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### Kinematic Equations

• 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.

### Kinematic Equations

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.

• Intro 0:00
• Objectives 0:07
• Problem-Solving Toolbox 0:42
• Graphs Are Not Always the Most Effective
• Kinematic Equations Helps us Solve for Five Key Variables
• Deriving the Kinematic Equations 1:29
• Kinematic Equations 7:40
• Problem Solving Steps 8:13
• Label Your Horizontal or Vertical Motion
• Choose a Direction as Positive
• Create a Motion Analysis Table
• Solve for Unknowns
• Example 1: Horizontal Kinematics 8:51
• Example 2: Vertical Kinematics 11:13
• Example 3: 2 Step Problem 13:25
• Example 4: Acceleration Problem 16:44
• Example 5: Particle Diagrams 17:56
• Example 6: Quadratic Solution 20:13
• Free Fall 24:24
• When the Only Force Acting on an Object is the Force of Gravity, the Motion is Free Fall
• Air Resistance 24:51
• Drop a Ball
• Remove the Air from the Room
• Analyze the Motion of Objects by Neglecting Air Resistance
• Acceleration Due to Gravity 25:22
• g = 9.8 m/s2
• Approximate g as 10 m/s2 on the AP Exam
• G is Referred to as the Gravitational Field Strength
• Objects Falling From Rest 26:15
• Objects Starting from Rest Have an Initial velocity of 0
• Acceleration is +g
• Example 7: Falling Objects 26:47
• Objects Launched Upward 27:59
• Acceleration is -g
• At Highest Point, the Object has a Velocity of 0
• Symmetry of Motion
• Example 8: Ball Thrown Upward 28:47
• Example 9: Height of a Jump 29:23
• Example 10: Ball Thrown Downward 33:08
• Example 11: Maximum Height 34:16

### Transcription: Kinematic Equations

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

There is our first kinematic equation.0152

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

There is our second kinematic equation.0258

Now the next step is I am going to come back up here and I am going to go from δx and pull that step back.0264

So we are going to write down δx = 1/2T, V - V initial + V initialT.0271

I am going to start there and just take a slightly different derivation to come up with our next kinematic equation -- our next tool in our toolbox for solving problems.0282

When I do that then, if I start multiplying this out, I get 1/2VT - 1/2V0T + V0T.0292

That implies then that δx = -- I have 1/2VT and -1/2V0T + V0T is just going to be +1/2V0T.0303

If I divide everything by T, I then determine that δx/T = V - V0/2.0316

But realize now δx/T -- that is your average velocity here.0328

So average velocity equals -- Oops that is a positive, that is a plus, plus, plus -- Vaverage = V + Vinitial/2.0332

The average velocity for something with constant acceleration, you take the initial, you take the final, you go halfway in between them.0343

There is another kinematic equation -- another useful tool.0350

Let us go back now to average velocity = δx/T and let us rearrange that to say that δx = Vaverage x T.0356

But we just said that Vaverage is V + V initial/2, so that is going to be V + V initial/2 x T.0369

Now in this next step, we are going to have to go back to some of the stuff we have derived previously.0380

We know V = V0 + AT, therefore we could rearrange this to say that T = V - V0/A.0385

So as I write this again, δx = V + V initial/2 and I am going to replace T here with V - V0/A.0396

This implies then that δx is going to be equal to, -- well now with a little bit of math -- we are going to have V2 - V0V + V0V -- that makes 0 - V02/2A.0415

With a little bit of rearrangement, V2 - V02 = 2Aδx or one more step, V final2 = V initial2 + 2Aδx -- V2 = V initial2 + 2Aδx.0433

So our kinematic equation is -- just to clean these up -- V final = V initial + AT, δx = V initialT+ 1/2AT2, V2 = V initial2 + 2Aδx.0456

And we also threw in there and mentioned that Vaverage = V final + V initial/2 -- the average of those two, and that one is oftentimes neglected because that is more of a common sense one.0471

But all of these work for situations of constant acceleration.0482

Now let us talk about some problem solving steps that will allow you to solve a ton of kinematic equations.0494

What we are going to do is we are going to label the axis for horizontal or vertical motion.0500

When we do that, we are going to pick a direction to be positive.0504

Typically, I like to pick the direction the object is moving initially, but it really does not matter as long as you are consistent within the problem.0507

We are going to create a motion analysis table, where we are going to have V initial, V, δx or y, A, T, and we are going to fill in all the information we are given.0514

When we know any three items in that table, we can solve for the unknowns.0524

Let us see how this works.0530

A race car starting from rest -- starting from rest -- V initial = 0, accelerates uniformly at a rate of 4.9 m/s2. There is our acceleration.0532

