## Kinematic Equations: Explanation, Review, and Examples

- The Albert Team
- Last Updated On: April 29, 2022

Now that you’ve learned about displacement, velocity, and acceleration, you’re well on your way to being able to describe just about any motion you could observe around you with physics. All that’s left is to learn how these values really play into each other. We know a few ways to move between them, but they’re all pretty limited. What happens if you need to find displacement, but only know acceleration and time? We don’t have a way to combine all of those values yet. Enter the four kinematic equations.

What We Review

## The Kinematic Equations

The following four kinematic equations come up throughout physics from the earliest high school class to the highest level college course:

Don’t let all of these numbers and symbols intimidate you. We’ll talk through each one – what they mean and when we use them. By the end of this post, you’ll be a master of understanding and implementing each of these physics equations. Let’s start with defining what all of those symbols mean.

## The First Kinematic Equation

This physics equation would be read as “the final velocity is equal to the initial velocity plus acceleration times time”. All it means is that if you have constant acceleration for some amount of time, you can find the final velocity. You’ll use this one whenever you’re looking at changing velocities with a constant acceleration.

## The Second Kinematic Equation

This one is read as “displacement equals final velocity plus initial velocity divided by two times time”. You’ll use this one whenever you don’t have an acceleration to work with but you need to relate a changing velocity to a displacement.

## The Third Kinematic Equation

This one may look a bit scarier as it is longer than the others, but it is read as “displacement equals initial velocity times time plus one half acceleration times time squared”. All it means is that our displacement can be related to our initial velocity and a constant acceleration without having to find the final velocity. You’ll use this one when final velocity is the only value you don’t know yet.

It is worth noting that this kinematic equation has another popular form: x=x_{0}+v_{0}t+\frac{1}{2}at^{2} . While that may seem even more intimidating, it’s actually exactly the same. The only difference here is that we have split up \Delta x into x-x_{0} and then solved to get x on its own. This version can be particularly helpful if you’re looking specifically for a final or initial position rather than just an overall displacement.

## The Fourth Kinematic Equation

Our last kinematic equation is read as “final velocity squared equals initial velocity squared plus two times acceleration times displacement”. It’s worth noting that this is the only kinematic equation without time in it. Many starting physicists have been stumped by reaching a problem without a value for time. While staring at an equation sheet riddled with letters and numbers can be overwhelming, remembering you have this one equation without time will come up again and again throughout your physics career.

It may be worth noting that all of these are kinematic equations for constant acceleration. While this may seem like a limitation, we learned before that high school physics courses generally utilize constant acceleration so we don’t need to worry about it changing yet. If you do find yourself in a more advanced course, new physics equations will be introduced at the appropriate times.

## How to Approach a Kinematics Problem

So now that we have all of these different kinematic equations, how do we know when to use them? How can we look at a physics word problem and know which of these equations to apply? You must use problem-solving steps. Follow these few steps when trying to solve any complex problems, and you won’t have a problem.

## Step 1: Identify What You Know

This one probably seems obvious, but skipping it can be disastrous to any problem-solving endeavor. In physics problems, this just means pulling out values and directions. If you can add the symbol to go with the value (writing t=5\text{ s} instead of just 5\text{ s} , for example), even better. It’ll save time and make future steps even easier.

## Step 2: Identify the Goal

In physics, this means figuring out what question you’re actually being asked. Does the question want you to find the displacement? The acceleration? How long did the movement take? Figure out what you’re being asked to do and then write down the symbol of the value you’re solving for with a question mark next to it ( t=\text{?} , for example). Again, this feels obvious, but it’s also a vital step.

## Step 3: Gather Your Tools

Generally, this means a calculator and an equation. You’ll want to look at all of the symbols you wrote down and pick the physics equation for all of them, including the unknown value. Writing everything down beforehand will make it easier to pull a relevant equation than having to remember what values you need while searching for the right equation. You can use the latter method, but you’re far more likely to make a mistake and feel frustrated that way.

## Step 4: Put it all Together

Plug your values into your equation and solve for the unknown value. This will usually be your last step, though you may find yourself having to repeat it a few times for exceptionally complex problems. That probably won’t come up for quite a while, though. After you’ve found your answer, it’s generally a good idea to circle it to make it obvious. That way, whoever is grading you can find it easily and you can easily keep track of which problems you’ve already completed while flipping through your work.

## Kinematic Equation 1: Review and Examples

To learn how to solve problems with these new, longer equations, we’ll start with v=v_{0}+at . This kinematic equation shows a relationship between final velocity, initial velocity, constant acceleration, and time. We will explore this equation as it relates to physics word problems. This equation is set up to solve for velocity, but it can be rearranged to solve for any of the values it contains. For this physics equation and the ones following, we will look at one example finding the variable that has already been isolated and one where a new variable needs to be isolated using the steps we just outlined. So, let’s jump into applying this kinematic equation to a real-world problem.

A car sits at rest waiting to merge onto a highway. When they have a chance, they accelerate at 4\text{ m/s}^2 for 7\text{ s} . What is the car’s final velocity?

We have a clearly stated acceleration and time, but there’s no clearly defined initial velocity here. Instead, we have to take this from context. We know that the car “sits at rest” before it starts moving. This means that our initial velocity in this situation is zero. Other context clues for an object starting at rest is if it is “dropped” or if it “falls”. Our other known values will be even easier to pull as we were actually given numerical values. Now it’s time to put everything into a list.

- v_{0}=0\text{ m/s}
- a=4\text{ m/s}^2
- t=7\text{ s}

Our goal here was clearly stated: find the final velocity. We’ll still want to list that out so we can see exactly what symbols we have to work with on this problem.

We already know which of the kinematic equations we’re using, but if we didn’t, this would be where we search our equation sheet for the right one. Regardless, we’ll want to write that down too.

## Step 4: Put it All Together

At this point, we’ll plug all of our values into our kinematic equation. If you’re working on paper, there’s no need to repeat anything we’ve put above. That being said, for the purposes of digital organization and so you can see the full problem in one spot, we will be rewriting things here.

