problem solving using kinematic equations

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 Practice Problems with Answers

Are you struggling with kinematics problems? Do you want to understand the principles of motion in a clear, concise manner? Look no further! Our comprehensive guide on “Kinematics Problems” is here to help.

Whether you’re studying for an exam or working on homework, these solutions offer a practical approach to understanding and applying kinematics equations.

All kinematics equations are summarized in the following expressions: \begin{gather*} \Delta x=\frac{v_1+v_2}{2}\times \Delta t \\\\ v=v_0+at \\\\ \Delta x=\frac 12 at^2+v_0t \\\\ v^2-v_0^2=2a\Delta x \end{gather*} In the rest of this long article, you will see how to apply these equations in the given problems.

By the way, if you’re preparing for the AP Physics 1 exam , feel free to download this comprehensive AP formula sheet . It’s the ultimate resource you need!

Kinematics Practice Problems:

Problem (1): A car slows down its motion from 10 m/s to 6 m/s in 2 seconds under constant acceleration.  (a) What is its acceleration? (b) How far did the car travel during this time interval?

Solution : This is a basic kinematics problem, so we will explain the steps in detail.

Step 1: Since all these problems are in one dimension, draw a horizontal axis (like the positive $x$ axis), and place the object on it, so that its motion matches the direction of the axis. 

Step 2:  Specify the known and wanted information. Here, in the elapsed time interval $2\,{\rm s}$, the initial and final velocities of the car are given as $v_i=10\,{\rm m/s}$ and $v_f=6\,{\rm m/s}$. The wanted quantity is the constant acceleration of the object (car), $a=?$. 

Step 3: Apply the kinematics equation that is appropriate for this situation. 

(a) To find the acceleration in this problem, we are given the time, initial, and final velocity. The kinematics equation $v=v_0+at$ is suitable for this situation, as the only unknown variable is the acceleration $a$. By rearranging the equation, we get \begin{gather*} v=v_0+at\\\\ 6=10+a(2) \\\\ 6-10=2a \\\\\Rightarrow \quad a=\frac{6-10}{2}=-2\quad {\rm \frac{m}{s^2}}\end{gather*} Since the problem states that the acceleration is constant, we could also use any of the other constant acceleration kinematics equations. The negative sign of the acceleration indicates that it is directed toward the negative $x$ axis.

(b)  "How far'' indeed refers to the distance traveled, denoted by $x$ in the kinematics equations. 

Here, the best equation that relates the known and unknown information is $x=\frac 12 at^2+v_0t$ or $v^2-v_0^2=2ax$. We choose the first, so \begin{align*} x&=\frac 12 at^2+v_0t \\\\&=\frac 12 (-2)(2)^2+(10)(2) \\\\&=16\quad {\rm m}\end{align*} 

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Problem (2): A moving object slows down from $12\,{\rm m/s}$ to rest at a distance of 20 m. Find the acceleration of the object (assumed constant).

Solution : In the diagram below, all known information along with the direction of the uniform motion is shown. 

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, $v_f=0$.

The kinematics equation that suits this problem is $v^2-v_0^2=2a(x-x_0)$, where the only unknown variable is the acceleration $a$. 

For simplicity, we can assume the initial position of the motion $x_0$ is zero in all kinematics problems, $x_0$. \begin{gather*} v^2-v_0^2=2ax\\\\0^2-(12)^2 =2a(20) \\\\ \rightarrow a=\frac{-144}{2\times 20}\\\\\Rightarrow \boxed{a=-3.6\quad {\rm \frac{m}{s^2}}}\end{gather*} As before, the negative sign indicates of the acceleration indicates that it is directed to the left .

Problem (3): A bullet leaves the muzzle of an 84-cm rifle with a speed of 521 m/s. Find the magnitude of the bullet's acceleration by assuming it is constant inside the barrel of the rifle.

Solution : The bullet accelerates from rest to a speed of 521 m/s over a distance of 0.84 meters. We have the following known quantities: the initial velocity $v_0$, ​ the final velocity $v$, and the displacement $x-x_0$. The unknown is acceleration $a$. The perfect kinematics equation that relates all these variables is $v^2-v_0^2=2a(x-x_0)$. Solving for $a$, we have: \begin{gather*}v^2-v_0^2=2a(x-x_0)\\\\ (521)^2-0=2(a)(0.84-0) \\\\ \Rightarrow \boxed{a=1.62\times 10^5\quad {\rm m/s^2}}\end{gather*} This calculation results in a very large acceleration. 

Problem (4): A car starts its motion from rest and uniformly accelerates at a rate of $4\,{\rm m/s^2}$ for 2 seconds in a straight line.  (a) How far did the car travel during those 2 seconds?  (b) What is the car's velocity at the end of that time interval?

Solution : "Start from rest'' means the initial object's velocity is zero, $v_0=0$. The known information are $a=4\,{\rm m/s^2}$, $t=2\,{\rm s}$ and wants the distance traveled $x=?$. 