What is the car's speed after it has traveled 200 m?0544

All right, well I am going to say it is going to the right and we will call that the positive x direction and I will make my motion table -- V initial, V, δx, A, and T and fill in what I know.0548

V initial = 0 -- I know acceleration is 4.9 m/s2.0563

What is the car's speed after it has traveled 200 m -- that must be our displacement.0570

I know three things in the table -- I can always solve for the other two using my kinematic equations.0575

So I look for a kinematic equation that has all or most of these in it.0581

I want to know the car's final speed.0585

I am going to choose then, V2 = V initial2 + 2Aδx because I want V -- that is there -- I want V initial, I have it -- I need A, I have A and δx.0590

It should work out pretty well.0606

Next, if I want to know its speed, I am going to substitute in my values, V initial, 02 + 2 x A, 4.9 m/s2, x δx, 200 m.0610

Therefore I can say that V2 = 2 x 4.9 x 200 or 1960 m2/s2.0626

But I want just velocity, not velocity2, so I take the square root and say that V must equal the square root of 1960 which is +/-44.3 m/s.0636

Now I have to use a little common sense to figure out which root I want, the positive or the negative, and it should be pretty obvious to see that I want the positive root.0648

The car is not going to accelerate in one direction and then have a velocity in the opposite direction in this case.0655

So my answer, final velocity must be 44.3 m/s.0662

Let us take a look at one that is traveling in the vertical direction.0674

An astronaut standing on a platform on the moon drops a hammer.0678

If the hammer falls 6 m vertically in 2.7 s, what is its acceleration?0680

All right, as we start again, we have our object.0687

It is going to travel down, so we will call down the positive y direction -- that is the direction that it is traveling initially.0691

Let us make our motion table.0697

V initial, V, δy, A, T, -- and if they drop the hammer -- initial velocity -- when you drop something is 0.0700

The hammer falls 6 m vertically, 6 m in a time of 2.7 s.0711

What is its acceleration?0719

So we are looking for A -- we know three things -- we should be able to get there.0720

We have to pick the right kinematic equation now, and I am going to pick δy = V initialT + 1/2AT2.0725

And what is really nice in this scenario is if you see V initial = 0, well 0 x anything is 0, so that whole term goes away and becomes 0.0738

So this term becomes δy = 1/2AT2.0747

I am solving for A so if I multiply both sides by 2, 2δy = AT2 and divide both sides by T2, that would imply then that A = 2δy/T2.0754

Substitute in my variables, 2 times δy, 6m over T2, 2.72, so 12/2.72 should give me about 1.65 m/s2 as the acceleration of the hammer dropped on the moon.0770

And the acceleration due to gravity on the moon is somewhere in the ballpark of 1.6 m/s2, about 1/6th that of Earth, so that makes sense. Great!0794

Let us take a look at the two-step problem.0805

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

A = δV/T which is V final - V initial/T.0916

That shall be 21 m/s - 15m/s / 12s or 0.5 m/s2.0923

We know one more thing, now we can use any equation we want to get δx.0934

Let us say we want to δx = V initialT + 1/2AT2.0940

That will be 15 m/s x 12 s + 1/2 x A, 0.5, T2, 12 s2) -- again it comes out to 216 m.0948

Or we could have even gone and done V2 = V initial2 + 2Aδx.0967

Therefore δx must be V2 - V initial2/2A or 212 - 152/2xA, 1/2, which again comes out to be 216 m.0976

Many different routes to the same conclusion, to the same answer.0998

Let us take a look at an acceleration problem.1003

How long must a 5 kg kitty cat accelerate at 3 m/s2 in order to change its velocity by 9 m/s?1007

Well to do this one we will start off by listing what we know.1015

V initial, V, δx, A and T and what we know here is δv, the change in velocity, is going to be 9 m/s.1020

We do not know specifically V initial or V, but it tells us that A = 3 m/s2, and we want to know the time.1030

Well right away, I would jump to acceleration = δV/T, which will be 9 m/s over -- pardon me --1041

I am looking for time, so I have to solve for T first, which implies then that T = δV/A which will be 9 m/s/3 m/s2 -- 9/3 = 3 m/s/m/s2 -- 3 s.1052

All right. Let us take a look at one that uses a particle diagram.1074

A spark timer -- one of those things that goes "ta-ta-ta-ta-ta" at regular intervals is used to record the position of a lab cart accelerating uniformly from rest.1080

Every 0.1 second the timer marks a dot on a recording tape to indicate the position of the cart at that instant, as shown.1090

We have a ruler and there is our tape. Find the displacement of the cart at time, T = 0.3 s.1097

To begin with, if we want the displacement of the cart, this must be 0.1 s -- 0.2 s -- 0.3 s.1103

