Kinematic equation is the backbone of the mechanics course and will be continually referenced throughout the course, so make sure to understand them conceptually and mathematically. It describes how objects move around the physical world and these equations are why people can calculate the distance someone travelled knowing their speed and time. Note that these equations work in 1 dimension at a time, so to work in multiple dimensions we need to have multiple sets of these equations, which would all be connected through time. Next well put these equations to use in multiple 2 dimensional problems.
The kinematic equations can be used to solve kinematics problems in which acceleration is constant.
When you know three of the five kinematic quantities for constant acceleration, you can use the kinematic equations to solve for the other two.
You can avoid quadratics by solving for an intermediate quantity, simplifying your calculations.
Free fall is a condition in which the only force acting on an object is the object's weight.
The acceleration due to gravity on the surface of the Earth, a constant known as g, is approximately 9.8 m/s^2, which can be approximated as 10 m/s^2 to simplify calculations. This is also known as the gravitational field strength.
The motion of an object in free fall which travels up and then back down is symmetric.
The motion of a free falling object which travels up and then back down can be broken down into motion on the way up, then motion on the way down, which may simplify calculations.
At the highest point of its path, a free falling object's vertical velocity is 0.
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
Hi everyone and welcome back to Educator.com. I'm Dan Fullerton.0000
And now I would like to talk about kinematic equations.0004
Kinematic equations are used to solve problems for objects moving at a constant acceleration in a straight line, or for multiple dimensional problems when we start adding different kinematic equations for different dimensions.0007
Our goal is going to be to solve problems for objects moving at a constant acceleration in free fall and in straight lines.0019
If it is not constant acceleration, we are not going to be able to use these equations.0026
However, just about every problem you are going to see on the AP exam has to do with the constant acceleration.0031
So these are very, very useful for the AP course.0037
Now what we are really doing is talking about another way to solve problems.0042
You can solve all of these problems by making motion graphs like we talked about previously.0046
But they are not always the most effective or efficient way of understanding motion.0051
The kinematic equations help us solve for the five key variables -- they are going to describe the motion of an object in a single dimension -- V0 or V initial, initial velocity; V or V final, the final velocity; displacement δx, or if it is in the y direction, δy; the acceleration, the change in velocity; and the time.0056
Those five key quantities are what we need to know to understand what is happening to the motion of an object in a constant acceleration scenario.0077
So let us see if we cannot derive these kinematic equations.0088
As I go through this, realize you do not have to repeat it, but I would like you to do everything you can to try and follow along with the derivation because I think it is important that you understand where these are coming from.0093
So we are going to take the motion of an object that starts at some initial velocity, V0, and accelerates with a constant acceleration to some final velocity, V, in some amount of time, t.0103
So there is our time, t, and what we are going to do first to get our first equation is we are going to start with the definition of acceleration.0119
Acceleration is the change in velocity divided by time, or that will be V final - V initial/time.0126
With a little bit of rearrangement we can say that the V - V0 = acceleration x time, or V final = V initial + acceleration x time.0136
All right, now what we are going to do next is looking at this graph -- we can find V -- change in an object's position that changes its displacement by taking the area under the velocity-time graph.0156
So what we are going to do there is we are going to say that the change in the object's position is the area.0167
To get the area, I am going to break this up into a rectangle and a triangle.0175
The area of our triangle is 1/2 base x height and we have the area of our rectangle, length x width.0181
So as we look at our triangle, that is going to be 1/2 x base, that is T, and the height of our triangle right here is going to be V - V initial, and we have to add in our rectangle, V0 x T.0190
This implies then, since we already know that V = V0 + AT -- we just learned it up there.0211
We could rearrange this to say that V - V initial = AT. 0222
I am going to replace V - V initial here with AT.0227
So I then get δx = 1/2T x AT + V0T, or with a little bit of Algebra, I can show that δx = V0T + 1/2AT2.0232
A car traveling on a street road at 15 m/s accelerates uniformly to a speed of 21 m/s in a time of 12 s. 0808
Find the total distance traveled by the car in this 12 s time interval.0816
Well, let us assume that it is going to the right again and we will call that our positive x direction.0821
But as we do that, let us go make our motion table again -- V initial, V, δx, A and T.0824
V initial = 15 m/s -- it accelerates to a speed of 21 m/s -- that is going to be our final velocity, in a time of 12 s.0836
Find the total distance traveled by the car in this 12 s interval.0848
Right away as I look at this -- do I have any equations that will allow me to get δx with both V's and T?0854
Well not right away, but there are a couple of things that we could do.0863
One thing is we could look at that kinematic equation, Vaverage = V initial + V final/2, which is going to be 15 + 21/2 or 18 m/s.0867
Since we can treat this as the average velocity for the entire time interval, we know that average velocity = distance/time, therefore, the distance traveled will just be Vaverage x time, or 18 m/s x our time of 12 s -- 18 x 12 = 216 m.0882
But there are other ways to solve this problem too.0908
We could go and solve for A first if we did not want to use that formula.0912
On the other hand, we can launch objects upward.1677
We have to examine the motion of the object on the way up and the way down though.1680
Since the object's initial motion is up, I am going to call that my positive y direction and if up is my positive y direction, my acceleration is the opposite direction of what I called positive so the acceleration would be negative g.1685
I know that at the highest point of motion, if I throw anything up and it comes back down -- at its highest point for the split second -- its velocity is 0.1698
Since I know V0 now -- it is 4 m/s -- that is going to be 4 x 0.4 + 1/2 x A(-10) x T2 (0.4)2.1896
So δy -- when I do the math -- plug that into my calculator, I come up with about 0.8 m. About that high.1910
Let us try an alternate path to the solution though.1921
Let us look and analyze the solution on the way down.1924
So on the path down, I am going to assume that this is the same as if we had dropped an object from rest up here at this point.1929
We will call down the positive y direction -- V initial now is 0 -- we do not know final velocity right before she hits the ground -- Well, we would if we used our symmetry of motion, but let us pretend we just took this path.1935
A ball is thrown straight downward with a speed of half a meter per second from a height of 4 meters. 1994
What is the speed of the ball, 0.7 s, after it is released?1997
Well, if we called down the positive y direction, we have V initial, V, δy, A and T -- where V initial is now 0.5 m/s -- that is positive because we said down was positive -- A = 10 m/s2 and we want the speed 0.7 s after it was released. 2002
I can jump right to V = V initial + AT which is going to be 0.5 + A(10) x T(0.7) -- 0.5 + 10 x 0.7 = 7.5 m/s2.2027
Let us try one last simple problem just to make sure we have got it. Find the maximum height.2055
A quarter-kilogram baseball is thrown upward with an initial speed of 30 m/s.2061
Neglecting friction, no air resistance, the maximum height reached by the baseball is approximately what?2067
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