(a) The kinematics equation that relates the given information is $x=\frac 12 at^2+v_0 t+x_0$ since the only unknown quantity is $x$. Given the known data, we can calculate $x$ as follows: \begin{align*} x&=\frac 12 at^2+v_0 t+x_0 \\\\&=\frac 12 (4)(2)^2+(0)(2) \\\\&=\boxed{8\quad {\rm m}}\end{align*} As before, we set $x_0=0$. 

(b) Now that we know the distance traveled by car in that time interval, we can use the following kinematics equation to find the car's final velocity $v$: \begin{align*} v^2-v_0^2 &=2a(x-x_0) \\\\v^2-(0)^2&=2(4)(8-0) \\\\v^2&=64\end{align*} Taking the square root, we get $v$: \[v=\sqrt{64}=\pm 8\quad {\rm \frac ms}\] We know that velocity is a vector quantity in physics and has both a direction and a magnitude. 

The magnitude of the velocity (speed) was obtained as 8 m/s, but in what direction? Or which sign should we choose? Because the car is uniformly accelerating without stopping in the positive $x$ axis, the correct sign for velocity is positive. 

Therefore, the car's final velocity is $\boxed{v_f=+8\,{\rm m/s}}$. 

Problem (5): We aim to design an airport runway with the following specifications: The lowest acceleration of a plane should be $4\,{\rm m/s^2}$, and its take-off speed should be 75 m/s. How long would the runway have to be to allow the planes to accelerate through it? 

Solution: The known quantities are acceleration $a=4\,{\rm m/s^2}$, and final velocity $v=75\,{\rm m/s}$. The wanted quantity is the runway length $\Delta x=x-x_0$. The ideal kinematics equation that relates these variables is $v^2-v_0^2=2a(x-x_0)$. \begin{align*} v^2-v_0^2&=2a\Delta x\\\\ (75)^2-0&=2(4) \Delta x\\\\ \Rightarrow \Delta x&=\boxed{703\quad {\rm m}}\end{align*} Thus, for the runway to be effective, its length must be at least approximately 703 meters.

Problem (6): A stone is dropped vertically from a high cliff. After 3.55 seconds, it hits the ground. How high is the cliff? 

Solution : There is another type of kinematics problem in one dimension but in the vertical direction. In such problems, the constant acceleration is that of free falling, $a=g=-10\,{\rm m/s^2}$.  

"Dropped'' or "released'' in free-falling problems means the initial velocity is zero, $v_0=0$.  In addition, it is always better to consider the point of release as the origin of the coordinate, so $y_0=0$. 

The most relevant kinematics equation for these known and wanted quantities is $y=-\frac 12 gt^2+v_0t+y_0$ \begin{align*} y&=-\frac 12 gt^2+v_0t+y_0 \\\\&=-\frac 12 (9.8)(3.55)^2+0+0\\\\&=\boxed{-61.8\quad {\rm m}}\end{align*} The negative indicates that the impact point is below our chosen origin . 

Problem (7): A car slows down uniformly from $45\,\rm m/s$ to rest in $10\,\rm s$. How far did it travel in this time interval?

Solution : List the data known as follows: initial speed $v_0=45\,\rm m/s$, final speed $v=0$, and the total time duration that this happened is $t=10\,\rm s$. The unknown is also the amount of displacement, $\Delta x$. 

The only kinematics equation that relates these together is $\Delta x=\frac{v_1+v_2}{2}\times \Delta t$, where $v_1$ and $v_2$ are the velocities at the beginning and end of that time interval. \begin{align*} \Delta x&=\frac{v_1+v_2}{2}\times \Delta t \\\\ &=\frac{45+0}{2}\times 10 \\\\ &=\boxed{225\,\rm m}\end{align*} Keep in mind that we use this formula when the object slows down uniformly, or, in other words when the object's acceleration is constant.

Problem (8):  A ball is thrown into the air vertically from the ground level with an initial speed of 20 m/s.  (a) How long is the ball in the air? (b) At what height does the ball reach?

Solution : The throwing point is considered to be the origin of our coordinate system, so $y_0=0$. Given the initial velocity $v_0=+20\,{\rm m/s}$ and the gravitational acceleration $a=g=-9.8\,{\rm m/s^2}$. The wanted time is how long it takes the ball to reach the ground again.

To solve this free-fall problem , it is necessary to know some notes about free-falling objects. 

Note (1): Because the air resistance is neglected, the time the ball is going up is half the time it is going down.  

Note (2): At the highest point of the path, the velocity of the object is zero. 

(a) By applying the kinematics equation $v=v_0+at$ between the initial and the highest ($v=0$) points of the vertical path, we can find the going up time. \begin{align*} v&=v_0+at \\0&=20+(-9.8)t\\\Rightarrow t&=2.04\quad {\rm s}\end{align*} The total flight time is twice this time \[t_{tot}=2t=2(2.04)=4.1\,{\rm s}\] Hence, the ball takes about 4 seconds to reach the ground. 