Now let’s get a bit trickier with a problem that will require us to rearrange our kinematic equation.

A ball rolls toward a hill at 3\text{ m/s} . It rolls down the hill for 5\text{ s} and has a final velocity of 18\text{ m/s} . What was the ball’s acceleration as it rolled down the hill?

Just like before, we’ll make a list of our known values:

- v_{0}=3\text{ m/s}
- t=5\text{ s}
- v=18\text{ m/s}

Again, our goal was clearly stated, so let’s add it to our list:

We already know which equation we’re using, but let’s pretend we didn’t. We know that we need to solve for acceleration, but if you look at our original list of kinematic equations, there isn’t one that’s set up to solve for acceleration:

This begs the question, how to find acceleration (or any value) that hasn’t already been solved for? The answer is to rearrange an equation. First, though, we need to pick the right one. We start by getting rid of the second equation in this list as it doesn’t contain acceleration at all. Our options are now:

- \Delta x=v_{0}t+\dfrac{1}{2}at^{2}
- v^{2}=v_{0}^{2}+2a\Delta x

Now we’ll need to look at the first list we made of what we know. We know the initial velocity, time, and final velocity. There’s only one equation that has all the values we’re looking for and all of the values we know with none that we don’t. This is the first kinematic equation:

In this case, we knew the kinematic equation coming in so this process of elimination wasn’t necessary, but that won’t often be the case in the future. You’ll likely have to find the correct equation far more often than you’ll have it handed to you. It’s best to practice finding it now while we only have a few equations to work with.

Like before, we’ll be rewriting all of our relevant information below, but you won’t need to if you’re working on paper.

Although you can plug in values before rearranging the equation, in physics, you’ll usually see the equation be rearranged before values are added. This is mainly done to help keep units where they’re supposed to be and to avoid any mistakes that could come from moving numbers and units rather than just a variable. We’ll be taking the latter approach here. Follow the standard PEMDAS rules for rearranging the equation and then write it with the variable we’ve isolated on the left. While that last part isn’t necessary, it is a helpful organizational practice:

For a review of solving literal equations, visit this post ! Now we can plug in those known values and solve:

## Kinematic Equation 2: Review and Examples

Next up in our four kinematics equations is \Delta x=\dfrac{v+v_{0}}{2} t . This one relates an object’s displacement to its average velocity and time. The right-hand side shows the final velocity plus the initial velocity divided by two – the sum of some values divided by the number of values, or the average. Although this equation doesn’t directly show a constant acceleration, it still assumes it. Applying this equation when acceleration isn’t constant can result in some error so best not to apply it if a changing acceleration is mentioned.

A car starts out moving at 10\text{ m/s} and accelerates to a velocity of 24\text{ m/s} . What displacement does the car cover during this velocity change if it occurs over 10\text{ s} ?

- v_{0}=10\text{ m/s}
- v=24\text{ m/s}
- t=10\text{ s}
- \Delta x=\text{?}
- \Delta x=\dfrac{v+v_{0}}{2} t

This time around we won’t repeat everything here. Instead, We’ll jump straight into plugging in our values and solving our problem:

A ball slows down from 15\text{ m/s} to 3\text{ m/s} over a distance of 36\text{ m} . How long did this take?

- v_{0}=15\text{ m/s}
- v=3\text{ m/s}
- \Delta x=36\text{ m}

We don’t have a kinematic equation for time specifically, but we learned before that we can rearrange certain equations to solve for different variables. So, we’ll pull the equation that has all of the values we need and isolate the variable we want later:

Again, we won’t be rewriting anything, but we will begin by rearranging our equation to solve for time:

Now we can plug in our known values and solve for time.

## Kinematic Equation 3: Review and Examples

Our next kinematic equation is \Delta x=v_{0}t+\frac{1}{2}at^{2} . This time we are relating our displacement to our initial velocity, time, and acceleration. The only odd thing you may notice is that it doesn’t include our final velocity, only the initial. This equation will come in handy when you don’t have a final velocity that was stated either directly as a number or by a phrase indicating the object came to rest. Just like before, we’ll use this equation first to find a displacement, and then we’ll rearrange it to find a different value.

A rocket is cruising through space with a velocity of 50\text{ m/s} and burns some fuel to create a constant acceleration of 10\text{ m/s}^2 . How far will it have traveled after 5\text{ s} ?

- v_{0}=50\text{ m/s}
- a=10\text{ m/s}^2
- \Delta x=v_{0}t+\frac{1}{2}at^{2}

At this point, it appears that these problems seem to be quite long and take several steps. While that is an inherent part of physics in many ways, it will start to seem simpler as time goes on. This problem presents the perfect example. While it may have been easy to combine lines 4 and 5 mathematically, they were shown separately here to make sure the process was as clear as possible. While you should always show all of the major steps of your problem-solving process, you may find that you are able to combine some of the smaller steps after some time of working with these kinematic equations.

Later in its journey, the rocket is moving along at 20\text{ m/s} when it has to fire its thrusters again. This time it covers a distance of 500\text{ m} in 10\text{ s} . What was the rocket’s acceleration during this thruster burn?

- v_{0}=20\text{ m/s}
- \Delta x=500\text{ m}

As usual, we’ll begin by rearranging the equation, this time to solve for acceleration.

Now we can plug in our known values to find the value of our acceleration.

## Kinematic Equation 4: Review and Examples

The last of the kinematic equations that we will look at is v^{2}=v_{0}^{2}+2a\Delta x . This one is generally the most complicated looking, but it’s also incredibly important as it is our only kinematic equation that does not involve time. It relates final velocity, initial velocity, acceleration, and displacement without needing a time over which a given motion occurred. For this equation, as with the others, let’s solve it as is and then rearrange it to solve for a different variable.

A car exiting the highway begins with a speed of 25\text{ m/s} and travels down a 100\text{ m} long exit ramp with a deceleration (negative acceleration) of 3\text{ m/s}^2 . What is the car’s velocity at the end of the exit ramp?

