Kinematics Part 1: Horizontal Motion

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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Wow. I can't believe I learned more in this one video than my physics teacher has even tried explaining in the whole term. I wrote a few pages of notes from this and I finally understand instead of feeling overwhelmed and stupid. I really wish more teachers could explain subjects like this, it seems so much easier now.

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[@jdorsett4157]

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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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

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[@tanjeeschuan4999]

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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[@harunosakura9798]

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 !

[@juanorozco973]

Thanks so much i have a test tommorow and i finally understand this

[@palesasimelane4322]

I have been reading on this topic in my textbook.Did not understand anything.I understood this in the duration of the video.Thanks Prof

[@egregiouslytalented2381]

super bro.i am 11 years old trying to learn calculus and this kinematics.I HAVE BEEN SELECTED BY MY STATE GOVERNMENT FOR AN EDUCATION PROGRAMME FOR THE BLESSED,SO I HAVE TO LEARN EVERYTHING EARLY.U HAVE HELPED ME A LOT .THANKQ BRO.I CANT UNDERSTAND HOW TO THANK U

[@johnguillen68]

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.

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How did your class go? I'm in the same situation. Right now, after first class covering ch 1-2 my head is swimming

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@@beans7622 SAME I have a test tmr and I’m so confused

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Chidinma Ebere I ended up just cheating the entire semester and got pretty decent marks lmao. Good luck tho!!

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im a new physics student & i been struggling w grasping the whole concept. thank you so much for helping me with such vital yet easy information!

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[@liaselleaeri]

shouldn't 2.25 be the answer? i don't understand how it became negative. you transposed the 225 outside which is why it became negative. 2(50) was transposed too but the sign remained positive.. im so confused im gonna fail

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Dave: …some simple equations… Me: my brains on fire 😂 In all seriousness though, thank you. I’m determined to wrap my head around all of this.

[@isazisempi3896]

Ill never understand why in academia we skip so many steps in explanations. if you wrote down the formula x=x0+vi(t)+1/2a(t)^2 and then saw him use x=vi(t)+1/2a(t)^2 you would be confused until you realized they took out the xo lol. the question never gave a distance therefore I guess you can set it to zero. For somebody who never studied instantaneous velocity or physics they would wonder how the car had velocity an no displacement. failure in communication is the biggest setback in science and engineering.

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This actually helped a good bit and I have a test tomorrow. Thanks!

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[@o5-1-formerlycalvinlucien60]

for anyone wondering how the kinematic equations are derived, here is what i think is the solution (i'm not that good at calculus, but i think this gives the general idea.) starting with the basics, velocity is the derivative of position, or v = dx/dt and acceleration is the derivative of velocity, a = dv/dt solving for v just involves integration, dv/dt = a dv = a dt v = at + C1 (C1 i think here is the initial velocity) solving for position is just integration again, but using the previous equation dx/dt = at + v(i) dx = at + v(i) dt x = (at^2)/2 + vt + C2 (and C2 here initial position.) solving for the final equation: Δx = (v + v(i))/2 * t = d solve the first kinematic equation for time: v = at + v(i) v - v(i) = at (v - v(i))/a = t Δx = (v + v(i))/2 * (v - v(i))/a Δx = (v^2 - v(i)^2)/2a 2aΔx = v^2 - v(i)^2 v^2 = 2aΔx + v(i)^2 v^2 = 2ad + v(i)^2

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thank u so much i've been looking for these more complex formulas for an hour. glad u have them

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saw the intro and already clicked subscribe

[@meloncraft514]

It’s crazy how I can actually understand the equations in your videos since you’re good at explanations, but I still can’t figure out why you would 1/2at^2.

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Professor Dave, could you please explain about the motion of system of particles and rigid body.

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This was a really informative video thanks Professor Dave

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[@jorgeluisvilelavelarde5689]

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.

[@stefanandrioskarsson3092]

how did you get 43 at 5:20? i put 1/2(-8.4^2)(3.2)^2 in the calculator and i get -361.26

[@beans7622]

Thank you, very straightforward and easy to understand

[@elliejohnson2786]

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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wait why couldn't t= V-U/a

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There were more than one equation to solve distance , how do we know which one is the best to use when solving questions?

[@ProfessorDaveExplains]

You just have to see which variables you have defined. Choose the equation where you can plug everything in except distance so you can solve for it.

[@nickpetrowski7182]

How did you get to (-225 m2/s2)/[2(50m)] Where do the numbers come from and how do you know to divide?

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Hats off! How do you know so much 😨

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Thank you Prof Dave.. you helped me in physics..

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Thanks for explaining with this perfection

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[@horizonbrave1533]

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?

[@ProfessorDaveExplains]

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.

[@genesispolanco7479]

Im confused on the comprehension question. If it was +2a(50), why are we suddenly dividing it in the 3rd step?

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He knows lot about science and stuff professor Dave explains

[@simonvandenbroek6384]

I don't get one thing. you say 2.5 m/s squared * 10s. but we don't do anything with the squared?

[@Comrade4213]

Could explain to me where did you get these kinematics equations from.

[@user-zq3bq7jk3r]

Thank you professor dave

[@finlaymma]

at 5:06 you use vf = vi + at but there's another equation t=vf-vi/a which gives the same answer just wondering if they're the same and I can use them in the same scenarios, anyone know?

[@Jxrdan_07]

I used the same thing before I found the answer, I’m guessing you can use multiple things but yeah idk

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Thanks for the help

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I low key really like the music at the end of the video haha

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beautiful explanation sir

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I love physics

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no im not ready

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I know the feeling. I'm just gonna keep trying. If you put in the work you get the grade right?