All I am going to do there is I am just going to find the displacement by looking up here at my ruler and saying that the displacement there must be pretty close to 9 cm.1114

Now to find the average speed of the cart from 0 - 0.3 s.1125

In that case, if we want the average speed of the cart, Vaverage = δx/T, Which is going to be 0.09 m/time of 0.3 s which is going to be 0.3 m/s.1131

What is the acceleration of the cart?1155

To find the acceleration of the cart, now we have to go back to our kinematics.1157

I would use δx = V0T + 1/2AT2, because we know that V initial, V0 right here is 0 so that term goes away.1162

This implies then that A = 2δx/T2 or 2 x 0.9 m/our time, 0.3 s2 which will give us an acceleration of 2 m/s2.1172

And finally, it asks us on the blank diagram, draw at least 4 dots indicating a cart moving at constant velocity.1193

All right, all I need to do is make sure that my dots are evenly spaced -- then it is moving at constant velocity.1202

Nice review of particle diagrams.1210

Let us take a look at a kinematics problem that may require a quadratic solution.1213

Arnie the aardvark accelerates at a constant 2 m/s2 from an initial velocity of 1 m/s.1217

How long does it take Arnie to cross a distance of 50 m?1224

Well we have V initial = 1 m/s.1228

We do not know final velocity -- δx is going to be 50 m -- it tells us that acceleration is 2 m/s2 and we are asked to find the time -- How long?1233

To pick a kinematic equation that has those four quantities in it, I am going to write δx = V initialT+1/2AT2.1249

Now δx = 50, so that is going to be equal to V initial 1T or T + 1/2 x A2 x T2, therefore 50 = T + T2.1260

That is starting to look like a quadratic equation.1279

What I am going to do is to rearrange that to fit the form of the quadratic formula T2 + T - 50 = 0.1282

Now we can apply the quadratic formula where T = -V+/- the square root of V2 -4AC/2A.1294

Or that is going to be -- well our B is going to be 1, so that is going to be -1 +/- the square root of 12-4 x A, which is 1 x C -- -50/2 x A which is 1.1300

With a little bit of math there, that will be -1 + 14.18/2 or -1 - 14.18/2, which gives us choices of T being equal to either 6.59 s or -7.59 s.1334

Which one makes sense? Well of course, it is that positive time.1360

Now having gone through that, that quadratic formula is a lot of work and very prone to error.1367

So what I would recommend here is if you get to the point where you have a quadratic formula solving for T, solve for the other variable first using kinematic equations, then go solve for T.1371

You can avoid using the quadratic.1381

So the alternate solution -- the way I would typically do this problem to avoid all of that extra math is I would solve for V first using V2 = V initial2 + 2Aδx.1386

V initial2 -- that will be 12 + 2 x A, 2, x δx -- 50 -- so we will find then that V2 = 201 m/s2, therefore V = 14.18 m/s.1399

Now with that, we can go back to another kinematic equation -- any of them we want to get T -- I am just going to use V = V0 + AT.1420

So, let us rearrange that to say that T = V - V0/A which is 14.18 m/s - V initial -- 1 m/s/acceleration -- 2m/s2 and of course I will come out with the same answer, 6.59 s.1431

It is the same thing we had before, but I was able to avoid that quadratic formula and those extra complications.1453

A nice way to simplify the math.1460

Let us talk about free fall.1464

When the only force acting on an object is the force of gravity -- the object's weight -- another name for the force of gravity -- we refer to the object's motion as free fall.1466

This includes objects that have a non-zero initial velocity.1476

So free fall can be something that I throw up in the air, could be something I drop or that I throw down.1479

It does not matter as long as the only force acting on it is gravity.1484

If you drop a ball and a sheet of paper, for example, it is obvious that they do not fall at the same rate.1496

If you could remove all the air from the room, however, you will find that they do fall at the same rate.1500

You will analyze the motion of objects in this course almost always by neglecting air resistance, which is a form of friction.1506

Later on we can add it back in, but for the purposes of this course keeping things simple for now, we are going to neglect air resistance.1513

Let us talk about the acceleration due to gravity.1521

Near the surface of the earth, objects accelerate downward at a rate of 9.8 m/s2.1524

This is such an important number, that in Physics, we call this acceleration little 'g'.1531

For the purposes of the AP exam, we can approximate that as 10 m/s2 to make the math more simple.1537

Save us some time, you will still get full credit for all of your solutions.1544

More accurately though, g is actually referred to as the gravitational field strength.1548

When we get into gravity, we will talk about how we could calculate that -- what that means in a little more depth.1553

Now as we move further away from earth, g decreases.1558

If you were on another planet, g would be different.1560

But on the surface of the Earth, it is 9.8 m/s2 and to make the math easy we can call it 10.1566

