一是T P。Six. C T square red plus five t see. So that's v. So t is zero. C equals two. Yeah that's two. This is zero equals T T. Plus two V T equals six, so six equals 2002 squot plus five t plus two, so zero equals two thirt squred plus five t minus four, so t equals. Time times it by to three t squred plus ten minus eight equals zero. And. Negative. 我是。Displacement of p from the origin after t second is given by s equals t squore t times t plus k meters given p is that instantaneous rest when t equals four. Find the acceleration of t when t equals ten. Okay, so. Displacement. Equal als t square times t plus k brcket. All to cool. Instant. Ously. Rest at t equals four. T equa. So displacement to acceleration, displacement velocity and velocity to acceleration differentiate twice. Say. T squt plus k equt cubed plus t squared k. K T squad three t squred plus two K T and then. This is velocity and velocity. The v equals zero, t equals four. Zero equals 34 square t zero equals 1640. T equal A A A equals minus six. This is locity. Than acceleration. Do one. Yes. Three t squad plus two K T equals 60. Plus two k. K equals minus six. So a equals 60 minus twelve. What do we need to find to find the acceleration when t equals c ten six times ten minus twelve equals 48m. Second, mintwo question. Twelve. Lost your part of the t second ds is given by v equals six t minus t squsecond when t equals zero partigoes up, Orio. Possible. V equals 60 minus two t squared. Okay. All starts. Brigin. Distance. 螺丝机。Velocity, interrogation、displacement okay. We need to find t with article. Rest so 60 minus two t squared equals zero. Two, 7 31t zero t equals three，seven, 13 equals three when possible is at rest and 60 squared over 23t squared and this is three t squared minus two thirt squared plus c. When t equals zero, the particle is the orizero. So zero equals c. So this is just going to be. Three t squared minus two thirds t squared, and then find this instantaneously at rest. You know, t is three, so three, three squared minus two thirds nine. That's cubed. 27. Course 27 miles or is 9:18 equals nine me is ers. Okay, so I am done. Okay, so see what you've here, what you've got. Okay, so for ten, it's two third seconds. Yeah. Eleven it's acceleration is 48m per second squared. Yep, good. And then for question twelve, it's 9m. Yes, 9m. This is good for that one. Okay, what we'll do then and we'll move on to next mechanics from as. So we've done sudone velocity, time graft and variable acceleration. So that's basically kinematics. Then you've got sections on forces. So 2D vectors and f equals ma. I think we can start with 2D vectors. I'll upload some to begin with. If you're not sure what to do or where to start, let me know and I'll try to give you some pointers in the right direction. I'd say 2D vectors is is I wouldn't say it's that difficult, but. Have you seen this notation before for I and J in de? So they just mean a unit vector acting in the x and y directions respectively. So like a vector, you can write vectors in different ways. Yeah we can write a vector with a little hat, and it might be written in column notation where it has a value of x in the x direction and y in the y direction. Other notation might be the value of x acting in the I direction plus a value of y acting in a unit vector, which should be written as J. The three dimensions, we're not going to look at it here, but the three dimensions, you would just easily add that on z and then your unit vector representing the z axis or z direction would be okay, but these aren't going to include that. So you want to apply laws, things like equilibrium. Equilibrium means that. All the horses. How balanced might e all the left equals the right and all the upwards equals the downwards. Yeah so you can balance iand J's separately is and J's direction separately. Magnitude of a vector is given by Pythagoras. Okay. Yeah so itbe the square root, x squared plus y squared, thatbe the magnitude direction of vector usually quoted with respect to the positive x axis that e my direction. So that might look something like you have your axes and you have your vector like this. Yeah, let's say that's your vector. Normally the angle would be quoted like this. Yeah, so it would be your J component. Yeah, divided by your x component and itbe the arc tangent of that. Yeah. So if this is like a value of y, this is a value of x. Then the. Thewould be artay over that x okay. And you might also need to use the fact that f equals ma in one of these questions. Newton, second law, the f equals m and that applies for vectors as well. So the force vector. Would be a constant MaaS multiplied by your acceleration vector, okay? Okay, all right, I'll let you have a go, but if you're stuck, just just just let me know and I'll give you a hint. The in the right direction. Okay, okay. Three I two P J and minus two I five J. And four Q I five J N. A is an equilibrium, find the values of p and Q. P and q. Article all the phone same. How are you getting on? I'm just thinking sir, like for the first question, like because it's in equilibrium, so like do I like so like the the what the the force is zero. So I add all the I forces. Yeah all the first. Yeah, Yeah yes. So the suof, the the sum of the I direction forces is equal to zero. And then also separately, the sum of the J forces is also equal to zero. So you do them separately. Three. Plus minus two plus 4K zero. So that's one plus four q equals zero. So four q equals minus one q equals minus one over four. And then it's two p plus five, plus five equals zero. Two p equals minus ten P5. Okay, possible p of us not 28 kilograms moves under was f and the acceleration p is minus three I five tonand the anles mean acceleration and vector I. Between the acceleration. Okay, why over the next. Three point. My like. All. I don't quite know how to do like question two. Which one question two a, Yeah. So find the angle between the acceleration and the vector. I so this is exactly what I was kind of writing here. The acceleration vector is the direction of this vector v, for example. Well, I mean, it would look a bit different, but you basically want the angle between the acceleration and the x axis. Okay Yeah maybe draw it is what this means this minus three I plus five J it means it moves in the negative x direction by three but it moves in the positive y direction by five and then you want that between the vector I so it's actually gonna to be. Just thinking about that. That should be. Larger. Yeah it should be it should be an obangle if you think about it because it wants it with respect to the positive x the positive x axis Yeah Yeah gonna be like this it's going this and like this so that there wants this angle minus three I plus five J and I have okay. Three squplus five square. 