U is the initial velocity, often zero, thank you. V is the final velocity and acceleration, and t is time, so those are fine. So again, these are five different equations. I think you're given three or four of them in the back of your booklet. So if we don't know any displacement, we can use this equation if we haven't any information about acceleration. And if we don't know time, we can use this equation if we don't know final velocity and if we don't know initial velocity. So again, depending on the information that we're given, we can choose which is the appropriate equation to use. I'll just get up a paper. And we have these preloads in any ways. So we should come up, we'll remind ourselves which of these. I just want the formula sheet. Okay, so s is V U plus V I over two times. T V is U plus at s is equal to U two plus a half at squared and v squared is U squared plus a half as so we're given four of them. Okay. So if you know three of these using the equations, you can find your unknown quantity. For example, a car traveling 10m per second accelerates 1m per second for 5s. What is its final speed? So v is what we want to find. U is 10m per second a. Is given one and time v is equal to U plus at. So our final velocity for the car accelerating for 5s would be 15m per second. How far does it travel during this time? So s is equal to U T plus a half at squared. So we're using a different equation now to find a distance traveled while it's accelerating. So U is ten, t is five, a half times acceleration times t squared. So we get distance. So v is equal to U plus at. So the final velocity is this initial velocity plus this radiant. The displacement is found by finding the area under the graph line. So it's ut that's your rectangular blue bit plus a half at squared. So U times t and then it's a half v minus U times t. So that's your acceleration. So it's U T plus a half at squared. If we integrate U is equal to U plus at with respect to time, we get displacement. So s is equal to U T plus one half at squared. So a is v minus U over t so instead of a, we can substitute there and we get s is equal to U T plus a half vt minus a half U T, which s is equal to v plus U over two times t. So deriving this version of the equation s is equal to a half U plus v times v. So rearranging for T V minus U over a so two, as is v squared minus U squared. So v plus U, V minus U is v squared minus U squared. So v squared minus U squared is two as. So each equation can be derived from another one. S is equal to U T plus one half A T squared, so we substitute U is v minus at. V minus A T. S is V T minus a half at t squared. So. Let's try this question. A child on a skateboard is traveling at 6m per second. So that's our you. He ignites a rockus which burns for 10s, giving a final velocity of 13m per second. What acceleration did the. Did he achieve? So which equation do you think would be most appropriate to use here? Saleration equal to b minus U divided by t. Yeah, so a equals 13 minus six over ten. What acceleration did you get? 0.7, no point seven. Yeah an insane father allowed his teenage daughter to ride a jet powered bike. Starting from rest, it accelerates at 2m per second. How long did it take to reach 12m per second? How far did it go in this time? So time and distance is what we want to find. Starting from rest. So we know you initial velocity is zero. Yes. Acceleration is two and the final velocity is twelve, so the delta velocity is is two and accelaries is two, so time is 6s. Yes. And then how far did it go? Of. As able v plus U divided by two times t so as equal to twelve six plus six times 6:30 36. Yeah good. A man is walking along, minding his own business. At 1.5m per second, a fairly benign terrier chases him. A dog, the man accelerates at 0.5m per second, squared for 30m, until the dog loses interest. What is his new velocity and how long did this embarrassing incident take place for? Salaries at 0.5m second square, or 30m. So initial velocity 1.5, acceleration 0.5s equal to 30. And what's his new velocity? And we do not know the time. As equal to you. No, no, no, not this as. 因。No T S equal to deep ness over t ed. And v square equal to U square plus two as. So v squared is equal to 1.5 squared plus. Two atwo atwo times 0.5, times 30Yeah so that's 5.68. Meters per second. Yeah how long did this embarrassing instant take place for? So how long did this instant last? Multiple five t equal to changing velonew velocity is five, one, 68, so 5.68 divided, but sorry, -1.5 equal to. 4.18 and this number. This number divided by 0.5. 8.36. Yeah. Okay. So then we have. A golf ball is struck with a velocity of 30m per second at an angle of 30 degrees. How long does it stay in the air? For what height does it reach? Struggle with a velotive 30m and an angle of 30 degrees. So it suggests you so the velocity is. But do we want to know about the vertical vertical direction? 30 degrees. Three measures per second. So how do we work out how high it goes? How? Yeah we should make this vertical side. So so basically, I think you have to work out the time it takes to get to its maximum height and then Yeah, Yeah, Yeah. So with a velocity of 30m per second. So. So acceleration equal to velocity divided by time, so time is equal to the velocity divided by the acceleration. So 30 divided by 9.81. 