(b) The kinematics equation $v^2-v_0^2=2a(y-y_0)$ is best for this part. \begin{align*} v^2-v_0^2&=2a(y-y_0) \\0-20^2&=2(-9.8)(y-0) \\ \Rightarrow y&=\boxed{20\quad {\rm m}}\end{align*} Hence, the ball goes up to a height of about 20 meters.

Problem (9): An object moving in a straight line with constant acceleration, has a velocity of $v=+10\,{\rm m/s}$ when it is at position $x=+6\,{\rm m}$ and of $v=+15\,{\rm m/s}$ when it is at $x=10\,{\rm m}$. Find the acceleration of the object.

Solution : Draw a diagram, put all known data into it, and find a relevant kinematics equation that relates them together. 

We want to analyze the motion in a distance interval of $\Delta x=x_2-x_1=10-6=4\,{\rm m}$, thus, we can consider the velocity at position $x_1=6\,{\rm m}$ as the initial velocity and at $x_2=10\,{\rm m}$ as the final velocity. 

The most relevant kinematics equation that relates these known quantities to the wanted acceleration $a$ is $v^2-v_0^2=2a(x-x_0)$, where $x-x_0$ is the same given distance interval. Thus, \begin{align*} v^2-v_0^2&=2a(x-x_0) \\\\ (15)^2-(10)^2&=2(a)(4) \\\\225-100&=8a\\\\\Rightarrow a&=\frac{125}{8}\\\\&=15.6\,{\rm m/s^2}\end{align*} 

Problem (10): A moving object accelerates uniformly from 75 m/s at time $t=0$ to 135 m/s at $t=10\,{\rm s}$. How far did it move at the time interval $t=2\,{\rm s}$ to $t=4\,{\rm s}$? 

Solution : Draw a diagram and implement all known data in it as below. 

Because the problem tells us that the object accelerates uniformly, we can infer that its acceleration is constant throughout its entire path. 

Given the initial and final velocities of the moving object, we can determine its acceleration using the definition of instantaneous acceleration as follows: \[a=\frac{v_2-v_1}{t_2-t_1}=\frac{135-75}{10}=6\,{\rm m/s^2}\] To analyze the motion between the requested times (referred to as stage II in the figure), we need some information for that time interval, such as their velocities or the distance between them. 

As you can see in the figure, the initial velocity of stage II is the final velocity of stage I. By using a relevant kinematics equation that relates these data to each other, we find that \begin{align*} v&=v_0+at\\\\&=75+(6)(2) \\\\&=87\,{\rm m/s}\end{align*} This velocity will be the initial velocity for stage II of the motion. Now, all the known information for stage II is as follows: initial velocity $v_0=87\,{\rm m/s}$, acceleration $a=6\,{\rm m/s^2}$, and time interval $\Delta t=2\,{\rm s}$. The unknown is the distance traveled, denoted by $x=?$

The appropriate equation that relates all these variables is $x=\frac 12 at^2+v_0t+x_0$. Substituting the known values, we get \begin{align*}x&=\frac 12 at^2+v_0t+x_0\\\\&=\frac 12 (6)(2)^2+(87)(2)+0\\\\&=186\quad {\rm m}\end{align*} Hence, our moving object travels a distance of 186 m between the instances of 2 s and 4 s. 

Problem (11): A fast car starts from rest and accelerates at a uniform rate of $1.5\,{\rm m/s^2}$ for 4 seconds. After a while, the driver applies the brakes for 3 seconds, causing the car to uniformly slow down (decelerate) at a rate of $-2\,{\rm m/s^2}$.  (a) How fast is the car at the end of the braking period? (b) How far has the car traveled after the braking period?

Solution : This motion is divided into two parts. First, draw a diagram and specify each section's known kinematics quantities. 

(a) In the first part, given the acceleration, initial velocity, and time interval, we can calculate the final velocity at the end of 4 seconds. \begin{align*} v&=v_0+at\\&=0+(1.5)(4) \\&=6\quad {\rm m/s}\end{align*} This velocity is considered as the initial velocity for the second part, where we want to find the final velocity.

In the next part, given the magnitude of acceleration and braking time interval, we can calculate the final velocity as follows: \begin{align*} v&=v_0+at\\&=6+(-2)(3) \\&=0\end{align*} The zero velocity here indicates that the car comes to a stop after the braking period.

(b) The distance traveled in the second part can now be calculated using the kinematics equation $x=\frac 12 at^2+v_0t+x_0$, because the only unknown quantity is distance $x$. \begin{align*} x&=\frac 12 at^2+v_0t+x_0\\\\&=\frac 12 (-2)(3)^2+(6)(3)+0\\\\&=+9\quad {\rm m}\end{align*} Therefore, after braking, the car travels a distance of 9 meters before coming to a stopped. 