9:30 4:34. Five is three. Hundred -15 over three. Nine. Going to be at 180-50 921 degrees. 啊。处理。Oh, I think goes I may okay to equals M A F equals 0.8 times what 3:34 equnot. 4.66. Okay. Parts p of MaaS two kilograms move with constant acceleration, and the action of force at an initial velocity of p is two minus two I plus six J. After plus 4s, velocity p is seven I plus two J. The magnitude of the acceleration, so MaaS equals two kilograms. U equals minus two J denomminus two I plus six J meters second, and v equals seven I plus two J meters per second. My God, constant acceleration is the formula a equals v minus U over t. The v minus U equals seven I plus two J minus minus two I plus six J she als nine I -40. 4s. A equals four. I minus jthis is acceleration. Magnitude, it's square root. Square root. They some magnitude. 94 squaminone square. Equals. 2.26. The angle f and. A equals nine zero four I minus J. F and effect I. So this is triangle and this is 904. This is minus one. Okay, so it equals nine Oh four squared minus. Just one squahead. Route 97 over four. Opposite face ent. One是空去four分之equals n -14 over nine。23.96. Angle and below the y axis. So it's going to be. Wait, it's a tangy. 统计统。Okay, so I'm done. Okay, let's let's say what you got. Okay, so for question one, q equals minus a quarter and p equals minus five. Yeah. Okay. And then for question two, the angle is 121 degrees. Yes, 121 degrees. Good. Okay. Two, two b. Yeah, two b is 4.66 newtons. Yeah, 4.66 newtons, lovely. Three A, I think the magnitude is 2.46m per second squared. Yes. And then for three b, it's 24 degrees. Yeah, that's great. That's all good. Okay, now we'll go at these. And then yes, you need a hint. Let the resultant of two forces, F1 and f two, is I -14 genuters. Other than that, F1 equals two pi minus four q jn and f two equals three qi plus four pj n you to define the value of p and. Q. I minus. 14. Q J. V3t minus four p. The results in forms is forces adding together. So these two adding together equals I -14J. Two p plus three q equals 14p. X plus minus four q plus four p minus four q equals 14 it's -14. Okay. P minus okay equminus 14 is A P minus q equals. -3.5. Phi equals q -3.5. He he wants Yeah he is Yeah is us so Oh my God. Also. Q times q -3.5 plus three q equals 12q minus seven plus three q equals 15q equals eight q equals. 1.6q equals 1.6p equals q -3.5 equals 1.6-3.5, -1.9. So that's p. Now we know do. Partip moves with the constant velocity of four I minus J find the speed of p. V equals four I minus J. Speed. Sir, does speed equal to magnitude? Yes, Yeah, it does. Speed is the magnitude of the velocity okay? The. Speed. It's going to four squared plus minus one squared eight. 16 plus 1:17 17 17m per second. Find the direction of motion e, giving you the answer as a befrom a long time ago. Oh, right, that is. Up. I is x so four to here and then. Minus one here. Go not 414 minus. Or is a minus one. No, no, no, God. -76 degrees. All. South to East West East. No sales. East West. My 70South East South. Of east southeast. So it's here 76 no, no, no, here south of east. Should here. 90 okay so it's 90 plus 70 666 degrees take of this. The particle moved with congniflict acceleration. Three I minfour jameters second. Equals three v equals minus five I plus a two J meters per second, equal speed is U V equals U plus 80. V so U equals v -18. So U equals v is minus five I plus two J minus three times three I minus four J U equals. Minus five I plus two J this is nine I'm. So. You, we gather the eyes and jys minus five, minus nine I, plus two, plus twelve J. -14I plus 14J. And then Pythagoras -14 squared plus 14 squared square root. Holding in rotwo meters per second. Okay. That's question six that Yeah that's false question seven. J okay, say force one equals three I minus two J. P I plus P J. What's the pretitwo forces is parallel to? Sir, for question seven, what does like the resultant of two forces, F1 and f two, is parallel to inj. It means that the resultant force of F1 and f two parallel to a vector which is acting at a 45 degree angle. 45 degree angle 45 degree angle with respect to the x axis. Yeah because I plus J getting these equal amounts of each, if you have equal amounts of I and equal amounts of J is acting in the middle somewhere. So it basically means that it's parallel to a vector which looks like this Yeah where this angle is 45 degrees okay. All right, thank you, sir. So by the value of me, so resulted fequals these two plus together, so three plus pi plus two plus no, minus two, minus two plus two pj. Parallel parallel so they equal one I think Yeah. Three plus p minus two plus two p plus one. Three plus p. Equals minus two plus two P P equals five. Perfect. This. 45 please I don't know means I am J I J. Okay, I think I'm done. So. Okay, let's let's hear what you got. Okay, so for question four, I got q equals 1.6 and p equals -1.9. Yeah question five, a is root 17m per second. Yes. And for question B, I think it is 166 degrees. I got 104. Because it's a bearing. So that means measured from due north in a clockwise direction. So you do arc engine of one over four to get the angle below the x axis, which is 14 degrees. And then you can just add that on 90 degrees. So that give you 104. All right. So that's 14. Okay so then for 68 you know for six for six is 14 rotwo meters per second, you hang on. Did you get a vector as well? Did you use a Suthat to work out a vector for U? Yeah. I used v equals U minus at. So U equals v minus at. You mean v equals U plus at, so therefore U is v minus at? Yeah. Yeah. Yeah. Well, and you got -14I, plus 14J. Yes. Yeah. Okay. And seven question seven. P equals five. Yeah. Okay, perfect. I think we've gotten to the end of end of that one. Okay, great. I think let's finish it there then and I'll see you later. Okay. All right. Thank you. So okay, bye bye.