Kinematics Part 1: Horizontal Motion 

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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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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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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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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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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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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

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.

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

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0² = 15² + 2(50)x x = -2.25

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

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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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.

for your question with bike, i got - > 2,25 acceleration. Correct?