Problem (12): A car moves at a speed of 20 m/s down a straight path. Suddenly, the driver sees an obstacle in front of him and applies the brakes. Before the car reaches a stop, it experiences an acceleration of $-10\,{\rm m/s^2}$.  (a) After applying the brakes, how far did it travel before stopping?  (b) How long does it take the car to reach a stop? 

Solution : As always, the first and most important step in solving a kinematics problem is to draw a diagram and input all known values into it, as shown below.

(a) The kinematics equation $v^2-v_0^2=2a(x-x_0)$ is the perfect equation to use here, as the only unknown quantity in it is the distance traveled, denoted by $x$. Thus, \begin{align*} v^2-v_0^2&=2a(x-x_0) \\\\ 0^2-(20)^2&=2(-10)(x-0) \\\\\Rightarrow \quad x&=\frac {-400}{-20}\\\\&=20\quad {\rm m}\end{align*} (b)  The phrase "How long does it take'' asks us to find the time interval. The initial and final velocities, as well as acceleration, are known, so the only relevant kinematics equation is $v=v_0+at$. Thus, \begin{align*} v&=v_0+at\\\\0&=20+(-10)t\\\\\Rightarrow t&=\frac{-20}{-10}\\\\&=2\quad {\rm s}\end{align*} Therefore, after braking, the car moves for 2 seconds before coming to a stop.

Problem (13): A sports car moves a distance of 100 m in 5 seconds with a uniform speed. Then, the driver brakes, and the car, comes to a stop after 4 seconds. Find the magnitude and direction of its acceleration (assumed constant). 

Solution : uniform speed means constant speed or zero acceleration for the motion before braking. Thus, we can use the definition of average velocity to find its speed just before braking as below \begin{align*} \bar{v}&=\frac{\Delta x}{\Delta t}\\\\&=\frac{100}{5}\\\\&=20\quad {\rm m/s}\end{align*} Now, we know the initial and final velocities of the car in the braking stage. Since the acceleration is assumed to be constant, by applying the definition of average acceleration, we would have \begin{align*} \bar{a}&=\frac{v_2-v_1}{\Delta t}\\\\&=\frac{0-20}{4}\\\\&=-5\quad {\rm m/s^2}\end{align*} The negative shows the direction of the acceleration, which is toward the negative $x$-axis. 

Hence, the car's acceleration has a magnitude of $5\,{\rm m/s^2}$ in the negative $x$ direction. 

Problem (14): A race car accelerates from rest at a constant rate of $2\,{\rm m/s^2}$ in 15 seconds. It then travels at a constant speed for 20 seconds, and after that, it comes to a stop with an acceleration of $2\,{\rm m/s^2}$.  (a) What is the total distance traveled by car? (b) What is its average velocity over the entire path?

Solution : To solve this kinematics question, we divided the entire path into three parts. 

Part I: "From rest'' means the initial velocity is zero. Thus, given the acceleration and time interval, we can use the kinematics equation $v=v_0+at$ to calculate the distance traveled by the car at the end of 15 seconds for the first part of the path. \begin{align*} x&=\frac{1}{2}at^2 +v_0 t+x_0\\\\&=\frac 12 (2)(15)^2 +(0)(15)+0\\\\&=\boxed{125\quad{\rm m}}\end{align*}   As a side calculation, we find the final velocity for this part as below \begin{align*}v&=v_0+at\\\\&=0+(2) (15) \\\\&=30\quad {\rm m/s}\end{align*} Part II: the speed in this part is the final speed in the first part because the car continues moving at this constant speed after that moment. 

The constant speed means we are facing zero acceleration. As a result, it is preferable to use the average velocity definition rather than the kinematics equations for constant (uniform) acceleration.

The distance traveled for this part, which takes 20 seconds at a constant speed of 30 m/s, is computed by the definition of average velocity as below \begin{align*} \bar{v}&=\frac{\Delta x}{\Delta t}\\\\30&=\frac{\Delta x}{20}\end{align*} Thus, we find the distance traveled as $\boxed{x=600\,{\rm m}}$. 

Part III: In this part, the car comes to a stop, $v=0$, so its acceleration must be a negative value as $a=-2\,{\rm m/s^2}$. Here, the final velocity is also zero. Its initial velocity is the same as in the previous part. 

Consequently, the best kinematics equation that relates those known to the wanted distance traveled $x$, is $v^2-v_0^2=2a(x-x_0)$. \begin{align*} v^2-v_0^2&=2a(x-x_0) \\\\0^2-(30)^2&=2(-2)(x-0) \\\\ \Rightarrow \quad x&=\boxed{225\quad {\rm m}}\end{align*} The total distance traveled by car for the entire path is the sum of the above distances \[D=125+600+225=\boxed{950\quad {\rm m}}\]

Problem (15): A ball is dropped vertically downward from a tall building of 30-m-height with an initial speed of 8 m/s. After what time interval does the ball strike the ground? (take $g=-10\,{\rm m/s^2}$.)