Let us talk about objects falling from rest.1575

Objects starting from rest have an initial velocity of V0 of 0 -- falling from rest -- drop it -- initial velocity is 0.1578

Since the object's initial motion is down, I like to call down the positive direction when I am setting out my kinematic equations.1587

The acceleration then, if we are calling down the positive direction, the object accelerates in the same direction, so the acceleration then would be positive g.1594

Let us take a look at an example for a falling object.1606

How far will a brick starting from rest fall freely in 3 s? Again we are going to neglect air resistance here.1611

Well if it is falling, I am going to call down the positive y direction.1617

I am going to set up my kinematic equations table -- V initial, V, δy, A and T.1621

If it is falling from rest, V initial = 0 -- It falls for 3 s and I need to know one more thing.1630

Acceleration due to gravity is +g which I am going to approximate as 10 m/s2.1637

So define how far I am looking for δy, I will use the equation δy = V initialT + 1/2AT2.1645

Then use that favorite trick where V initial = 0 -- that term goes away.1655

So δy = 1/2xA, 10, x time -- 32. That is going to give us 45 m.1660

Objects falling from rest.1675

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

And we have some symmetry of motion there.1706

If I throw up an object for 2 s, it will take 2 s to come back down.1710

If I throw it up with an initial velocity of 10 m/s up -- when it gets right down to the same height, it will be going 10 m/s down.1715

There is a symmetry of motion involved here.1723

So let us take a look at another example. Now we are going to throw a ball upward.1727

A ball thrown vertically upward reaches a maximum height of 30 m above the surface of the earth.1733

Find the speed of the ball at its maximum height.1736

Well right away I am going to call up the positive y direction.1739

Before I get any further it says find the speed of the ball at its maximum height -- a trick question.1743

For the highest point for any object going up and down for a split second, its speed is 0.1750

So the answer at that point is V = 0. You did not even have to do any math there.1765

How about the height of a jump?1774

A basketball player jumps straight up to grab a rebound.1775

If she is in the air for 0.8 s, how high does she jump?1778

Well, here is our basketball player -- jumps up for a split second at highest point -- stops -- comes back down.1782

That entire trip, up and down, took 0.8 s and we know time to go up is 0.4 s and the time to go down is 0.4 s, so that time is 0.4 s.1791

We also know at the highest point that the velocity of the basketball player for a split second is 0.1795

So we can analyze this in a couple of different ways.1802

Let us analyze the basketball player's jump on the way up.1806

Let us just look at that part of the motion, the way up.1810

So on the way up, let us call up the positive direction.1814

We have V initial, V, δy, A and T.1817

We do not know initial velocity, but we know that final velocity on the way up is 0 because we are only looking for this 0.4 s time interval -- so half the time, 0.4 s to go up.1824

We also know if we called up the positive y direction that A is -10 m/s2.1835

Now if we want to figure out how high she jumped -- to what her displacement was at this 0.4 s, I could do this in a number of ways.1841

But why don't I start with finding her initial velocity -- V = V0 + AT, that will save me a quadratic, which implies that V initial = V - AT.1850

This implies that V initial then is going to be V(0) - A(-10) x T(0.4) -- Negative x negative = positive. 10 x 0.4 = 4 m/s.1866

Now I can use δy = V0T + 1/2AT2.1887

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

Δy is what we are trying to find.1949

A is now +10, since we called down the positive direction and our time is 0.4 s.1952

We can use δy = V initialT + 1/2AT2.1960

V initial is 0 so that term goes away -- this becomes 1/2 x A(10), x T2 (0.4)2 or again, 0.8 m.1969

And of course we should get the same answer because we have that symmetry of motion.1983

How about a ball thrown downward?1989

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

So we are looking for a displacement here.2071

All right, we are going to call up the positive direction because that is the way the ball goes initially.2074

Our initial velocity is 30 m/s -- final velocity when it reaches its highest point must be 0 -- another known that we have there.2080

We are trying to find δy.2090

Acceleration is -10 m/s2. Negative because we called up positive and the acceleration due to gravity is down, and we do not know time.2094

Let us choose a kinematic equation that is going to help us out.2104

I want δy so I am going to use V2 = V initial2 + 2Aδy.2107

Let us rearrange this for δy and I get V2 - V initial2/2A.2116

Therefore, δy = V2, that is going to be 0 - V initial2(30)2/2 x A(-10).2125

That is going to be -900/-20, or 45 m.2141

Kinematic equations are terrific tools for problem solving, but you have to remember that they can only be used in situations of constant acceleration.2151

Make that table with V initial, V, δx or δy, A and T, and when you know three of those quantities you can always solve for the other two.2158

Hopefully that gets you a great start.2168

Thanks for watching Educator.com. Make it a great day!2170