simple, horizontal motion only. The math is really easy, I promise. You're being a big baby, just watch this!
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thank you SOOO much for making this. I've been struggling with kinematics really badly in class, and this explained it in such an easy and understandable way
Our physics teacher shows your videos during class but I always have something to do thats why Im distracted every class but when i go home i watch ur videos again and I understand the lesson. Thank you!!
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I love how he talks about the history of physics which shows there’s many discoveries/discrepancies in modern science physics at that point of time.
ik this is pretty late but what i did is I rearranged the equation to get "a" (acceleration) by itself. Which is (vf^2 - vi^2)/2d. All i did from here was plug in the values which we know are [vf=0] cuz it skids to a stop, [vi=15] cuz that's our initial velocity before the skid, [d=50] cuz that's the total displacement traveled during the skid, then with a trusty calculator it comes out to -2.25m/s^2. If you're confused on why it's negative it's because it's going in the opposite direction of the velocity, thus slowing the bike down until a full stop! 🛑
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There's also a method where you plot a graph with the x axis being time and y axis being velocity, the area of the polygon in the graph is the displacement of the time, velocity and acceleration of the problem.
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This video just helped me catch up in my Mechanics class. I've been so lost until now. Thanks!
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Thank you for showing the working at the end! I followed along and just... Didn't really know how to get the correct units, I wasn't paying attention to how the squared velocity applies the squared aspect to the units as well.
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Professor Dave, could you please explain about the motion of system of particles and rigid body.
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Sorry i have a question in 4:03 pls someone explains me, im not good at english for sure so forgive me 'bout my grammar. Now i want to ask. I thought that the acceleration was 2.5m/s^2 so each second the point would probably move a distance equal to its instantaneous velocity. Let take an example: in the first second, the velocity became from 0 to 2.5m/s because of the acceleration so the distance which the point had ride must be 2.5m. The next second ( about 1s ) the velocity continiously changed its instantaneous velocity from 2.5m/s to 5m/s so propably the distance the point had ride were also equal to its instantaneous velocity at that time and its equal to 5m. So all of the distance which the point had ride = 2,5 + 5 = 7,5m. But if we use the principle in this video, the distance will be: 1/2at^2 =1/2 x 2,5 × 2^2 = 6,25m. So if we follow this principle, the point will ride a distance = 6,25m, but like i've talked about, the distance must be 7,5m.
I definitely thinking about it and i dont know what exactly the true question, so i ask and hope that somebody will answer me. Thank all of you, first.
Oh, oh so im talking about uniform variable motion so the displacement probably = the distance the point had ride in Ox
Tks for reading, and pls help my by answering this ques. Have a good day !
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Hello, I have this question. I have data of horses. vertical acceleration. If these values are registered by ms and I have -0.4, 1.2, 3.1 and -0.08 m/s2. How can I get a measure of each horse? may I take just the average? or Should I do another calculation? Thanks a lot for your help.
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Could explain to me where did you get these kinematics equations from.
saw the intro and already clicked subscribe
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Excellent
My teacher couldn’t explain a thing! He just talks for himself, thank you for saving me from failing mechanical science!
My question is after you are done with these and rounding to appropriate sig figs, do you use the 2 used as part of the formula or only the measurements in the equation to calculate how many sig figs should be in your answer? My teacher is confusing me on that part.
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Hey Proffy Dave, I have a question...When I am getting more into physics I see a lot of talk about "Deriving an equation"...or something like "Derive the kinematic equations" etc... Is this the same meaning as literally taking the derivative of the equation? Or does "deriving" in this sense mean something else? From what i see...when someone says they "Derive an equation" it doesn't seem like the same idea as taking the derivative of a function or equation? How are they releated/different?
Oh no no, it has nothing to do with derivatives! It just means to formulate or generate the equation, like someone had to first develop the equation, and we can use existing equations and plug things in and manipulate them algebraically to get new equations, so that's what that means.
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I never noticed before, but the wheels are not turning. Very well explained thanks so much
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How did you get to (-225 m2/s2)/[2(50m)] Where do the numbers come from and how do you know to divide?
best work
wow, this was a great video. So basically we have to find out first what we have and what we need to find out in order to know which equation we have to use? The first day of Physics was so confusing and now that we're on the 3 chapter (Freefalling) we have an exam on the first 3 chapters but I still feel somewhat lost. I'm gonna have to watch videos and apply it to what I'm reading to get a better understanding.
How did your class go? I'm in the same situation. Right now, after first class covering ch 1-2 my head is swimming
@@everardomezajr3495 aye yo im in the same situation rn 😭
@@beans7622 SAME I have a test tmr and I’m so confused
Chidinma Ebere I ended up just cheating the entire semester and got pretty decent marks lmao. Good luck tho!!
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