Solution : This is a free-falling kinematics problem. As always, choose a coordinate system along with the motion and the origin as the starting point. 

Here, the dropping point is considered the origin, so in all kinematics equations, we set $y_0=0$. By this choice, the striking point is 30 meters below the origin, so in equations, we also set $y=-30\,{\rm m}$. 

Remember that velocity is a vector in physics whose magnitude is called speed. In this problem, the initial speed is 8 m/s downward. This means that the velocity vector is written as $v=-8\,{\rm m/s}$. 

Now that all necessary quantities are ready, we can use the kinematics equation $y=\frac 12 at^2+v_0t+y_0$, to find the wanted time that the ball strikes the ground. \begin{align*} y&=\frac 12 at^2+v_0t+y_0\\\\-30&=\frac 12 (-10)t^2+(-8)t+0\end{align*} After rearranging, a quadratic equation like $5t^2+8t-30=0$ is obtained, whose solutions are given as below: \begin{gather*} t=\frac{-8\pm\sqrt{8^2-4(5)(-30)}}{2(5)}\\\\ \boxed{t_1=1.77\,{\rm s}} \quad , \quad t_2=-3.37\,{\rm s}\end{gather*} $t_1$ is the accepted time because the other is negative, which is not acceptable in kinematics. Therefore, the ball takes about 1.7 seconds to hit the ground. 

Note: The solutions of a quadratic equation like $at^2+bt+c=0$, where $a,b,c$ are some constants, are found by the following formula: \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] 

Problem (16): The acceleration versus time graph for an object that moves at a constant speed of 30 m/s is shown in the figure below. Find the object's average velocity between instances $t_1=10\,{\rm s}$ and $t_2=30\,{\rm s}$. 

Solution : The best and shortest approach to solving such a kinematics problem is to first draw its velocity-vs-time graph. Next, the area under the obtained graph gives us the total displacement, which is divided by the total time interval to yield the average velocity.

The path consists of three parts with different accelerations. 

In the first part, the object slows down its motion at a constant rate of $-2\,{\rm m/s^2}$ in 10 seconds. Its initial velocity is also 30 m/s. With these known quantities in hand, the kinematics equation $v=v_0+at$ gives us the velocity at the end of this time interval. \begin{align*} v&=v_0+at\\&=30+(-2)(10) \\&=10\quad {\rm m/s}\end{align*} This calculation corresponds to a straight line between the points $(v=30\,{\rm m/s},t=0)$ and $(v=10\,{\rm m/s},t=10\,{\rm s})$ on the $v-t$ graph as shown below.

Next, the object moves with zero acceleration for 5 seconds, which means the velocity does not change during this time interval. This implies that we must draw a horizontal line in the $v-t$ graph.

In the last part, the object accelerates from 10 m/s with a constant rate of $+2\,{\rm m/s^2}$ in 15 seconds. Thus, its final velocity at the end of this time interval is determined as below \begin{align*} v&=v_0+at\\&=10+(2)(15) \\&=40\quad {\rm m/s}\end{align*} Now, it's time to draw the velocity-vs-time graph. As an important point, note that all these motions have a constant acceleration, so all parts of a velocity-time graph, are composed of straight-line segments with different slopes.

For part I, we must draw a straight-line segment between the velocities of 30 m/s and 10 m/s. 

Part II is a horizontal line since its velocities are constant during that time interval, and finally, in Part III, there is a straight line between velocities of 10 m/s and 40 m/s. 

All these verbal phrases are illustrated in the following velocity-vs-time graph . 

Recall that the area under a velocity vs. time graph always gives the displacement. Hence, the area under the $v-t$ graph between 10 s and 30 s gives the displacement. Therefore, the areas of rectangle $S_1$ and trapezoid $S_2$ are calculated as below \begin{gather*} S_1 =10\times 5=50\quad {\rm m} \\\\S_2=\frac{10+40}{2}\times 15=375\quad {\rm m}\end{gather*} Therefore, the total displacement in the time interval $[15,30]$ is \[D=S_{tot}=S_1+S_2=425\,{\rm m}\] From the definition of average velocity, we have \[\bar{v}=\frac{displacement}{time}=\frac{425}{20}=21.25\,{\rm m/s}\] 

Challenging Kinematics Problems

In the following, some challenging kinematics problems are presented for homework.

A driver is moving along at $45\,\rm m/s$ when she suddenly notices a roadblock $100\,\rm m$ ahead. Can the driver stop the vehicle in time to avoid colliding with the obstruction if her reaction time is assumed to be $0.5\,\rm s$ and her car's maximum deceleration is $5\,\rm m/s^2$?

Solution : The time between seeing the obstacle and taking action, such as slamming on the brake, is defined as the reaction time. During this time interval, the moving object travels at a constant speed. 

Thus, in all such questions, we have two phases. One is constant speed, and the other is accelerating with negative acceleration (deceleration). 