- v_{0}=25\text{ m/s}
- \Delta x=100\text{ m}
- a=-3\text{ m/s}^2

Note that our acceleration here is a negative value. That is because our problem statement gave us a deceleration instead of an acceleration. Whenever you have a deceleration, you’ll make the value negative to use it as an acceleration in your problem-solving. This also tells us that our final velocity should be less than our initial velocity so we can add that to the list of what we know as well.

- Final velocity will be less than initial.

Being able to know something to help check your answer at the end is what makes this subject a bit easier than mathematics for some students.

While we generally try to not have any operations going on for the isolated variable, sometimes it’s actually easier that way. Having your isolated variable raised to a power is generally a time to solve before simplifying. This may seem like an arbitrary rule, and in some ways it is, but as you continue through your physics journey you’ll come up with your own practices for making problem-solving easier.

Now that we have both sides simplified, we’ll take the square root to eliminate the exponent on the left-hand side:

If we remember back at the beginning, we said that our final velocity would have to be less than our initial velocity because the problem statement told us that we were decelerating. Our initial velocity was 25\text{ m/s} which is, indeed, greater than 5\text{ m/s} so our answer checks out.

A ghost is sliding a wrench across a table to terrify the mortal onlooker. The wrench starts with a velocity of 2\text{ m/s} and accelerates to a velocity of 5\text{ m/s} over a distance of 7\text{ m} . What acceleration did the ghost move the wrench with?

- v_{0}=2\text{ m/s}
- v=5\text{ m/s}
- \Delta x=7\text{ m}

We can also make an inference about our acceleration here – that it will be positive. Not every problem will tell you clearly the direction of the acceleration, but if your final velocity is greater than your initial velocity, you can be sure that your acceleration will be positive.

- Positive acceleration

You’ll get better at picking up on subtle hints like this as you continue your physics journey and your brain starts naturally picking up on some patterns. You’ll likely find this skill more and more helpful as it develops and as problems get more difficult.

We’ll start by rearranging our equation to solve for acceleration.

As usual, now that we’ve rearranged our equation, we can plug in our values.

Again, we can go back to the beginning when we said our acceleration would be a positive number and confirm that it is.

## Problem-Solving Strategies

At this point, you’re likely getting the sense that physics will be a lot of complex problem-solving. If so, your senses are correct. In many ways, physics is the science of explaining nature with mathematical equations. There’s a lot that goes into developing and applying these equations, but at this point in your physics career, you’ll find that the majority of your time will likely be spent on applying equations to word problems. If you feel that your problem-solving skills could still use some honing, check out more examples and strategies from this post by the Physics Classroom or through this video-guided tutorial from Khan Academy.

That was a lot of equations and examples to take in. Eventually, whether you’re figuring out how to find a constant acceleration or how to solve velocity when you don’t have a value for time, you’ll know exactly which of the four kinematic equations to apply and how. Just keep the problem-solving steps we’ve used here in mind, and you’ll be able to get through your physics course without any unsolvable problems.

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## Kinematics in Two Dimensions

Practice problem 1.

- the total distance of the entire trip
- the total displacement of the entire trip
- the average speed of the entire trip
- the average velocity of the entire trip
- the average acceleration of the entire trip

Distance? No problem. First I walked 6.0 km and then I walked 10 km for a total of 16 km. Distance is a scalar quantity, so the individual distances add just like regular numbers.

Displacement is a bit more challenging. Displacement is a vector and vectors have direction, so it's best to diagram this problem (a procedure that's remarkably useful in general). The resultant displacement is the vector sum of the two displacements experienced during the trip. Since they're perpendicular to one another, the resultant is the hypotenuse of a right triangle. Its magnitude can be found using pythagorean theorem and its direction can be found using the tangent function.

r = 11.7 km at 59° west of north

The speed was 6.0 km/h for the first 6.0 km and 5 km/h for the last 10 km. The naive solution is to average the speeds using the add-and-divide method taught in junior high school. This method is wrong, not because the method itself is wrong, but because it doesn't apply to this situation.

The weights of the two segments are not equal. The second segment lasted twice as long as the first (as you will soon see).

Go back to the definition to solve this problem. Average speed is the total distance (which we've already found) divided by the total time (which we need to find). Since time is a scalar, add the times for each leg of the journey to get the total time.

The average speed is then…

The velocity was 6.0 km/h north over the first 6.0 km and 5 km/h west over the last 10 km. Average velocity is the total displacement divided by the total time. Both of these quantities have already been determined.

Acceleration in this context is relatively meaningless. It would be better to illustrate acceleration in two dimensions with a different problem (like the one below).

## practice problem 2

- the speed of the current
- the magnitude of the swimmer's resultant velocity
- the direction of the swimmer's resultant velocity
- the time it takes the swimmer to cross the river

Since distance and velocity are directly proportional, this begins as a similar triangles problem.

Since speed and distance are directly proportional, the ratio of the downstream distance to the width of the river is the same as the ratio of the current speed to the swimmer's speed.

Determining the resultant velocity is a simple application of Pythagorean theorem.

Direction angles are often best determined using the tangent function. This problem is no exception. The only thing open to discussion is our choice of angle. I suggest using the angle between the resultant velocity and the displacement vector that points directly across the river, but this is just my preference. Be sure to indicate that the resultant lies on a particular side of this vector for clarity.

This is where it gets interesting. By now you should understood that time is the ratio of displacement to velocity. This is a vector problem, so direction matters. This is why we should probably use the words displacement and velocity instead of distance and speed. The only question is which distance and which speed should we use? The simple answer is pick the pair you like the best, just be sure they point in the same direction . It works along either of the component directions…

It also works along the resultant direction…

There's an interesting sideline to this question that astute readers might have noticed when looking at the first ratio in the chain of three shown above. The time it takes to cross a river by a swimmer swimming straight across is independent of the speed of the river. The only factors that matter are the speed of the swimmer and the width of the river. This swimmer will always cross the river in 50 s regardless of the speed of the river. 1 m/s, 10 m/s, 100 m/s, it doesn't matter. This example is a perfect illustration of an idea to be presented in the next section of this book. Motion in two dimensions can be thoroughly described with two independent one-dimensional equations. This idea is central to the field of analytical geometry.