3.1. What if we worked out 3.1? So winthere is no measures per second square. I think that is the velocity on time, the time second 3.1s to get to the maxiimal height. Okay, good. So we know the time it takes to get the maximum height. S then we can probably use s is equal to ut. Plus a half gt squared or at squared. So the distance so U will be zero. So. S is equal to serum plus 0.5 times 9.81. Times. 3.1 squared. So what do you get for that? What distance, what height does your golf pole reach? The distance is equal to. Just going to get something. Yep. So 46.5m, maybe I should tell us two. Oh no, what height, what height does it reach? Good. Okay. I'm trying to find some more questions. Okay, so. Let's try some of these. The particle is accelerated uniformly from rest, so that after 10s, it has achieved a speed of 15m per second. Find its acceleration and the distance it has covered. 1.5m per second square. So what we know here, we know time. We know initial velocity, we know final velocity. We need to find acceleration and distance. Okay. A equals b minus U and rout. So we know. So it's. 15 minus zero over ten. Which is 1.5m per second squrored. 1.5. Issues per second screw. And the acceleration and the distance it has covered. So which equation do you want to use for distance? Distance. Well, we know the final and the initial velocity and the acceleration, but we do not know the time, which. Yeah, we do windows a ton. For 10s Yeah. So we know it's been accelerating for 10s. So we know that. So. We know v, we know U two, we know a. So we could use that equation or we could use that one. Which one are you going to use? Which one? Is it still in still in question one? Yeah so find its excel or we we the distance it has covered so we could and distance we can use as equal to v plus U divided by two times t yes. So 15 divided by two and times 1075. 75m good. Let's try question three. A train is uniformly retarded from 35m per second to 21m per second. So that means it's decelerating. Over a distance of 350m, calculate the rate of deceleration and the time taken to come to rest from 35m per second. So remember, acceleration is negative decelnegative acceleration is deceleration. I don't normally see it called retardation, but at the slowing down. Appreciate. Presentation dation, okay, 35m per second to 21m per second and a distance of 35, 350m. Taken to come to rest from 35m per second. So we know 21 squared is equal to U squared, 35 squared. This is two. Times. Times 350m. So I've chosen to use this equation. Should work. What do you get for the deceleration? The telvision will the desanderation there's no time. So that is v square are equal to Yep. V square equal equal to U square plus two A S. So. 21s 21 square equal could use 35 square. That's two A S. 700. Hey. So 2121 square minus. 75 square. And is then divided by 700. I equal to -1.12 that is the. Reputation? Yes, good. And take arrest. To rest wrong. To come to thst. So the change in velocity is -35. And deccenter and itation is -1.12 minus per second squared. Oh. So we've found the rate of deceleration 1.12m per second. So that is 35 divided by 1.12. Study 1.25. Seconds maybe not maybe not correct. Anyone point ult, because that is correct. So it raised this speed, and then it comes to rest. So this becomes initial, and then zero is the final to come to rest. Okay, let's try five. A car accelerates uniformly from 5m per second to 15m per second, taking 7.5s. How far did it travel during this period? 7.5s up to 15. That's equal to the sum of velocity that is 20 divided by two. That is ten, ten times 7.5. If so, that is 75. You. Five times 7.5 plus. 75m good. Okay, let's try question eleven. It's drihundred at 25, the driver suddenly noticthat there's a fallen tree p blocking the road 65m, and hethe driver immediately applithe brakes, giving the car constant rotation of 5m second square. How far from of the freest cover comes to rest? Comes to rest. How far? Right? So that's its deceleration, which is negative acceleration. So we want to know the distance, and we know that the final velocity is zero and the initial velocity is 25 and acceleration is minus five. So we square. With zero equal to 625 minus two a ten s. That's equal to 62.5m. Yes, so 65-62.5 gives you how far in front of the tree does the car come to the rest. So is 2.5m good? Okay. We'll do one more projectile question before four, okay? And. Yes, his. 26m per second, that is mingle 45 degrees. To the branch. Hey, ignore. Air existence. Now long is. How far? This is truful pretty sumptially. And what is the little that is reached? So projectile motion is when you have to consider the vertical motion independently of the horizontal motion. How long? So here is air. So it's. Protify Yeah, it's a maybe it's a. I always like to draw diagrams of what I'm looking at. Okay, Yeah. So first of all. 2626 divided by 9.81. The first 12.65. How did you guess? How did you get 2.65? Have you taken into account