Here, between the time of seeing the barrier and the time of braking, the driver covers a distance of \begin{align*} x_1&=vt_{reac} \\\\ &=25\times 0.5 \\\\ &=12.5\,\rm m\end{align*} In the decelerating phase, the car moves a distance, which is obtained using the following kinematics equation: \begin{gather*} v^2-v_0^2=2ax_2 \\\\ (0)^2-(25)^2=2(-5) x_2 \\\\ \Rightarrow x_2=62.5\,\rm m\end{gather*} Summing these two distances gives a total distance that is a good indication of whether the moving object hits the obstacle or not. \begin{gather*} \Delta x_{actual}=x_1+x_2=75\,\rm m \\\\ \Rightarrow \Delta x_{covered}<\Delta x_{actual} \end{gather*} As a result, because the distance covered by the car is less than the actual distance between the time of seeing the barrier and the obstacle itself, the driver has sufficient time to stop the car in time to avoid a collision.

For a moving car at a constant speed of $90\,\rm km/h$ and a human reaction time of $0.3\,\rm s$; find the stopping distance if it slows down at a rate of $a=3\,\rm m/s^2$. 

Solution : We use SI units, so first convert the given speed in these units as below \begin{align*} v&=90\,\rm km/h \\\\ &=\rm 90\times \left(\frac{1000\,m}{3600\,s}\right) \\\\ &=25\,\rm m/s\end{align*} As we said previously, during the reaction time, your car moves at a constant speed and covers a distance of \begin{align*} x_1&=vt_{react} \\\\ &=(25)(0.3) \\\\ &=7.5\,\rm m \end{align*} Deceleration means the moving object slows down, or a decrease per second in the velocity of the car occurs. In this case, we must put the acceleration with a negative sign in the kinematics equations. 

During the second phase, your car has negative acceleration and wants to be stopped. Thus, the distance covered during this time interval is found as follows \begin{align*} v^2-v_0^2=2a\Delta x \\\\ (0)^2-(25)^2=2(3)\Delta x \\\\ \Rightarrow \quad \boxed{\Delta x=104.17\,\rm m}\end{align*} 

Assume you are traveling $35\,\rm m/s$ when suddenly you see red light traffic $50\,\rm m$ ahead. If it takes you $0.456\,\rm s$ to apply the brakes and the maximum deceleration of the car is $4.5\,\rm m/s^2$,  (a) Will you be able to stop the car in time?  (b) How far from the time of seeing the red light will you be? 

Solution : When you see the red light until you apply the brakes, your car is moving at a constant speed. This time interval is defined as the reaction time, $\Delta t_{react}=0.456\,\rm s$. After you get the brakes on, the car starts to decelerate at a constant rate, $a=-4.5\,\rm m/s^2$. Pay attention to the negative signs of such problems. The negative tells us that the car is decreasing its speed. 

(a) In the first phase, the car moves a distance of \begin{align*} x_1&=v\Delta t_{react} \\\\ &=35\times 0.455 \\\\ &=15.96\,\rm m\end{align*} In the phase of deceleration, the car is moving a distance whose magnitude is found using the time-independent kinematics equation as below \begin{gather*} v^2-v_0^2=2ax_2 \\\\ (0)^2-(35)^2=2(4.5)x_2 \\\\ \Rightarrow \quad x_2=136.11\,\rm m\end{gather*} The sum of these two distances traveled gives us the total distance covered by the car from the time of seeing the red traffic light to the moment of a complete stop. \[x_{tot}=x_1+x_2=152.07\,\rm m \] Because the total distance traveled is greater than the actual distance to the red light, the driver will not be able to stop the car in time. 

(b) As previously calculated, the total distance traveled by the car is nearly $152\,\rm m$ or the car is about $102\,\rm m$ past the red light traffic. 

A person stands on the edge of a $60-\,\rm m$-high cliff and throws two stones vertically downward, $1$ second apart, and sees they both reach the water simultaneously. The first stone had an initial speed of $4\,\rm m/s$.  (a) How long after dropping the first stone does the second stone hit the water? (b) How fast was the second stone released? (c) What is the speed of each stone at the instant of hitting the water?

Solution:  Because all quantities appearing in the kinematics equation are vectors, we must first choose a positive direction. Here, we take up as a positive $y$ direction.

Both stones arrived in the water at the same time. Thus, calculate the time the first stone was in the air. Next,  use the time interval between the two drops to find the duration the second stone was in the air. (a) The first stone is released downward at a speed of $4\,\rm m/s$, thus, its initial velocity is $v_0=-4\,\rm m/s$. The minus sign is for moving in the opposite direction of the chosen direction. 