## practice problem 3

Finding the change in velocity is complicated in this problem by the change in direction. A diagram is indispensable. Let's assume that the initial direction of the car is 0° (to the right in standard position) and that the final velocity will be 90° (toward the top of the page in standard position). The difference of two vectors drawn this way would then connect the the head of the initial vector to the head of the final vector. Use Pythagorean Theorem for magnitude and tangent for direction as usual. Only after we have done all of this can we then plug numbers into the definition.

a = 20 m/s 2 at 143°

## practice problem 4

Start with a diagram.

Strip it down to its essence.

Two sides of this triangle are given ( v asteroid and v impact ). None of the angles are known. The third side ( v earth ) can be determined from basic knowledge. The average speed of the Earth is the distance covered in one orbit (the circumference) divided by the time it takes to complete that orbit (one year). We could do this on a hand held calculator…

or use an online calculator that knows the Earth-Sun distance…

We now have three sides of a triangle and can find the desired angle using the law of cosines.

a 2 = b 2 + c 2 − 2 bc cos A

Solve algebraically, substitute numerical values, and compute the answer.

- 2.6 Problem-Solving Basics for One-Dimensional Kinematics
- Introduction to Science and the Realm of Physics, Physical Quantities, and Units
- 1.1 Physics: An Introduction
- 1.2 Physical Quantities and Units
- 1.3 Accuracy, Precision, and Significant Figures
- 1.4 Approximation
- Section Summary
- Conceptual Questions
- Problems & Exercises
- Introduction to One-Dimensional Kinematics
- 2.1 Displacement
- 2.2 Vectors, Scalars, and Coordinate Systems
- 2.3 Time, Velocity, and Speed
- 2.4 Acceleration
- 2.5 Motion Equations for Constant Acceleration in One Dimension
- 2.7 Falling Objects
- 2.8 Graphical Analysis of One-Dimensional Motion
- Introduction to Two-Dimensional Kinematics
- 3.1 Kinematics in Two Dimensions: An Introduction
- 3.2 Vector Addition and Subtraction: Graphical Methods
- 3.3 Vector Addition and Subtraction: Analytical Methods
- 3.4 Projectile Motion
- 3.5 Addition of Velocities
- Introduction to Dynamics: Newton’s Laws of Motion
- 4.1 Development of Force Concept
- 4.2 Newton’s First Law of Motion: Inertia
- 4.3 Newton’s Second Law of Motion: Concept of a System
- 4.4 Newton’s Third Law of Motion: Symmetry in Forces
- 4.5 Normal, Tension, and Other Examples of Forces
- 4.6 Problem-Solving Strategies
- 4.7 Further Applications of Newton’s Laws of Motion
- 4.8 Extended Topic: The Four Basic Forces—An Introduction
- Introduction: Further Applications of Newton’s Laws
- 5.1 Friction
- 5.2 Drag Forces
- 5.3 Elasticity: Stress and Strain
- Introduction to Uniform Circular Motion and Gravitation
- 6.1 Rotation Angle and Angular Velocity
- 6.2 Centripetal Acceleration
- 6.3 Centripetal Force
- 6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
- 6.5 Newton’s Universal Law of Gravitation
- 6.6 Satellites and Kepler’s Laws: An Argument for Simplicity
- Introduction to Work, Energy, and Energy Resources
- 7.1 Work: The Scientific Definition
- 7.2 Kinetic Energy and the Work-Energy Theorem
- 7.3 Gravitational Potential Energy
- 7.4 Conservative Forces and Potential Energy
- 7.5 Nonconservative Forces
- 7.6 Conservation of Energy
- 7.8 Work, Energy, and Power in Humans
- 7.9 World Energy Use
- Introduction to Linear Momentum and Collisions
- 8.1 Linear Momentum and Force
- 8.2 Impulse
- 8.3 Conservation of Momentum
- 8.4 Elastic Collisions in One Dimension
- 8.5 Inelastic Collisions in One Dimension
- 8.6 Collisions of Point Masses in Two Dimensions
- 8.7 Introduction to Rocket Propulsion
- Introduction to Statics and Torque
- 9.1 The First Condition for Equilibrium
- 9.2 The Second Condition for Equilibrium
- 9.3 Stability
- 9.4 Applications of Statics, Including Problem-Solving Strategies
- 9.5 Simple Machines
- 9.6 Forces and Torques in Muscles and Joints
- Introduction to Rotational Motion and Angular Momentum
- 10.1 Angular Acceleration
- 10.2 Kinematics of Rotational Motion
- 10.3 Dynamics of Rotational Motion: Rotational Inertia
- 10.4 Rotational Kinetic Energy: Work and Energy Revisited
- 10.5 Angular Momentum and Its Conservation
- 10.6 Collisions of Extended Bodies in Two Dimensions
- 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum
- Introduction to Fluid Statics
- 11.1 What Is a Fluid?
- 11.2 Density
- 11.3 Pressure
- 11.4 Variation of Pressure with Depth in a Fluid
- 11.5 Pascal’s Principle
- 11.6 Gauge Pressure, Absolute Pressure, and Pressure Measurement
- 11.7 Archimedes’ Principle
- 11.8 Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action
- 11.9 Pressures in the Body
- Introduction to Fluid Dynamics and Its Biological and Medical Applications
- 12.1 Flow Rate and Its Relation to Velocity
- 12.2 Bernoulli’s Equation
- 12.3 The Most General Applications of Bernoulli’s Equation
- 12.4 Viscosity and Laminar Flow; Poiseuille’s Law
- 12.5 The Onset of Turbulence
- 12.6 Motion of an Object in a Viscous Fluid
- 12.7 Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes
- Introduction to Temperature, Kinetic Theory, and the Gas Laws
- 13.1 Temperature
- 13.2 Thermal Expansion of Solids and Liquids
- 13.3 The Ideal Gas Law
- 13.4 Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature
- 13.5 Phase Changes
- 13.6 Humidity, Evaporation, and Boiling
- Introduction to Heat and Heat Transfer Methods
- 14.2 Temperature Change and Heat Capacity
- 14.3 Phase Change and Latent Heat
- 14.4 Heat Transfer Methods
- 14.5 Conduction
- 14.6 Convection
- 14.7 Radiation
- Introduction to Thermodynamics
- 15.1 The First Law of Thermodynamics
- 15.2 The First Law of Thermodynamics and Some Simple Processes
- 15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency
- 15.4 Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated
- 15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators
- 15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
- 15.7 Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation
- Introduction to Oscillatory Motion and Waves
- 16.1 Hooke’s Law: Stress and Strain Revisited
- 16.2 Period and Frequency in Oscillations
- 16.3 Simple Harmonic Motion: A Special Periodic Motion
- 16.4 The Simple Pendulum
- 16.5 Energy and the Simple Harmonic Oscillator
- 16.6 Uniform Circular Motion and Simple Harmonic Motion
- 16.7 Damped Harmonic Motion
- 16.8 Forced Oscillations and Resonance
- 16.10 Superposition and Interference
- 16.11 Energy in Waves: Intensity
- Introduction to the Physics of Hearing
- 17.2 Speed of Sound, Frequency, and Wavelength
- 17.3 Sound Intensity and Sound Level
- 17.4 Doppler Effect and Sonic Booms
- 17.5 Sound Interference and Resonance: Standing Waves in Air Columns
- 17.6 Hearing
- 17.7 Ultrasound
- Introduction to Electric Charge and Electric Field
- 18.1 Static Electricity and Charge: Conservation of Charge
- 18.2 Conductors and Insulators
- 18.3 Coulomb’s Law
- 18.4 Electric Field: Concept of a Field Revisited
- 18.5 Electric Field Lines: Multiple Charges
- 18.6 Electric Forces in Biology
- 18.7 Conductors and Electric Fields in Static Equilibrium
- 18.8 Applications of Electrostatics
- Introduction to Electric Potential and Electric Energy
- 19.1 Electric Potential Energy: Potential Difference
- 19.2 Electric Potential in a Uniform Electric Field
- 19.3 Electrical Potential Due to a Point Charge
- 19.4 Equipotential Lines
- 19.5 Capacitors and Dielectrics
- 19.6 Capacitors in Series and Parallel
- 19.7 Energy Stored in Capacitors
- Introduction to Electric Current, Resistance, and Ohm's Law
- 20.1 Current
- 20.2 Ohm’s Law: Resistance and Simple Circuits
- 20.3 Resistance and Resistivity
- 20.4 Electric Power and Energy
- 20.5 Alternating Current versus Direct Current
- 20.6 Electric Hazards and the Human Body
- 20.7 Nerve Conduction–Electrocardiograms
- Introduction to Circuits and DC Instruments
- 21.1 Resistors in Series and Parallel
- 21.2 Electromotive Force: Terminal Voltage
- 21.3 Kirchhoff’s Rules
- 21.4 DC Voltmeters and Ammeters
- 21.5 Null Measurements
- 21.6 DC Circuits Containing Resistors and Capacitors
- Introduction to Magnetism
- 22.1 Magnets
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- Introduction to Radioactivity and Nuclear Physics
- 31.1 Nuclear Radioactivity
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- Introduction to Applications of Nuclear Physics
- 32.1 Diagnostics and Medical Imaging
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- Introduction to Particle Physics
- 33.1 The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited
- 33.2 The Four Basic Forces
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- Introduction to Frontiers of Physics
- 34.1 Cosmology and Particle Physics
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- 34.3 Superstrings
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- 34.7 Some Questions We Know to Ask
- A | Atomic Masses
- B | Selected Radioactive Isotopes
- C | Useful Information
- D | Glossary of Key Symbols and Notation

