Learn where the kinematic equations come from, and how you can use them to analyze scenarios involving constant acceleration.

## What are the kinematic equations?

The kinematic equations relate the five kinematic variables listed below.

The

We just have to make sure we know that by writing

If we know three of these five kinematic variables for an object undergoing **constant acceleration**, we can use a kinematic equation to solve for one of the unknown variables.

The **kinematic equations** are listed below.

In this section, we're just stating the kinematic equations. In the sections below, we'll derive each kinematic equation individually so we can see where they come from and why they're only true assuming constant acceleration.

In the solved examples section below, we will also solve an example problem using each individual equation in order.

Since the kinematic equations *are only accurate if the acceleration is constant during the time interval considered*, we have to be careful to not use them when the acceleration is changing. Also, the kinematic equations assume all variables are referring to the same direction: horizontal

If you plug in a horizontal initial velocity

If you plug in a vertical initial velocity

In other words, to be really clear, the kinematic formulas for a given direction—for example,

However, including all these subscripts can start to get messy. So textbooks, courses, and professors often leave off the subscripts and just remember that the kinematic equations will work for any direction—even diagonal—in which the acceleration is constant, as long as you include only variables in that particular direction.

## What is a free falling object—i.e., a projectile?

It might seem like the fact that the kinematic equations only work for time intervals of constant acceleration would severely limit the applicability of these equations. However one of the most common forms of motion, free fall, just happens to be constant acceleration.

All free falling objects—also called projectiles—on Earth, regardless of their mass, have a constant downward acceleration due to gravity of magnitude

A free falling object is defined as any object that is accelerating only due to the influence of gravity. We often assume the effect of air resistance is small enough to ignore, in which case an object that is dropped, thrown, or otherwise falling freely has a constant downward acceleration of magnitude

Note that

**Warning:** Forgetting to include a negative sign is one of the most common sources of error when using kinematic equations.

## How do you select and use a kinematic equation?

We choose the kinematic equation that includes *both* the unknown variable we're looking for and three of the kinematic variables we already know. This way, we can solve for the unknown we want to find, which will be the only unknown in the equation.

For instance, say we knew a book on the ground was kicked forward with an initial velocity of

**Problem solving tip:** Note that each kinematic equation is missing one of the five kinematic variables—

To choose the kinematic equation that's right for your problem, figure out *which variable you are not given and not asked to find*. For example, in the problem given above, the final velocity

Yes, there is.

The fifth kinematic equation looks just like the third kinematic equation

This fifth kinematic equation often isn't as useful since the initial velocity is typically known/given in many situations.

## How do you derive the first kinematic equation, $v={v}_{0}+at$ ?

This kinematic equation is the easiest to derive since it is really just a rearranged version of the definition of acceleration. We can start with the definition of acceleration,

Yes. If the time interval

So, the derivation we are going through below really assumes that the

So, for the kinematic equation we will derive to be accurate, the acceleration should be constant. Or else the

Now we can replace

Finally if we just solve for

And if we agree to just use **first kinematic equation**.

## How do you derive the second kinematic equation, $\mathrm{\Delta}x=\left({\displaystyle \frac{v+{v}_{0}}{2}}\right)t$ ?

A cool way to visually derive this kinematic equation is by considering the velocity graph for an object with constant acceleration—in other words, a constant slope—and starts with initial velocity

*The area under any velocity graph gives the displacement *. So, the area under this velocity graph will be the displacement

We can conveniently break this area into a blue rectangle and a red triangle as seen in the graph above.

The height of the blue rectangle is

The base of the red triangle is

The total area will be the sum of the areas of the blue rectangle and the red triangle.

If we distribute the factor of

We can simplify by combining the

And finally we can rewrite the right hand side to get the second kinematic equation.

This equation is interesting since if you divide both sides by *average velocity* *average of the final and initial velocities*

## How do you derive the third kinematic equation, $\mathrm{\Delta}x={v}_{0}t+{\displaystyle \frac{1}{2}}a{t}^{2}$ ?

There are a couple ways to derive the equation

Consider an object that starts with a velocity

Since the area under a velocity graph gives the displacement

The term

The term

The base of the red triangle is given by

But this height can be rewritten as

So,

That's it. The equation

Here's the alternative plugging-and-chugging derivation. The third kinematic equation can be derived by plugging in the first kinematic equation,

If we start with second kinematic equation

and we use

We can expand the right hand side and get

Combining the

And finally multiplying both sides by the time

Again, we used other kinematic equations, which have a requirement of constant acceleration, so this third kinematic equation is also only true under the assumption that the acceleration is constant.