the angle? The angle is. So the initial velocity vertically is the cost 45. So U equals 26 times cos 4518 point Yeah. Now going up. When it's rising with Jackson itreach its maximum height when the final velocity is zero, so it goes up, it reaches its maximum height, then it starts to go down, so the fact maximum velocity in the vertical direction will be zero. So this is its initial velocity vertique. Taking into account of the angle, its final velocity vertically will be that we know a is 9.81. So we can use. B equals U plus 80. So v is zero because at this point it stopped increasing vertically. U is 18.4. Plus 9.81. Times t. So 18.3a is divided by 9.81 gives me t equals 1.87. Okay. How long is the ball in the air? So this takes 1.87s to get to the midpoint, and then it also takes 1.87s to land. So we multiply this by two to get the time. Three points. 75s. Okay, that means why we use v equal to U plus 80. What have you got here? The final velocity, basically, we're comparing vertical motion initially initially the in the first question, why we use v plus U with equal to U plus 80, why we shouldn't use equal to change in velocity divided by t. These are the same equation, aren't they? This is exactly these are identical. What you made a mistake with is you got confused with that's your initial velocity. That's your initial velocity vertically. Your final velocity vertically at its highest point is zero. Okay. So if we know the acceleration is 9.81. Is equal to zero -18.4. Over t. We should get the same answer. T is 18.4 divided by 9.81. And you get 1.87. To get to this point. But we're assuming that a parabola path is symmetrical, Jackson. We're assuming that it takes this 1.87s to get to this. Midpoint of the journey and 1.87s to get to the end of the journey. Okay. So the total is 3.75s. How far does it travel horizontally? Remember, horizontal motion is independent of vertical motion. Ph. Vertical motion. So we know the time at the same time, the distance in motion. Yeah. So that is cosine 26, 26 times. Cosine 45 same. 18.4 and times the same time is. So landing reccalculator system the time. We've calculation seven Yeah 1.87Yeah I see 1.871 point four. Why did I double the time? The time? I didn't double the time. No, okay, Yeah, I know. I know. Yeah, I know. Now one point 1.87 times two. Yes, no, not plus times 18.4. 38.8. I get 68m. 60 83.75. Miss that is 16 nine. Yeah 69. You're right. Okay. What is the maximum height reached so horshes. That's. What we know already that's equal to v plus U divided by t times divided by two times t we can use this equation. And yes, good. And we can know about the maximum height. So 26, I know 18.4, 18, 18.4 and two zero, so 18.4 divided by two. Times T T equal to 1.87. 1.87 so. And we know that Yeah that is a distance 17.2. Good. Yeah. Okay. So this is the type of question that comes up on paper one bminton, a shutlecock is hit by a racket. The graph shows how the vertical displacement and horizontal displacement. Very from the moment it leaves the racket. So this is displacement in the vertical direction and in the horizontal direction. From the moment it's hit, the positions are recorded every tenth of a second. The graph shows how the velocity of the shutlecock in the vertical direction varies with time. Explain how the velocities es have been calculated from the success essive vertical positions of the shuttlecock. So. How the philosophies are changing with time. So how does the y displacement and x displacement give. How do we get this value from this? So the time is still every tenth of a second. So that's okay. But the velocities. Calculate the difference in positions on the y axes, okay? And divided by ten th of a second, so three minus two over so v minus U. So this change in distance with time is velocity. Statewide, these velocities have been plotted at the mid range of the time interval. So why instead of at no point 1s it's been traveled at this time interval? I suppose it's average. It's an average velhy, so this velocity is average. Yeah. I'm not understanding. Standable that see the velocity of the displacement is changing. That's the y direction displacement. The x direction is changing. So we're getting the displacement after 0.2s compared to after 1.1s, but it's changing. So the the velocity is only an average velocity. And you can see that the displacement is changing all the time. So let me see the question again, why these velocities have been plotted at mid range over the time interval mid range. Is that a mid range help? If you look at the other graph, it's at every point one of a second, not every point zero five of a second. Yeah. Mid range at the mid range. So it's there it's not there. That's mid range Yeah the middle of so that is the average. Yes, okay. Shall we continue with this question tomorrow? It's from 2020. Okay, how do you feel about this topic? Jackson, I think just for me, it's no problem for me to use the equation to calculate the results, but maybe it's difficult for me to explain it. Yes, we look at some past papers tomorrow, some how this topic comes up in past paper questions. Okay, Jackson, bye bye.