## Learning Objectives

By the end of this section, you will be able to:

- Apply problem-solving steps and strategies to solve problems of one-dimensional kinematics.
- Apply strategies to determine whether or not the result of a problem is reasonable, and if not, determine the cause.

Problem-solving skills are obviously essential to success in a quantitative course in physics. More importantly, the ability to apply broad physical principles, usually represented by equations, to specific situations is a very powerful form of knowledge. It is much more powerful than memorizing a list of facts. Analytical skills and problem-solving abilities can be applied to new situations, whereas a list of facts cannot be made long enough to contain every possible circumstance. Such analytical skills are useful both for solving problems in this text and for applying physics in everyday and professional life.

## Problem-Solving Steps

While there is no simple step-by-step method that works for every problem, the following general procedures facilitate problem solving and make it more meaningful. A certain amount of creativity and insight is required as well.

Examine the situation to determine which physical principles are involved . It often helps to draw a simple sketch at the outset. You will also need to decide which direction is positive and note that on your sketch. Once you have identified the physical principles, it is much easier to find and apply the equations representing those principles. Although finding the correct equation is essential, keep in mind that equations represent physical principles, laws of nature, and relationships among physical quantities. Without a conceptual understanding of a problem, a numerical solution is meaningless.

Make a list of what is given or can be inferred from the problem as stated (identify the knowns) . Many problems are stated very succinctly and require some inspection to determine what is known. A sketch can also be very useful at this point. Formally identifying the knowns is of particular importance in applying physics to real-world situations. Remember, “stopped” means velocity is zero, and we often can take initial time and position as zero.

Identify exactly what needs to be determined in the problem (identify the unknowns) . In complex problems, especially, it is not always obvious what needs to be found or in what sequence. Making a list can help.

Find an equation or set of equations that can help you solve the problem . Your list of knowns and unknowns can help here. It is easiest if you can find equations that contain only one unknown—that is, all of the other variables are known, so you can easily solve for the unknown. If the equation contains more than one unknown, then an additional equation is needed to solve the problem. In some problems, several unknowns must be determined to get at the one needed most. In such problems it is especially important to keep physical principles in mind to avoid going astray in a sea of equations. You may have to use two (or more) different equations to get the final answer.

Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete with units . This step produces the numerical answer; it also provides a check on units that can help you find errors. If the units of the answer are incorrect, then an error has been made. However, be warned that correct units do not guarantee that the numerical part of the answer is also correct.

Check the answer to see if it is reasonable: Does it make sense? This final step is extremely important—the goal of physics is to accurately describe nature. To see if the answer is reasonable, check both its magnitude and its sign, in addition to its units. Your judgment will improve as you solve more and more physics problems, and it will become possible for you to make finer and finer judgments regarding whether nature is adequately described by the answer to a problem. This step brings the problem back to its conceptual meaning. If you can judge whether the answer is reasonable, you have a deeper understanding of physics than just being able to mechanically solve a problem.

When solving problems, we often perform these steps in different order, and we also tend to do several steps simultaneously. There is no rigid procedure that will work every time. Creativity and insight grow with experience, and the basics of problem solving become almost automatic. One way to get practice is to work out the text’s examples for yourself as you read. Another is to work as many end-of-section problems as possible, starting with the easiest to build confidence and progressing to the more difficult. Once you become involved in physics, you will see it all around you, and you can begin to apply it to situations you encounter outside the classroom, just as is done in many of the applications in this text.

## Unreasonable Results

Physics must describe nature accurately. Some problems have results that are unreasonable because one premise is unreasonable or because certain premises are inconsistent with one another. The physical principle applied correctly then produces an unreasonable result. For example, if a person starting a foot race accelerates at 0 . 40 m/s 2 0 . 40 m/s 2 for 100 s, his final speed will be 40 m/s (about 150 km/h)—clearly unreasonable because the time of 100 s is an unreasonable premise. The physics is correct in a sense, but there is more to describing nature than just manipulating equations correctly. Checking the result of a problem to see if it is reasonable does more than help uncover errors in problem solving—it also builds intuition in judging whether nature is being accurately described.

Use the following strategies to determine whether an answer is reasonable and, if it is not, to determine what is the cause.

Solve the problem using strategies as outlined and in the format followed in the worked examples in the text . In the example given in the preceding paragraph, you would identify the givens as the acceleration and time and use the equation below to find the unknown final velocity. That is,

Check to see if the answer is reasonable . Is it too large or too small, or does it have the wrong sign, improper units, …? In this case, you may need to convert meters per second into a more familiar unit, such as miles per hour.

This velocity is about four times greater than a person can run—so it is too large.

If the answer is unreasonable, look for what specifically could cause the identified difficulty . In the example of the runner, there are only two assumptions that are suspect. The acceleration could be too great or the time too long. First look at the acceleration and think about what the number means. If someone accelerates at 0 . 40 m/s 2 0 . 40 m/s 2 , their velocity is increasing by 0.4 m/s each second. Does this seem reasonable? If so, the time must be too long. It is not possible for someone to accelerate at a constant rate of 0 . 40 m/s 2 0 . 40 m/s 2 for 100 s (almost two minutes).

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- Authors: Paul Peter Urone, Roger Hinrichs
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- Book title: College Physics 2e
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## How to solve kinematics problems, part 2

This article is the second chapter in a series on how to understand and approach kinematics problems. The first chapter covered position, velocity, and acceleration. Now that we understand these quantities, we are going to use them to solve problems in one dimension.

## Kinematics Equations for Constant Acceleration

The four horsemen of the kinematics apocalypse are:

x f – x i = (v f – v i )*t/2

v f – v i = a*t

v f 2 = v i 2 + 2*a*(x f – x i )

x f = x i + v i *t + ½a*t 2

Note: the little f stands for final (as in the final velocity or position) while the little i stands for initial.

Note: These equations only work for constant acceleration, but nearly all problems have constant acceleration.

## 1-Dimensional Problem Solving Steps

For every one dimensional kinematics problem, the steps are pretty much the same.

- Write down every quantity the problem gives you (initial and final position, initial and final velocity, acceleration, time, etc)
- Write down which quantity you are trying to find
- Find the kinematic equation (or sometimes two equations) to relate these quantities.
- Solve the algebra.

Yes, it really is that simple. (In fact, most physics problems work the same way. For more details on the physics problem solving algorithm, check out this article .)

## Avoiding Common Mistakes: Hidden Quantities

Sometimes the problem may tell you a quantity secretly; you may not even realize you got it. For example, if they tell you the displacement (how far something travelled) but not positions, you can treat the displacement as x f and set x i to 0. Likewise, if the problem doesn’t say anything special about acceleration, then the acceleration is probably just gravity, a = g = 9.8m/s 2 . These hidden quantities are as valid as regular quantities, they are just a little harder to spot.

## Avoiding Common Mistakes: Top of Flight

A special example of the hidden quantity is when they tell you an object is at “the top of its flight/motion/path/etc.” This means they are secretly telling you that v f is 0 because an object traveling in one dimension always has a velocity of 0 at the top of its path. Wondering why? Well, if the velocity was going up, then a millisecond later, the object would be higher (and thus it can’t be at the top of its path). Likewise, if the object had a velocity downward, then it would have been higher a millisecond before (so it can’t be at the top either).

## Avoiding Common Mistakes: Positive and Negative Numbers

It can be tricky to keep track of your negatives. The key is direction; down is always negative. So if an object is going down, it will have a negative velocity. If the acceleration is going down (which it almost always is) then the acceleration is negative. And don’t forget from our first chapter, it is possible to have positive velocity and negative acceleration at the same time!

## Example: A Woman and Her Ball

A woman is holding a ball at 1 meter, and throws it upward at 5m/s. a) How high does the ball reach? b) How long does the ball take to hit the ground? c) How fast is the ball going when it hits the ground?

Let’s find out!

## Part A: How high does the ball reach?

What do we know?

The initial position x i = 1 m The initial velocity v i = 5 m/s Secret Quantity: a = -9.8 m/s 2 (gravity) Secret Quantity: At the top of the ball’s arc (i.e. when its at its highest) v f = 0 m/s

What are we trying to find?

The position at the top of the throw, x f

What equation relates these quantities?

