# What are the kinematicÂ formulas?

Here are the main equations you can use to analyze situations with constant acceleration.

## What are the kinematic formulas?

The kinematic formulas are a set of formulas that relate the five kinematic variables listed below.

$\Delta x\quad\text{Displacement}$

$t\qquad\text{Time interval}~$

$v_0 ~~\quad\text{Initial velocity}~$

$v\quad ~~~\text{Final velocity}~$

$a \quad~~ \text{ Constant acceleration}~$

$t\qquad\text{Time interval}~$

$v_0 ~~\quad\text{Initial velocity}~$

$v\quad ~~~\text{Final velocity}~$

$a \quad~~ \text{ Constant acceleration}~$

If we know three of these five kinematic variablesâ€”$\Delta x, t, v_0, v, a$â€”for an object under

**constant acceleration**, we can use a kinematic formula, see below, to solve for one of the unknown variables.The

**kinematic formulas**are often written as the following four equations.Since the kinematic formulas

*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 formulas assume all variables are referring to the same direction: horizontal $x$, vertical $y$, etc.## What is a freely flying objectâ€”i.e., a projectile?

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

All freely flying objectsâ€”also called projectilesâ€”on Earth, regardless of their mass, have a constant downward acceleration due to gravity of magnitude $g=9.81\dfrac{\text{m}}{\text{s}^2}$.

A freely flying object is defined as any object that is accelerating only due to the influence of gravity. We typically assume the effect of air resistance is small enough to ignore, which means any object that is dropped, thrown, or otherwise flying freely through the air is typically assumed to be a freely flying projectile with a constant downward acceleration of magnitude $g=9.81\dfrac{\text{m}}{\text{s}^2}$.

This is both strange and lucky if we think about it. It's strange since this means that a large boulder will accelerate downwards with the same acceleration as a small pebble, and if dropped from the same height, they would strike the ground at the same time.

It's lucky since we don't need to know the mass of the projectile when solving kinematic formulas since the freely flying object will have the same magnitude of acceleration, $g=9.81\dfrac{\text{m}}{\text{s}^2}$, no matter what mass it hasâ€”as long as air resistance is negligible.

Note that $g=9.81\dfrac{\text{m}}{\text{s}^2}$ is just the magnitude of the acceleration due to gravity. If upward is selected as positive, we must make the acceleration due to gravity negative $a_y=-9.81\dfrac{\text{m}}{\text{s}^2}$ for a projectile when we plug into the kinematic formulas.

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

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

We choose the kinematic formula 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 formula.For instance, say we knew a book on the ground was kicked forward with an initial velocity of $v_0=5\text{ m/s}$, after which it took a time interval $t=3\text{ s}$ for the book to slide a displacement of $\Delta x=8\text{ m}$. We could use the kinematic formula $\Delta x=v_0 t+\dfrac{1}{2}at^2$ to algebraically solve for the unknown acceleration $a$ of the bookâ€”assuming the acceleration was constantâ€”since we know every other variable in the formula besides $a$â€”$\Delta x, v_0, t$.

**Problem solving tip:**Note that each kinematic formula is missing one of the five kinematic variablesâ€”$\Delta x, t, v_0, v, a$.

To choose the kinematic formula 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 $v$ of the book was neither given nor asked for, so we should choose a formula that does not include $v$ at all. The kinematic formula $\Delta x=v_0 t+\dfrac{1}{2}at^2$ is missing $v$, so it's the right choice in this case to solve for the acceleration $a$.## How do you derive the first kinematic formula, $v=v_0+at$ ?

This kinematic formula is probably 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,

$a=\dfrac{\Delta v}{\Delta t}$ $\quad$

Now we can replace $\Delta v$ with the definition of change in velocity $v-v_0$.

Finally if we just solve for $v$ we get

And if we agree to just use $t$ for $\Delta t$, this becomes the

**first kinematic formula**.## How do you derive the second kinematic formula, ${\Delta x}=(\dfrac{v+v_0}{2})t$?

A cool way to visually derive this kinematic formula is by considering the velocity graph for an object with constant accelerationâ€”in other words, a constant slopeâ€”and starts with initial velocity $v_0$ as seen in the graph below.

*The area under any velocity graph gives the displacement $\Delta x$*. So, the area under this velocity graph will be the displacement $\Delta x$ of the object.

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 $v_0$ and the width is $t$, so the area of the blue rectangle is $v_0t$.

The base of the red triangle is $t$ and the height is $v-v_0$, so the area of the red triangle is $\dfrac{1}{2}t(v-v_0)$.

The base of the red triangle is $t$ and the height is $v-v_0$, so the area of the red triangle is $\dfrac{1}{2}t(v-v_0)$.

