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# The kinematic equations

Learn what the kinematic equations are and how you can use them.

## What are the kinematic equations?

The kinematic equations relate the five kinematic variables listed below.
$\mathrm{\Delta }x\phantom{\rule{1em}{0ex}}\text{Displacement}$

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.
$1.\phantom{\rule{1em}{0ex}}v={v}_{0}+at$
$2.\phantom{\rule{1em}{0ex}}\mathrm{\Delta }x=\left(\frac{v+{v}_{0}}{2}\right)t$
$3.\phantom{\rule{1em}{0ex}}\mathrm{\Delta }x={v}_{0}t+\frac{1}{2}a{t}^{2}$
$4.\phantom{\rule{1em}{0ex}}{v}^{2}={v}_{0}^{2}+2a\mathrm{\Delta }x$
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 dimension: horizontal $\left(x\right)$, vertical $\left(y\right)$, etc.

## What is a free falling object (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 $g=9.81\frac{\text{m}}{{\text{s}}^{2}}$.
$g=9.81\frac{\text{m}}{{\text{s}}^{2}}\phantom{\rule{1em}{0ex}}\text{(Magnitude of acceleration due to gravity)}$
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 $g=9.81\frac{\text{m}}{{\text{s}}^{2}}$.
Note that $g=9.81\frac{\text{m}}{{\text{s}}^{2}}$ 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 $\left({a}_{y}=-9.81\frac{\text{m}}{{\text{s}}^{2}}\right)$ for a projectile when we plug into the 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 ${v}_{0}=5\frac{\text{m}}{\text{s}}$, after which it took a time interval for the book to slide a displacement of . We could use the kinematic equation $\mathrm{\Delta }x={v}_{0}t+\frac{1}{2}a{t}^{2}$ to algebraically solve for the unknown acceleration $a$ of the book—assuming the acceleration was constant—since we know every other variable in the equation besides $a.$
Problem solving tip: Note that each kinematic equation is missing one of the five kinematic variables—$\mathrm{\Delta }x,t,{v}_{0},v,a$.
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 $v$ of the book was neither given nor asked for, so we should choose a equation that does not include $v$ at all. The kinematic equation $\mathrm{\Delta }x={v}_{0}t+\frac{1}{2}a{t}^{2}$ is missing $v$, so it's the right choice in this case to solve for the acceleration $a$.

## More problem solving tips

• Don't 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. 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 free falling projectiles is assumed to be $g=9.81\frac{\text{m}}{{\text{s}}^{2}}$, so this acceleration will usually not be given explicitly in a problem but will just be implied for a free falling object.
• Don't forget that all the kinematic variables, 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 free falling object must be negative: ${a}_{g}=-9.81\frac{\text{m}}{{\text{s}}^{2}}$.
• Solving a problem involving the third kinematic equation, $\mathrm{\Delta }x={v}_{0}t+\frac{1}{2}a{t}^{2}$, might require the use of the quadratic formula.
• Don't 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 ${v}_{0}$ 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.

## Try it!

### Example 1

A water balloon is dropped from the top of a very tall building. Assume air resistance is negligible. What is the velocity of the water balloon after falling for ?

### Example 2

A leopard is running at $6.20\frac{\text{m}}{\text{s}}.$ The leopard then speeds up to $23.1\frac{\text{m}}{\text{s}}$ in a time of . How much ground did the leopard cover in going from $6.20\frac{\text{m}}{\text{s}}$ to $23.1\frac{\text{m}}{\text{s}}?$

### Example 3

A student throws her pencil straight upward at $18.3\frac{\text{m}}{\text{s}}.$ How long does it take the pencil to first reach a point higher than where it was thrown?

### Example 4

A motorcyclist starts with a speed of $23.4\frac{\text{m}}{\text{s}}$ and, seeing traffic up ahead, decides to slow down over a length of with a constant deceleration of magnitude . What is the new velocity of the motorcyclist after slowing down over

## Want to join the conversation?

• With equation 3, using the quadratic formula often gives 2 solutions. If both are physically possible (no negative time, anything that doesn't make sense in context of the question), how do I know which solution is correct?
(1 vote)
• You need to apply the solutions to the problem. For example if you are using the equations for ballistic trajectories and you get two answers for when the projectile is at a specific height and you are looking for the velocity it heats a raised platform you are going to want the answer where the projectile has traveled far enough to reach the platform and is traveling down.