Kinematic Equations - Math Steps, Examples & Questions

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Kinematic equations

Here you will learn about the kinematic equations, including what they are and how to use them to solve kinematics problems.

Students will first learn about kinematic equations as part of algebra in high school.

What are kinematic equations?

Kinematic equations are used to solve problems involving moving objects.

Kinematics is the math concerned with the movement of objects.

To do this, you need to know the $5$ different kinematic formulas. They involve $5$ different variables of motion. Velocity is the speed in a given direction. Displacement is the distance from the original position. In all kinematics formulas, the kinematic variables represent:

$s=$ displacement

$u=$ initial velocity

$v=$ final velocity

$a=$ acceleration

$t=$ time

The kinematics formulas are,

v=u+at

s=ut+\cfrac{1}{2} \, at^2

s=vt-\cfrac{1}{2} \, at^2

s=\cfrac{1}{2} \, (u+v)t

v^2=u^2+2as

Kinematic equations can derive one or more variables when the others are known. These equations describe motion under constant velocity or constant acceleration.

Since they only apply to situations with unchanging acceleration or speed, they cannot be used when either is varying.

The kinematic equations or kinematic formulas are sometimes known as the suvat equations or equations of motion.

These kinematic formulas are used when there is constant acceleration. You do not need to worry about other factors affecting the motion of an object such as air resistance.

The kinematic formulas are based on Newton’s laws of motion. You may also come across these formulas in high school physics problems.

When an object is slowing down, it has a negative acceleration. This is known as deceleration.

What are kinematic equations?

Common Core State Standards

How does this relate to high school math?

High School – Algebra – Creating Equations (HS.A.CED.A.4)
Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. For example, rearrange Ohm’s law $V=IR$ to highlight resistance $R.$

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How to work with kinematic equations

In order to work with kinematic equations:

Use the formula given in the question.
Work carefully to answer the question, one step at a time.
Write the final answer clearly.

Kinematic equations examples

Example 1: substituting values

Given that $s=\cfrac{1}{2} \, (u+v)t,$ find $s,$ when $u=5, \, v=20$ and $t=4.$

Use the formula given in the question.

The formula is $s=\cfrac{1}{2} \, (u+v)t$ .

2Work carefully to answer the question, one step at a time.

Substitute the known values into the formula.

\begin{aligned}&s=\cfrac{1}{2} \, (u+v)t \\\\ &s=\cfrac{1}{2}\times(5+20)\times{4} \\\\ &s=50 \end{aligned}

3Write the final answer clearly.

The final answer is $s=50.$

Example 2: rearranging a kinematic formula

Rearrange $v=u+at$ to make $a$ the subject.

Use the formula given in the question.

The formula is $v=u+at.$

Work carefully to answer the question, one step at a time.

You need to rearrange the formula, one step at a time using inverse operations.

$\begin{aligned}&v=u+at \\\\ &v-u=at \\\\ &\cfrac{v-u}{t}=a \end{aligned}$

Write the final answer clearly.

The final answer is $a=\cfrac{v-u}{t}.$

Example 3: solving kinematic equations

For the following question, use the formula $v^2=u^2+2as.$

Calculate $s$ when $v$ is $10, \, u$ is $0$ and $a$ is $20.$

Use the formula given in the question.

The formula given in the question is $v^2=u^2+2as.$

Work carefully to answer the question, one step at a time.

First, substitute the values into the kinematic formula:

$\begin{aligned}v^2&=u^2+2as \\\\ 10^2&=0^2+2\times{20}\times{s} \\\\ 100&=0+40s \end{aligned}$

Then you can rearrange to solve for $s.$

$\begin{aligned}100&=0+40s \\\\ 100&=40s \\\\ \cfrac{100}{40}&=s \\\\ 2.5&=s \end{aligned}$

Write the final answer clearly.

The final answer is $s=2.5.$

Example 4: solving a kinematic problem

For the following question, you may use the formula $s=ut+\cfrac{1}{2} \, at^{2}$ , where

$t=$ time $\hspace{1.5cm} u=$ initial velocity

$a=$ acceleration $\hspace{0.5cm} s=$ displacement

A car is stopped at a traffic light. When the light goes green, the car immediately accelerates at $5\mathrm{~m/s}^{2}.$ How far does the car travel in $10$ seconds?

Use the formula given in the question.

The formula given is $s=ut+\cfrac{1}{2} \, at^2.$

Work carefully to answer the question, one step at a time.

The initial velocity of the object $u$ is $0$ as the car is stationary at the start of the kinematics situation. You substitute the values into the kinematic formula.

$\begin{aligned}s&=ut+\cfrac{1}{2} \, at^{2} \\\\ s&=0\times{10}+\cfrac{1}{2}\times{5}\times{10}^{2} \\\\ s&=250 \end{aligned}$

Write the final answer clearly.

The final answer is, the car has traveled $250\mathrm{~m}$ in $10$ seconds.

Example 5: solving a kinematic problem

For the following question, you may use the formula $s=vt-\cfrac{1}{2} \, at^{2}$ , where

$t=$ time $\hspace{1.5cm} v=$ final velocity

$a=$ acceleration $\hspace{0.5cm} s=$ displacement

A ball was dropped from stationary and accelerated at $10\mathrm{~m/s}^{2}.$ The ball passed a motion sensor after $15\mathrm{~s},$ travelling at $90\mathrm{~m/s}.$ How far was the motion sensor away from the initial position of the ball?

Use the formula given in the question.