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

If we distribute the factor of $\dfrac{1}{2}t$ we get

We can simplify by combining the $v_0$ terms to get

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

This formula is interesting since if you divide both sides by $t$, you get $\dfrac{\Delta x}{t}=(\dfrac{v+v_0}{2})$. This shows that the

*average velocity*$\dfrac{\Delta x}{t}$ equals the*average of the final and initial velocities*$\dfrac{v+v_0}{2}$. However, this is only true assuming the acceleration is constant since we derived this formula from a velocity graph with constant slope/acceleration.## How do you derive the third kinematic formula, $\Delta x=v_0 t+\dfrac{1}{2}at^2$?

There are a couple ways to derive the equation $\Delta x=v_0 t+\dfrac{1}{2}at^2$. There's a cool geometric derivation and a less exciting plugging-and-chugging derivation. We'll do the cool geometric derivation first.

Consider an object that starts with a velocity $v_0$ and maintains constant acceleration to a final velocity of $v$ as seen in the graph below.

Since the area under a velocity graph gives the displacement $\Delta x$, each term on the right hand side of the formula $\Delta x=v_0 t+\dfrac{1}{2}at^2$ represents an area in the graph above.

The term $v_0t$ represents the area of the blue rectangle since $A_{rectangle}=hw$.

The term $\dfrac{1}{2}at^2$ represents the area of the red triangle since $A_{triangle}=\dfrac{1}{2}bh$.

That's it. The formula $\Delta x=v_0 t+\dfrac{1}{2}at^2$ has to be true since the displacement must be given by the total area under the curve. We did assume the velocity graph was a nice diagonal line so that we could use the triangle formula, so this kinematic formulaâ€”like all the rest of the kinematic formulasâ€”is only true under the assumption that the acceleration is constant.

Here's the alternative plugging-and-chugging derivation. The third kinematic formula can be derived by plugging in the first kinematic formula, $v=v_0+at$, into the second kinematic formula, $\dfrac{\Delta x}{t}=\dfrac{v+v_0}{2}$.

If we start with second kinematic formula

and we use $v=v_0+at$ to plug in for $v$, we get

We can expand the right hand side and get

Combining the $\dfrac{v_0}{2}$ terms on the right hand side gives us

And finally multiplying both sides by the time $t$ gives us the third kinematic formula.

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

## How do you derive the fourth kinematic formula, $v^2=v_0^2+2a\Delta x$?

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

We want to eliminate the time $t$ from this formula. To do this, we'll solve the first kinematic formula, $v=v_0+at$, for time to get $t=\dfrac{v-v_0}{a}$. If we plug this expression for time $t$ into the second kinematic formula we'll get

Multiplying the fractions on the right hand side gives

And now solving for $v^2$ we get the fourth kinematic formula.

## What's confusing about the kinematic formulas?

People often forget that the

*kinematic formulas 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 $v_0=0$, "dropped" often means $v_0=0$, and "comes to a stop" means $v=0$. Also, the magnitude of the acceleration due to gravity on all freely flying projectiles is assumed to be $g=9.81\dfrac{\text{m}}{\text{s}^2}$, so this acceleration will usually not be given explicitly in a problem but will just be implied for a freely flying object.People forget that all the kinematic variablesâ€”$\Delta x, v_o, v, a$â€”except for $t$ can be negative. A

*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 freely flying object must be negative: $a_g=-9.81\dfrac{\text{m}}{\text{s}^2}$.The third kinematic formula, $\Delta x=v_0 t+\dfrac{1}{2}at^2$, might require the use of the

**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 formula*must be consistent*with that time interval. In other words, the initial velocity $v_0$ has to be the velocity of the object at the initial position and start of the time interval $t$. Similarly, the final velocity $v$ must be the velocity at the final position and end of the time interval $t$ being analyzed.## What do solved examples involving the kinematic formulas look like?

### Example 1: First kinematic formula, $v=v_0+at$

A water balloon filled with Kool-Aid is dropped from the top of a very tall building.

**What is the velocity of the water balloon after falling for $t=2.35 \text{ s}$?**

Assuming upward is the positive direction, our known variables are

$v_0=0 \quad$ (Since the water balloon was dropped, it started at rest.)

$t=2.35\text{ s} \quad$ (This is the time interval after which we want to find the velocity.)

$a_g=-9.81\dfrac{\text{m}}{\text{s}^2} \quad$(This is implied since the water balloon is a freely falling object.)

$t=2.35\text{ s} \quad$ (This is the time interval after which we want to find the velocity.)

$a_g=-9.81\dfrac{\text{m}}{\text{s}^2} \quad$(This is implied since the water balloon is a freely falling object.)

The motion is vertical in this situation, so we'll use $y$ as our position variable instead of $x$. The symbol we choose doesn't really matter as long as we're consistent, but people typically use $y$ to indicate vertical motion.

Since we don't know the displacement $\Delta y$ and we weren't asked for the displacement $\Delta y$, we'll use the first kinematic formula $v=v_{0}+at$, which is missing $\Delta y$.