## How do you derive the fourth kinematic equation, ${v}^{2}={v}_{0}^{2}+2a\mathrm{\Delta}x$ ?

To derive the fourth kinematic equation, we'll start with the second kinematic equation:

We want to eliminate the time

Multiplying the fractions on the right hand side gives

And now solving for

## What's confusing about the kinematic equations?

People often forget that the *kinematic equations are only true assuming the acceleration is constant* during the time interval considered.

Sometimes a known variable will not be explicitly given in a problem, but rather implied with **codewords**. For instance, "starts from rest" means

People forget that all the kinematic variables—*missing negative sign* is a very common source of error. If upward is assumed to be positive, then the acceleration due to gravity for a free falling object must be negative:

The third kinematic equation, **quadratic formula**, see solved example 3 below.

People forget that even though you can choose any time interval during the constant acceleration, *the kinematic variables* you plug into a kinematic equation *must be consistent* with that time interval. In other words, the initial velocity

## What do solved examples involving the kinematic equations look like?

### Example 1: First kinematic equation, $v={v}_{0}+at$

A water balloon is dropped from the top of a very tall building.

**What is the velocity of the water balloon after falling for **

Assuming upward is the positive direction, our known variables are

If you wait long enough, yes, but we won't use that in our equation for two reasons.

First, we are only considering the time interval of

Second, we can only apply the kinematic equations for a time interval of constant acceleration. When the water balloon hits the ground the acceleration changes dramatically from the collision. So if we are going to claim that the acceleration in our kinematic equation is

If we knew the acceleration upon impact, and if that acceleration was constant we could apply the kinematic equations during the time interval of impact. But that's not the question we're considering here, and in most free fall problems we'll only consider the time after release and the time before impact.

The motion is vertical in this situation, so we'll use

Since we don't know the displacement

**Note:** The final velocity was negative since the water balloon was heading downward.

Yes, we're free to call downward the positive direction for any problem, but we have to be consistent with that decision. If we call downward positive, then every vector pointing downward has to be considered positive.

So the acceleration would have to be

### Example 2: Second kinematic equation, $\mathrm{\Delta}x=\left({\displaystyle \frac{v+{v}_{0}}{2}}\right)t$

A leopard is running at

**How much ground did the leopard cover in going from **

Assuming the initial direction of travel is the positive direction, our known variables are

Since we do not know the acceleration

### Example 3: Third kinematic equation, $\mathrm{\Delta}x={v}_{0}t+{\displaystyle \frac{1}{2}}a{t}^{2}$

A student throws her pencil straight upward at

**How long does it take the pencil to first reach a point **

Assuming upward is the positive direction, our known variables are

Since we don't know the final velocity

Normally we would just solve our expression algebraically for the variable we want to find, but this kinematic equation cannot be solved algebraically for time if none of the terms are zero. That's because when none of the terms are zero and

To put this into a more solvable form of the quadratic equation, we move everything onto one side of the equation. Subtracting

At this point, we solve the quadratic equation for time

So, plugging into the quadratic formula, we get

Since there is a plus or minus sign in the quadratic formula, we get two answers for the time

There are two positive solutions since there are two times when the pencil was

So, to answer our question of "How long does it take the pencil to first reach a point

If the quadratic formula gives a negative time and a positive time, then the object in question only reaches the specified displacement *once* after being thrown. In that case, we can simply choose the positive time and neglect the negative time.

The negative time is referring to a time before the object was thrown when it would have been at the specified displacement. In other words, assume the object had been on its trajectory before being thrown and just happened to reach the point of being thrown with the initial velocity specified.

### Example 4: Fourth kinematic equation, ${v}^{2}={v}_{0}^{2}+2a\mathrm{\Delta}x$

A motorcyclist starts with a speed of

**What is the new velocity of the motorcyclist after slowing down through the **

Assuming the initial direction of travel is the positive direction, our known variables are

Since we don't know the time

Note that in taking a square root, you get two possible answers: positive or negative. Since our motorcyclist will still be going in the direction of motion it started with and we assumed that direction was positive, we'll choose the positive answer

Now we can plug in values to get