The only relevant kinematics equation that relates this known information is $\Delta y=-\frac 12 gt^2+v_0t$, where $\Delta y=-60\,\rm m$ is the vertical displacement, and the negative indicates that the stone hit a point below the chosen origin. Substituting the numerical values into this and solving for the time duration $t$ gives \begin{gather*} \Delta y=-\frac 12 gt^2+v_0t \\\\ -60=-\frac 12 (10)t^2+(-4)t \\\\ 5t^2+4t-60=0 \\\\ \Rightarrow \boxed{t=3.0\,\rm s} \, , \, t'=-3.8\,\rm s \end{gather*} The second answer is not acceptable. 

(b) The second stone was released $1$ second after throwing the first one and arrived at the same time as the first stone. Therefore, the time interval that the second stone was in the air is found to be \begin{align*} t_2&=t_1-1 \\ &=3.0-1\\ &=2\,\rm s\end{align*} 

(c) It is better to apply the time-independent kinematics equation $v^2-v_0^2=-2g\Delta y$ to find the stone's velocity at the moment it hit the water. For the first stone, we have \begin{gather*} v^2-v_0^2=-2g\Delta y \\\\ v^2-(-4)^2=-2(10)(-60) \\\\ \Rightarrow \quad \boxed{v=34.8\,\rm m/s} \end{gather*} The second stone's velocity is left to you as an exercise.

In a tennis game, the ball leaves the racket at a speed of $75\,\rm m/s$ whereas it is in contact with the racket for $25\,\rm ms$, and starts at rest. Assume the ball experiences constant acceleration.  (a) What was the ball's acceleration during this serve? (b) How far has the ball traveled on this serve?

Solution : In this question, we are asked to find the ball's acceleration and distance traveled during that pretty small time interval.  (a) In a time interval of $\Delta t=25\times 10^{-3}\,\rm s$, we are given the beginning velocity $v_1=0$ and the end velocity $v_2=85\,\rm m/s$. Because it is assumed the acceleration is constant, the average acceleration definition, $a=\frac{\Delta v}{\Delta t}$, is best suited for these known quantities. \begin{align*} a&=\frac{v_2-v_1}{\Delta t} \\\\ &=\frac{75-0}{25\times 10^{-3}} \\\\ &=3000\,\rm m/s^2 \end{align*} A huge acceleration is given to the tennis ball.  (b) Here, we are asked to find the amount of distance traveled by the ball during the time the ball was in contact with the racket. Because we have a constant acceleration motion, it is best to use the following equation to find the distance traveled. \begin{align*} \Delta x&=\frac{v_1+v_2}{2}\times \Delta t \\\\ &=\frac{0+75}{2}\times (25\times 10^{-3}) \\\\ &=937.5\times 10^{-3}\,\rm m\end{align*} In millimeters, $\Delta x=937.5\,\rm mm$, and in centimeters $\Delta x=93.75\,\rm cm$. Therefore, during this incredibly short time interval, the ball moves about $94\,\rm cm$ along with the racket.

Starting from rest and ending at rest, a car travels a distance of $1500\,\rm m$ along the $x$-axis. During the first quarter of the distance, it accelerates at a rate of $+1.75\,\rm m/s^2$, while for the remaining distance, its acceleration is $-0.450\,\rm m/s^2$.  (a) What is the time travel of the whole path?  (b) What is the maximum speed of the car over this distance?

None of the time-dependent kinematics equations give us the time travel $t'$ without knowing the initial speed at the instant of the start of this second path. 

We can find it using the equation $v^2-v_0^2=2a\Delta x$, setting $v=0$ at the end of the path, and solving for $v_0$ \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ (0)^2-v_0^2=2(-0.450)(1125) \\\\ v_0=\sqrt{2\times 0.45\times 1125} \\\\ \Rightarrow v_0=31.82\,\rm m/s\end{gather*} Given that, one can use the simple equation $v=v_0+at$ and solve for the time travel in this part of the path. \begin{gather*} v=v_0+at \\\\ 0=31.82+(-0.450)t' \\\\ \Rightarrow t'=70.71\,\rm s\end{gather*} Therefore, the total time traveled over the entire path is the sum of these two times. \begin{align*} T&=t+t' \\\\ &=20.70+70.71 \\\\ &=\boxed{91.41\,\rm s} \end{align*}

A train that is $75$ meters long starts accelerating uniformly from rest. When the front of the train reaches a railway worker who is standing $150$ meters away from where the train started, it is traveling at a speed of $20\,\rm m/s$. What will be the speed of the last car as it passes the worker?

Solution : The front of the train is initially $150\,\rm m$ away from the worker, and when it passes him, it has a speed of $25\,\rm m/s$. From this data, we can find the acceleration of the front of the train (which is the same acceleration as the whole train) by applying the following kinematics equation \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ (25)^2-(0)^2=2a\times 150 \\\\ \Rightarrow \quad a=2.08\,\rm m/s^2\end{gather*} Given the train's acceleration, now focus on the last car.

