High school physics
Choosing kinematic equations
Kinematic equations help solve for an unknown in a problem when an object has either a constant velocity or constant acceleration. This video will help you choose which kinematic equations you should use, given the type of problem you're working through.
Want to join the conversation?
- For the first question, don't we have to convert the 80km/h to m/s before calculating?(47 votes)
- Yes. At4:49, there is a note that says that the conversion must be done first, and the instructor messed up. If you didn't convert, at best you'd have a convoluted answer containing a ridiculous fraction with units everywhere.(18 votes)
- So, what are the actual answers to these questions? I solved them myself as well, but I'm new to this and may have done something wrong along the way.
- you will learn it soon enough!
for the first problem x= 182.87m and t=16.46s
for the second problem x=172.8m and final v is 28.8m/s(15 votes)
- How does one derive the 4th formula (v^2 = (v0)^2 +2aΔx)?(3 votes)
- Looking at the equation you notice there is no t. So, we need to use the equations we know and replace t. The easiest one to do that for is the first equation. Then use that information in another equation to get our 4th,
Vf = Vo + at to solve for t, subtract Vo from both sides Vf - Vo = at
now divide both sides by a and we get Vf -Vo/a = t
Now just use equation 3 replacing t with Vf - Vo/a
I don't know how to type the triangle so I'll say delta x
deltaX = Vf + Vo/2 (t). --- changes to --- delta X = (Vf + Vo/2)(Vf - Vo/a)
Now just solve. Multiply both sides by 2a to get rid of denominator on right side
2a delta X = (Vf + Vo)(Vf - Vo). Vf^2 + VfVo - VfVo - Vo^2
2a delta X = Vf^2 - Vo^2 add Vo^2 to both sides
Vo^2 + 2a delta X = Vf^2 or rewrite Vf^2 = Vo^2 + 2a delta X(13 votes)
- Aren't there five equations, not four?(4 votes)
- Yes there is a fifth equation for initial velocity,(Displacement=vt-0.5at squared). Its not as useful as the other formulas since initial velocity is typically known.(5 votes)
- what is the difference between v=at and v=x/t?(3 votes)
- the first one tells you what your velocity will be if you start from rest and accelerate at a rate of a for time t
the second one tells you what your average velocity is when you have displacement x during a time t(4 votes)
- I got a question on the test like this: "A chef throws pizza dough in the air and catches it at the same height, the dough travels in the air for 1.0 second. What is the velocity of the pizza dough before it is in the air?" I understand that the known values are the time, acceleration and displacement, and the initial velocity is the one we are trying to find, but why is the final velocity assumed to be unknown? (this is stated in the hint) shouldn't the final velocity be 0 since catching it effectively stops the motion of the dough? I was thinking the final velocity might be referring to the velocity right before impact, but other questions say "coming to a stop" to signal a final velocity of 0, shouldn't it be the same for this question as well?(3 votes)
- In kinematic equations we are trying to understand motion, not just state the obvious fact that after an object is done moving it comes to rest. Final velocity refers to the velocity at the end of the fall. When the dough has been caught, that's not the end of the fall, that's after the end of the fall, and we don't care that the hand made the dough stop falling.
The question is basically asking you how fast do you have to throw something upward to get it to be in the air for 1 second.(3 votes)
- Do you have to know these equations for the AP Physics Exam?(4 votes)
- Hold up my physics teacher told me that there was an equations sheet given on the AP physics exams and that you needed to have to know your concepts well and how to use the equations but as for memorizing them was not necessary because an equation sheet was provided?(0 votes)
- "These kinematic formulas can be used only when acceleration is constant." Is this statement true??(1 vote)
- Yes this statement is true. When we derive these kinematic formulas, we assumed the acceleration is constant. That's why they are valid only when acceleration is constant. You may way to review the derivation of the formulas here: https://www.khanacademy.org/science/physics/one-dimensional-motion/kinematic-formulas/a/what-are-the-kinematic-formulas(6 votes)
- why is it delta x but not delta y because delta x usually means change in time(2 votes)
- We use the letter t to talk about time. x usually stands for position in Physics, even though it’s often graphed on the vertical axis, if that’s what you were thinking. Often, x means position in the x direction, and y means position in the y direction. It would just be confusing to have Δx mean change in time and Δy mean change in position.
