Kinematics Part 1: Horizontal Motion
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- Опубликовано: 16 окт 2024
- 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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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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For the last word problem, I used d=(Vi+Vf/2)t to solve for the time. Then, I calculated the acceleration since I already have Vi, Vf and time. I arrived on the same answer which is a=-2.25 m/s^2. :)
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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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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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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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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
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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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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
I never noticed before, but the wheels are not turning. Very well explained thanks so much
My teacher couldn’t explain a thing! He just talks for himself, thank you for saving me from failing mechanical science!
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Dave: …some simple equations…
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In all seriousness though, thank you. I’m determined to wrap my head around all of this.
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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.
Listen by me clearly. Thaks.
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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.
beautiful explanation sir
Hi Sir I have been puzzled on how to solve this problem are you willing to help me?
Question 1. A 35 g bullet was fired horizontally from a gun with a velocity of 320.0 m/s. If the barrel is 1.1 meter from the ground, how long will it take for the bullet to reach the ground?
Question 2. A 35 g bullet was fired horizontally from a gun with a velocity of 315.0 m/s. If the barrel is 1.5 meter from the ground, how far will the bullet reach as it first hit the ground?
Thank you so much
Ready to see you help me Dave
PLEASE SOLVE THIS
A coin is pushed from the edge of a laboratory bench with a horizontal velocity of 15m/s of the height of the bench is from the floor is 1.5m. calculate the distance from the foot of the bench of the point of impact with the floor. (g = 10m/s2)
I ALSO NEED EXPLANATIONS PLEASE
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
I think it's because you put the exponent to the -8.4 instead of the m/s
the thing is, you put the whole expression (?) into the calculator except for the units, but the squared (^2) is supposed to just be for the m/s, but you left the exponent there after removing the m/s even though you're meant to also remove the exponent (^2) after removing the m/s itself, which in turn, squared the -8.4 instead.
@@lightbulb129 i didn't understand what you just wrote, pretend im a 5 yo child please
i figured it out, you are putting -8.2² and it's just -8.2
the ² on ms doesn't count because it's just there to represent acceleration
Who knew Jesus would also be saving my Science Test
Thanks for the help
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.
He knows lot about science and stuff professor Dave explains
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All you need to knows are the 3 ecuations (third one is actually derived from the other 2)
you slayyyeddd! Thank you! I finally understand
How did you get to (-225 m2/s2)/[2(50m)] Where do the numbers come from and how do you know to divide?
I was wondering if my answer is correct by accident. I got the same answer but I used other formulas. Vf=Vi+(a*t) and Xf=Xi+(Vi*t)+(0,5*a*t^2). I used the first formula to write a in terms of t so: 0=15+(a*t) -> a=-15/t substitute in second formula: 50=0+15*t+(0,5*-(15/t)*t^2) solve for t gives t=50/7,5=6,67 seconds. Substitute back in first formula: 0=15+(a*6,67) solve for a: a=-15/6,67=-2,25 m/s^2.
I finally understand what I'm learning :D
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
lemme try to do it step by step
vf^2 = vi^2 + 2ad
we know vf = 0 bcs its complete stop
vi = 15 m/s
and d= 50
so the question wanna know the acceleration. just substitute
vf^2 = vi^2 + 2ad
0= (15)^2 + 2a(50)
0= 225 +100a
let a be the subject
-225= 100a
-225/100 =a
a=-2.25 m/s^2
it should be negative bcs the bike is slowing down to stop. thats y
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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.