The formula given is $s=vt-\cfrac{1}{2} \, at^2.$

Work carefully to answer the question, one step at a time.

The final velocity of the object $v=90\mathrm{~m/s}$ with $t=15\mathrm{~s}$ and $a=10\mathrm{~m/s}^{2}.$ You substitute the values into the kinematic formula.

$\begin{aligned}s&=vt-\cfrac{1}{2} \, at^{2} \\\\ s&=90\times{15}-\cfrac{1}{2}\times{10}\times{15}^{2} \\\\ s&=225 \end{aligned}$

Write the final answer clearly.

The motion sensor is $225\mathrm{~m}$ below the initial position of the ball.

Teaching tips for kinematic equations

Use position-time and velocity-time graphs to illustrate the relationships between the variables. This helps students see how changes in one variable affect another. Graphs make the abstract set of equations more concrete. Look for online interactive graphs to make this more engaging.

Guide students through the derivation of kinematic equations by starting with the basic definitions of velocity and acceleration. Start with $v=u+at,$ and then progress to the equations for displacement and average velocity.

Once students are comfortable with basic kinematic problem solving, introduce free-fall or projectile motion problems. Show how the same equations can be applied, just with specific values for acceleration due to the gravitation on Earth.

Focus on one-dimensional motion, where objects move along a straight path, either horizontally or vertically, without lateral movement. This simplicity helps students grasp more complex motion later, such as two-dimensional motion or rotational motion around an axis.

For a more advanced tip, introduce angular acceleration after students understand linear acceleration. Explain how angular acceleration relates to rotational motion, where the rate of change of angular velocity mirrors linear acceleration in circular paths.

Easy mistakes to make

Not knowing acceleration can be negative
You may be asked to work out the acceleration of a situation involving the motion of an object. Acceleration can be negative. When this happens the object is slowing down. This is known as deceleration.

Mixing up $\textbf{v}$ and $\textbf{u}$
The variables $v$ and $u$ can easily be muddled as they look similar and they both represent velocity. $u$ is the initial velocity; the velocity of the object at the beginning of the situation. $v$ is the final velocity; the velocity at the end of the situation.

Inconsistent units
The units that are mostly used in kinematics are $m$ (meters) for displacement, $s$ (seconds) for time, $m/s$ (meters per second) for velocity and $m/s^{2}$ (meters per second squared) for acceleration. Make sure the units in your answer are consistent with those of the given values.

Not knowing initial velocity of an object at rest
Sometimes you are told the object’s initial position is at rest, this means that the initial velocity $u$ is $0\mathrm{~m/s}.$

Practice kinematic equations questions

10

90

30

70

Substitute the values into the equation and solve.

\begin{aligned}v&=u+at \\\\ v&=50+(- \, 10)\times{4} \\\\ v&=50-40 \\\\ v&=10 \end{aligned}

540

15

30

270

Substitute the values into the equation and solve.

\begin{aligned}s&=\cfrac{1}{2} \, (u+v)t \\\\ s&=\cfrac{1}{2}\times(10+80)\times{6} \\\\ s&=\cfrac{1}{2}\times{90}\times{6} \\\\ s&=270 \end{aligned}

s=\cfrac{v^{2}+2a}{u^2}

s=\cfrac{v^{2}-u^{2}}{2a}

s=\cfrac{v^{2}+u^{2}}{2a}

s=\cfrac{v^{2}-2a}{u^{2}}

Rearrange, taking one step at a time.

\begin{aligned}&v^{2}=u^{2}+2as \\\\ &v^{2}-u^{2}=2as \\\\ &\cfrac{v^{2}-u^{2}}{2a}=s \end{aligned}

180

300

140

360

Substitute the known values into the formula, then solve to find the unknown variable $u.$

\begin{aligned}v&=u+at \\\\ 200&=u+20\times{3} \\\\ 200&=u+60 \\\\ u&=140 \end{aligned}

$40$ seconds

$5$ seconds

$40$ minutes

$10$ minutes

Substitute the known values into the formula, then rearrange to solve to find the unknown variable $t.$

\begin{aligned}v&=u+at \\\\ 50&=0+10\times{t} \\\\ 50&=10t \\\\ t&=5 \end{aligned}

The answer will be $</span><span style="font-weight: 400;">5</span><span style="font-weight: 400;">$ seconds, as seconds is consistent with the other units.

67.5\mathrm{~m}

112.5\mathrm{~m}

82.5\mathrm{~m}

97.5\mathrm{~m}

Substitute the known values into the formula.

\begin{aligned}s&=ut+\cfrac{1}{2} \, at^2 \\\\ s&=30\times 3+\cfrac{1}{2}\times(- \, 5)\times{3^2} \\\\ s&=90-22.5 \\\\ s&=67.5 \end{aligned}

The answer will be $67.5$ meters, as meters is consistent with the other units.

Kinematic equations FAQs

What are kinematic equations?

Kinematic equations describe the motion of objects under constant acceleration. They relate variables like displacement, velocity, time intervals, and acceleration.

What is the difference between displacement and distance in kinematic equations?

Displacement is the straight-line distance between the starting and ending point of an object’s motion, considering direction. Distance is the total path length traveled by the object, regardless of direction. Kinematic equations deal with displacement, not distance.

How are the kinematic equations related to each other?

Each equation is derived from the others using basic algebra and the definition of acceleration. For example, you can derive the second equation from the first kinematic equation by substituting velocity into the formula for displacement.

The next lessons are

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