Does that help?(3 votes)
- I might be totally mistaken, but to find the final velocity for the second problem, couldn't you just multiply the acceleration by the time and add it to the initial velocity?(2 votes)
- Correct, that's the last thing he said in the video.(2 votes)
- [Instructor] In this video we're going to go through a few examples of setting up some problems with constant acceleration. We're not going to solve them, we're just going to look at what we know and what the question is asking for and then identify which one of these equations over here will be the most useful for helping us solve it. So before we jump into the examples, I wanna say that it is very important to understand where these equations come from to really develop a strong understanding of position and velocity and acceleration and time and how they're all related to one another. And Sal has a lot of videos that go through that and can help you build that understanding. But once you have that understanding, these equations are kind of like using a calculator where they help you save time. It's important to know how to add, subtract, to multiply and divide, so you know what a calculator is doing when you use it. But once you understand that, the calculator is a really valuable tool. And that's what these equations are like, they're tools that we can use when we understand where they come from. With that out of the way, let's dive into our example here. So the question says, "A light rail commuter train "accelerates at a rate of 1.35 meters per second squared. "How long does it take to reach its top speed "of 80 kilometers per hour starting from rest?" All right, so let's unpack this and see what it's saying. So let's just start at the beginning. A light rail commuter train accelerates at a rate of 1.35 meters per second squared. So that's pretty direct, it's just telling us what the acceleration is. So let's write that down. The acceleration is 1.35 meters per second squared. All right, there's one thing. Okay, so let's keep going. How long does it take to reach its top speed of 80 kilometers per hour starting from rest? So that one's a little bit more complicated, there's more going on in here. But if we just start at the beginning and say, "How long does it take?" That by itself is a question. There's some more stuff afterward, but that by itself is just asking about the time, how long does it take? All right, so let's note that by circling this time here. We don't know what it is yet, but this is our question, we're being asked about the time. All right, so if we keep going here, we get, all right, how long does it take to reach its top speed of 80 kilometers per hour? That's saying its top speed is 80 kilometers per hour. When it's done speeding up, it'll be going at 80 kilometers per hour. So that's the final velocity or just our velocity at the time here. So that's 80 kilometers per hour. 80.0 kilometers per hour. And sometimes you'll see this right here written as V sub f to really explicitly say it's the final velocity. So the notation might vary in your physics class. Whatever notation you use, that's fine, but make sure to ask yourself what is this symbol really talking about. So anyway, 80 kilometers per hour, per hour. Okay. And so we have that. And then if we keep going, it says, "Starting from rest." So that's saying that at the beginning when it starts, it's at rest, which means it has zero for it's initial velocity, it's just sitting there. So let's fill that in, that's zero meters per second. Okay. So we had analyzed this question and we see that, actually, the change in distance didn't appear anywhere in this question, it's just these four values right here. So let's see if we can look at these equations and identify one that has all of these things and doesn't have delta x, since we don't know delta x and we're not looking for it either. So we see these two have delta x here so we can rule those out. And this one also has a delta x so we can rule that out. So that leaves this one. Let's see, it has velocity, the final velocity here, it has the initial velocity here, it has acceleration here, and then it also has time which is what we're looking for. So we can use this. We've figured out that for this question right here, we can just use that top equation. All right, and then to continue this we would plug in numbers and then solve for t and see what that time indeed is. So for this video, we're not going to have to go through that, we're just going to go through another example of setting things up. So let's go down to this question right here. While entering a freeway, a car accelerates from rest at a rate of 2.40 meters per second squared for 12.0 seconds. How far does the car travel during those 12 seconds and what is the car's final velocity? So this one actually asks two questions. Let's just focus on the first one to begin with, and also let's see what we know. So while entering a freeway, a car accelerates from rest at a rate of 2.40 meter per second for 12 seconds. So there's a lot of information here in this first sentence. So let's see, it says, "Accelerates from rest," right here, accelerates from rest. That tells us that the initial velocity is zero, right? From rest, it wasn't moving initially, it was at rest. So let's write that down. This also, just like the previous example, had an initial velocity of zero meters per second. That's not always the case, but it happens that in these two examples it was. So let's keep going. So it accelerates from rest at a rate of 2.40 meters per second squared. So that's telling us what the acceleration is, 2.40 meters per second squared. So let's write that down, 2.40 meters per second squared, meters per second squared. And in all of these problems, it's important to think about whether something is moving in the positive direction or in the negative direction or if it's accelerating in the positive direction or the negative direction. It actually looks like for these examples we've chosen everything is moving in the positive direction and accelerating in that same direction. So we won't have any negative signs pop up, but it is important to think about that when you're doing these kinds of things, just in case something is negative and that does affect your answer and what's going on. But anyway, so it's accelerating forward, everything is going in the same direction. So we're thinking of the forward as the positive direction and they're accelerating, this car, is accelerating at 2.40 meters per second squared. All right, and it says also that it's doing that for 12.0 seconds. So we can write that down as well, 12.0 seconds. All right, so we know three things here. And you'll find that in problems like this, that's the magic number. Once you have three of these, you can find out the other pieces using these equations. Anyway, so the first question says, "How far does the car travel during those 12.0 seconds?" Well, it's asking how far, so that's going to be the delta x here, this change in position. How far does it travel during those 12 seconds? It's going to be this value right here. So this is our question, circle it here. So now notice we don't have this, so we can look for an equation that doesn't have this final velocity value in it and then that will be an equation that will hopefully have these other four, but we'll have to check that. This one has that final velocity in it, so we can rule that out. I see this one also has the final velocity. This one doesn't have the final velocity. So let's look at that. So it has the position, the change in position, that's what we're looking for, it has it right there. The initial velocity is right here, so we have that as well. Let's see, it has time in it and we have time, so we will be able to plug that in. And then finally, it also has acceleration. So it has everything we need, it has the thing we're looking for, and it doesn't include anything that we just don't know, that we're not even looking for. So we found that for this problem right here. This second equation here is going to be the one that is most useful for answering the question, "How far does the car travel during those 12 seconds?" So that's for that question, but this block here actually has a second question. Maybe I'll switch to green. Let's look at this question. What is the car's final velocity? All right, so that means we're looking for this now. Right here, what's this? Well, we've already identified these different pieces that we know, and if we did the first part of the problem, we would actually have a value for delta x here. But assuming we didn't find that yet, we could look at this and say, "What's the car's final velocity? "I want to look for something that has V in it "and has these other three things that I know." And if you look and you check through here, let's see, what is it? So, yeah, it would actually end up being this one again. This is that equation that doesn't have the change in position in it. And so just using the green color, I'll underline this here to say that this equation is useful for this question. And actually, now that we have this value, we could've used really any of them that had our V in it. So our options open up once we know more than three different things. But anyway, hopefully going through a couple of these examples will be helpful for you when you run into your own questions and have to think through, "Okay, what do I have, "What am I looking for, "and which equation will help me "move forward and solve this question?"