We’re looking for an equation that includes x f , x i , v f , v i , and a v f 2 = v i 2 + 2*a*(x f – x i ) seems to fit nicely!

Plug in and solve

v f 2 = v i 2 + 2*a*(x f – x i ) (0 m/s) 2 = (5 m/s) 2 + 2*(-9.8 m/s 2 )*(x f – 1m) 0 = 25 (m 2 /s 2 ) – (19.6 m/s 2 )*(x f – 1m) (19.6 m/s 2 )*(x f – 1m) = 25 m 2 /s 2 x f –1m = (25 m 2 /s 2 )/(19.6 m/s 2 ) x f –1m = 1.28 m x f = 2.28 m

Tada! The final height = 2.28 meters

## Part B: How long does the ball take to hit the ground?

We still know x i = 1 m, v i = 5 m/s and a = -9.8 m/s 2 but now we also know that x f = 0 m (because the height of the ground is 0m)

Note: Since the ball is now on the ground instead of at the top of it’s flight, v f ≠ 0 so that is off the table.

The time, t

We’re looking for an equation that includes x f , x i , v i , a and t Looks like we’re going to need x f = x i + v i *t + ½a*t 2

x f = x i + v i *t + ½a*t 2 0 m = 1 m + (5 m/s)*t + ½ (-9.8 m/s 2 )* t 2 -(4.9 m/s 2 )* t 2 + (5 m/s)*t + 1 m = 0

This is a quadratic equation (ax 2 + bx + c = 0) and can be solved using the quadratic formula.

t = t = t = or t = t = or t = t = -.17 s or 1.2 s

We can ignore t=-.17 because we are not allowed to have a negative time (we call this a non-physical answer), which leaves us with time = 1.2 seconds!

## Part C: Put it all together.

We still know x i = 1 m, v i = 5 m/s, a = -9.8 m/s 2 and x f = 0 m but now we also know t = 1.2 seconds because we just solved it.

The velocity, v f

We have so many quantities we could use any of the equations, but let’s go with v f - v i = a*t because it’s simple and we haven’t used it yet.

v f – v i = a*t v f – 5 m/s = (-9.8 m/s 2 ) * 1.2 s v f = 5 m/s – 11.8 m/s v f = -6.8 m/s

Boom, the velocity when the ball hits the ground is 6.8 m/s downward (thus the negative sign).

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## 2.4: Problem-Solving for Basic Kinematics

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## Applications

There are four kinematic equations that describe the motion of objects without consideration of its causes.

learning objectives

- Choose which kinematics equation to use in problems in which the initial starting position is equal to zero

Kinematics is the branch of classical mechanics that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without consideration of the causes of motion. There are four kinematic equations when the initial starting position is the origin, and the acceleration is constant:

- \(\mathrm{v=v_0+at}\)
- \(\mathrm{d=\frac{1}{2}(v_0+v)t}\) or alternatively \(\mathrm{v_{average}=\frac{d}{t}}\)
- \(\mathrm{d=v_0t+(\frac{at^2}{2})}\)
- \(\mathrm{v^2=v^2_0+2ad}\)

Notice that the four kinematic equations involve five kinematic variables: \(\mathrm{d,v,v_0,a}\) and \(\mathrm{t}\). Each of these equations contains only four of the five variables and has a different one missing. This tells us that we need the values of three variables to obtain the value of the fourth and we need to choose the equation that contains the three known variables and one unknown variable for each specific situation.

Here the basic problem solving steps to use these equations:

Step one – Identify exactly what needs to be determined in the problem (identify the unknowns).

Step two – Find an equation or set of equations that can help you solve the problem.

Step three – Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete with units.

Step four – Check the answer to see if it is reasonable: Does it make sense?

Problem-solving skills are obviously essential to success in a quantitative course in physics. More importantly, the ability to apply broad physical principles, usually represented by equations, to specific situations is a very powerful form of knowledge. It is much more powerful than memorizing a list of facts. Analytical skills and problem-solving abilities can be applied to new situations, whereas a list of facts cannot be made long enough to contain every possible circumstance. Such analytical skills are useful both for solving problems in a physics class and for applying physics in everyday and professional life.

## Motion Diagrams

A motion diagram is a pictorial description of an object’s motion and represents the position of an object at equally spaced time intervals.

- Construct a motion diagram

A motion diagram is a pictorial description of the motion of an object. It displays the object’s location at various equally spaced times on the same diagram; shows an object’s initial position and velocity; and presents several spots in the center of the diagram. These spots reveal whether or not the object has accelerated or decelerated. For simplicity, the object is represented by a simple shape, such as a filled circle, which contains information about an object’s position at particular time instances. For this reason, a motion diagram is more information than a path diagram. It may also display the forces acting on the object at each time instance.

is a motion diagram of a simple trajectory. Imagine the object as a hockey puck sliding on ice. Notice that the puck covers the same distance per unit interval along the trajectory. We can conclude that the puck is moving at a constant velocity and, therefore, there is no acceleration or deceleration during the motion.

Puck Sliding on Ice : Motion diagram of a puck sliding on ice. The puck is moving at a constant velocity.

One major use of motion diagrams is the presentation of film through a series of frames taken by a camera; this is sometimes called stroboscopic technique (as seen in ). Viewing an object on a motion diagram allows one to determine whether an object is speeding up or slowing down, or if it is at constant rest. As the frames are taken, we can assume that an object is at a constant rest if it occupies the same position over time. We can assume that an object is speeding up if there is a visible increase in the space between objects as time passes, and that it is slowing down if there is a visible decrease in the space between objects as time passes. The objects on the frame come very close together.

Bouncing Ball : A bouncing ball captured with a stroboscopic flash at 25 images per second.