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

### Example 2: Second kinematic formula, ${\Delta x}=(\dfrac{v+v_0}{2})t$

A leopard is running at 6.20 m/s and after seeing a mirage that's taken the form of an ice cream truck; the leopard then speeds up to 23.1 m/s in a time of 3.3 s.

**How much ground did the leopard cover in going from 6.20 m/s to 23.1 m/s?**

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

$v_0= 6.20\text{ m/s} \quad$ (The initial speed of the leopard)

$v= 23.1\text{ m/s} \quad$ (The final speed of the leopard)

$t=3.30\text{ s} \quad$ (The time it took for the leopard to speed up)

$v= 23.1\text{ m/s} \quad$ (The final speed of the leopard)

$t=3.30\text{ s} \quad$ (The time it took for the leopard to speed up)

Since we do not know the acceleration $a$ and were not asked for the acceleration, we'll use the second kinematic formula for the horizontal direction ${\Delta x}=(\dfrac{v+v_{0}}{2})t$, which is missing $a$.

### Example 3: Third kinematic formula, $\Delta x=v_0 t+\dfrac{1}{2}at^2$

A student is fed up with doing her kinematic formula homework, so she throws her pencil straight upward at 18.3 m/s.

**How long does it take the pencil to first reach a point 12.2 m higher than where it was thrown?**

Assuming upward is the positive direction, our known variables are

$v_0=18.3 \text { m/s} \quad$ (The initial upward velocity of the pencil)

$\Delta y=12.2\text{ m} \quad$ (We want to know the time when the pencil moves through this displacement.)

$a=-9.81\dfrac{\text{ m}}{\text{ s}^2} \quad$ (The pencil is a freely flying projectile.)

$\Delta y=12.2\text{ m} \quad$ (We want to know the time when the pencil moves through this displacement.)

$a=-9.81\dfrac{\text{ m}}{\text{ s}^2} \quad$ (The pencil is a freely flying projectile.)

Since we don't know the final velocity $v$ and we weren't asked to find the final velocity, we will use the third kinematic formula for the vertical direction $\Delta y=v_{0y} t+\dfrac{1}{2}a_yt^2$, which is missing $v$.

Normally we would just solve our expression algebraically for the variable we want to find, but this kinematic formula can not be solved algebraically for time if none of the terms are zero. That's because when none of the terms are zero and $t$ is the unknown variable, this equation becomes a quadratic equation. We can see this by plugging in known values.

To put this into a more solvable form of the quadratic equation, we move everything onto one side of the equation. Subtracting 12.2 m from both sides we get

At this point, we solve the quadratic equation for time $t$. The solutions of a quadratic equation in the form of $at^2+bt+c=0$ are found by using the quadratic formula $t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$. For our kinematic equation $a=\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})$, $b=18.3\text{ m/s}$, and $c=-12.2\text{ m}$.

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 $t$: one when using the $+$ and one when using the $-$. Solving the quadratic formula above gives these two times:

$t=0.869\text{ s}$ and $t=2.86\text{ s}$

There are two positive solutions since there are two times when the pencil was 12.2 m high. The smaller time refers to the time required to go upward and first reach the displacement of 12.2 m high. The larger time refers to the time required to move upward, pass through 12.2 m high, reach a maximum height, and then fall back down to a point 12.2 m high.

So, to find the answer to our question of "How long does it take the pencil to first reach a point 12.2 m higher than where it was thrown?" we would choose the smaller time $t=0.869\text{ s}$.

### Example 4: Fourth kinematic formula, $v^2=v_0^2+2a\Delta x$

A European motorcyclist starts with a speed of 23.4 m/s and, seeing traffic up ahead, decides to slow down over a length of 50.2 m with a constant deceleration of magnitude $3.20\dfrac{\text{ m}}{\text{ s}^2}$. Assume the motorcycle is moving forward for the entire trip.

**What is the new velocity of the motorcyclist after slowing down through the 50.2 m?**

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

$v_0=23.4 \text { m/s} \quad$ (The initial forward velocity of the motorcycle)

$a=-3.20\dfrac{\text{ m}}{\text{ s}^2} \quad$ (Acceleration is negative since the motorcycle is slowing down and we assumed forward is positive.)

$\Delta x=50.2\text{ m} \quad$ (We want to know the velocity after the motorcycle moves through this displacement.)

$a=-3.20\dfrac{\text{ m}}{\text{ s}^2} \quad$ (Acceleration is negative since the motorcycle is slowing down and we assumed forward is positive.)

$\Delta x=50.2\text{ m} \quad$ (We want to know the velocity after the motorcycle moves through this displacement.)

Since we don't know the time $t$ and we weren't asked to find the time, we will use the fourth kinematic formula for the horizontal direction $v_x^2=v_{0x}^2+2a_x\Delta x$, which is missing $t$.

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 $v_x=+\sqrt{v_{0x}^2+2a_x\Delta x}$.

Now we can plug in values to get