- The four kinematic equations involve five kinematic variables: \(\mathrm{d,v,v_0,a}\) and \(\mathrm{t}\).
- Each equation contains only four of the five variables and has a different one missing.
- It is important to choose the equation that contains the three known variables and one unknown variable for each specific situation.
- Motion diagrams represent the motion of an object by displaying its location at various equally spaced times on the same diagram.
- Motion diagrams show an object’s initial position and velocity and presents several spots in the center of the diagram. These spots reveal the object’s state of motion.
- Motion diagrams contain information about an object’s position at particular time instances and is therefore more informative than a path diagram.
- kinematics : The branch of physics concerned with objects in motion.
- stroboscopic : Relating to an instrument used to make a cyclically moving object appear to be slow-moving, or stationary.
- diagram : A graph or chart.
- motion : A change of position with respect to time.

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The Physics Classroom has been devoted to helping students, teachers, and classrooms since the 1990s. We are as passionate about that mission now as we have ever been. If you are a teacher of Physics or Physical Science, we encourage you to use our Video Tutorial with your students. And we also encourage you to consider the use of other resources on our website that coordinate with the video. We have listed a few below to help you get started.

Curriculum Corner: 1D-Kinematics Section

Calculator Pad, Kinematics Section, Problems #18-35

The Review Session, Kinematics Section, Questions #43-50

The Physics Classroom Tutorial, 1D-Kinematics Chapter, Lesson 6

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The acceleration of gravity on the moon is 1.67 m/s 2. Determine the time for the feather to fall to the surface of the moon. See Answer See solution below. Rocket-powered sleds are used to test the human response to acceleration.

A common phrase in kinematics problems is "ending or coming to a rest", which means the final velocity of the object in the time interval we are considering is zero, v_f=0 vf = 0. The kinematics equation that suits this problem is v^2-v_0^2=2a (x-x_0) v2 −v02 = 2a(x −x0), where the only unknown variable is the acceleration a a .

If we know three of these five kinematic variables— Δ x, t, v 0, v, a —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. [Where did these formulas come from?] 1. v = v 0 + a t 2.

Solution: The given data are initial velocity, v0 = 0 final velocity, v = 101 Km/h = 101 1000 m 3600 s = 28.06 m/s time interval , t = 8 s acceleration , =? The relevant kinematic equation which relates those together is v = v0 + a t. So v = v0 + a t 28.06 = 0 + a(8) 28.06 − 0 ⇒ a = 8 = 3.51 m/s2 2.

Problem-Solving Strategies Conclusion The Kinematic Equations The following four kinematic equations come up throughout physics from the earliest high school class to the highest level college course: Kinematic Equations v=v_ {0}+at v = v0 + at \Delta x=\dfrac {v+v_ {0}} {2} t Δx = 2v +v0t \Delta x=v_ {0}t+\frac {1} {2}at^ {2} Δx = v0t + 21at2

Kinematics is the branch of classical mechanics that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without consideration of the causes of motion. There are four kinematic equations when the initial starting position is the origin, and the acceleration is constant: v = v0 + at v = v 0 + a t.

So, we need to use the equations we know and replace t. The easiest one to do that for is the first equation. Then use that information in another equation to get our 4th, Vf = Vo + at to solve for t, subtract Vo from both sides Vf - Vo = at. now divide both sides by a and we get Vf -Vo/a = t.

practice problem 2. A swimmer heads directly across a river swimming at 1.6 m/s relative to still water. She arrives at a point 40 m downstream from the point directly across the river, which is 80 m wide. Determine…. the speed of the current. the magnitude of the swimmer's resultant velocity.

Motion along a curved path on a flat surface or a plane (such as that of a ball on a pool table or a skater on an ice rink) is two-dimensional, and thus described by two-dimensional kinematics. Motion not confined to a plane, such as a car following a winding mountain road, is described by three-dimensional kinematics.

I've seen it a thousand times. Students understand everything during class, but then when it comes time to try the problems on a test, they draw a blank. You...

Kinematic equations: numerical calculations. A bumblebee is flying to the right when a breeze causes the bee to slow down with a constant leftward acceleration of magnitude 0.50 m s 2 . After 2.0 s , the bee is moving to the right with a speed of 2.75 m s . What was the velocity of the bumblebee right before the breeze?

Sample Problems and Solutions Kinematic Equations and Kinematic Graphs As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be "in a state of free fall ."

Solving Problems with Kinematic Equations Kinematic Equations and Free Fall Sample Problems and Solutions Kinematic Equations and Kinematic Graphs The goal of this first unit of The Physics Classroom has been to investigate the variety of means by which the motion of objects can be described.

Learn Kinematics Equations with free step-by-step video explanations and practice problems by experienced tutors. ... guys, For this video, we're gonna check out how we solve problems where an object is moving under constant acceleration in one or multiple parts. And what we're gonna find is that it's exactly like how we solve any motion ...

Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns). In complex problems, especially, it is not always obvious what needs to be found or in what sequence. Making a list can help. Step 4. Find an equation or set of equations that can help you solve the problem.

Step 1 Examine the situation to determine which physical principles are involved. It often helps to draw a simple sketch at the outset. You will also need to decide which direction is positive and note that on your sketch.

Solve the algebra. Yes, it really is that simple. (In fact, most physics problems work the same way. For more details on the physics problem solving algorithm, check out this article .) Avoiding Common Mistakes: Hidden Quantities Sometimes the problem may tell you a quantity secretly; you may not even realize you got it.

Kinematics is the branch of classical mechanics that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without consideration of the causes of motion. There are four kinematic equations when the initial starting position is the origin, and the acceleration is constant: v = v0 + at v = v 0 + a t.

This kinematics calculator will help you solve the uniform acceleration problems by using kinematics equations of physics. You can use our free kinematic equations solver to solve the equations that is used for motion in a straight line with constant acceleration. ... After reading, you will be able to completely solve different kinematics ...

The Using the Kinematic Equations - Part 2 Video Tutorial is the third of three lessons on the Kinematic Equations. The purpose of this video is to demonstrate through three examples an effective strategy for solving physics word problems using the kinematic equations. The video ends with a Learning Action Plan that will help make the learning ...

The purpose of this video is to demonstrate through three examples an effective strategy for solving physics word problems using the kinematic equations. The video ends with a Learning Action Plan that will help make the learning stick.