The last car is initially at rest and placed at a distance of $150+80=230\,\rm m$ away from the person. When it passes the person, it has traveled $\Delta x= 230\,\rm m$ and its speed is determined simply as below \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ v^2-(0)^2=2(2.08)(230) \\\\ \Rightarrow \quad \boxed{v=30.93\,\rm m/s}\end{gather*}

A wildcat moving with constant acceleration covers a distance of $100\,\rm m$ apart in $8\,\rm s$. Assuming that its speed at the second point is $20\,\rm m/s$,  (a) What was its speed in the first place?  (b) At what rate does its speed change over this distance? 

Solution : First of all, list all known data given to us. Time interval $\Delta t=8\,\rm s$, the horizontal displacement $\Delta x=100\,\rm m$, speed at second point $v_2=20\,\rm m/s$. 

We are asked to find the speed at the second point. To solve this kinematics problem, we use the following kinematics equation because the acceleration is constant and this is the most relevant equation that relates the known to the unknown quantities. \begin{gather*} \Delta x=\frac{v_1+v_2}{2}\times \Delta t \\\\ 100=\frac{v_1+20}{2}\times 8 \\\\ \Rightarrow \quad \boxed{v_1=5\,\rm m/s} \end{gather*} Therefore, the wildcat's speed in the first place is $5\,\rm m/s$. 

In this part, we should find the wildcat's acceleration because acceleration is defined as the time rate of change of the speed of a moving object. Given the first place speed, $v_1=5\,\rm m/s$, found in the preceding part, we can use the following time-independent kinematics equation to find the wanted unknown. \begin{gather*} v_2^2-v_1^2=2a\Delta x \\\\ (20)^2-(5)^2=2a(100) \\\\ \Rightarrow  \quad \boxed{a=1.875\,\rm m/s^2}\end{gather*}

In this tutorial, all concepts about kinematics equations are taught in a problem-solution strategy. 

We can also find these kinematic variables using a position-time  or velocity-time graph. Because slopes in those graphs represent velocity and acceleration, respectively, and the concavity of a curve in a position vs. time graph shows the sign of its acceleration in an x-t graph  as well.

These multiple-choice questions on kinematics for AP Physics 1 are also available to review for students enrolled in AP Physics courses.

Author : Dr. Ali Nemati Date published : 8-7-2021 Updated : June 12, 2023

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Table of content

Full table of contents

When analyzing one-dimensional motion with constant acceleration, the problem-solving strategy involves identifying the known quantities and choosing the appropriate kinematic equations to solve for the unknowns. Either one or two kinematic equations are needed to solve for the unknowns, depending on the known and unknown quantities. Generally, the number of equations required is the same as the number of unknown quantities in the given example. Two-body pursuit problems always require two equations to be solved simultaneously to derive the value of the unknown.

In complex problems, it is not always possible to identify the unknowns or the order in which they should be calculated. In such scenarios, it is useful to make a list of unknowns and draw a sketch of the problem to identify the directions of motion of an object. To solve the problem, substitute the knowns along with their units into the appropriate equation. This step provides a numerical answer, and also provides a check on units that can help find errors. If the units are incorrect, then an error has been made. However, correct units do not necessarily guarantee that the numerical part of the answer is also correct.

The final step in solving problems is to check the answer to see if it is reasonable. This final step is crucial as the goal of physics is to describe nature accurately. To see if the answer is reasonable, check both its magnitude and its sign, in addition to its units. This enables us to get a conceptual understanding of the problems that are being solved. Sometimes the physical principle may be applied correctly to solve the numerical problem, but produces an unreasonable result. For example, if an athlete starting a foot race accelerates at 0.4 m/s² for 100 seconds, their final speed will be 40 m/s (about 150 km/h). This result is unreasonable because a person cannot run at such a high speed for 100 seconds. Here, the physics is correct in a sense, but there is more to describing nature than just manipulating equations correctly.

This text is adapted from Openstax, University Physics Volume 1, Section 3.4: Motion with Constant Acceleration .

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Kinematic Equations

Learn about the kinematic equations., kinematic equations lesson, the four kinematic equations.

• Equation 1:      v = v 0 + at
• Equation 2:      v 2 = v 0 2 + 2a(Δx)
• Equation 3:      x = x 0 + v 0 t
• Equation 4:      x = x 0 + v 0 t + 1 / 2 at 2

Kinematic Variables

• x - Displacement
• v - Velocity
• a - Acceleration

These are the four variables at play with the kinematic equations. The equations describe the motion of an object that is subject to constant acceleration. By leveraging the equations, we can solve for the initial and final values of these variables.

Important: Directionality of the Variables

Displacement, velocity, and acceleration are all directional, whereas time is non-directional. Directional variables will have a positive value when their vector points in the positive direction, and a negative value when their vector points in the negative direction. Since time is non-directional, it will always have a positive value.

When using the kinematic equations to solve problems, it is helpful to choose a direction for positive x , v , and a . Sometimes this direction is given in the problem statement. The opposite of the positive direction will be the negative direction.

When to use the Kinematic Equations

Kinematics is the study of object motion without reference to the forces that cause motion. The kinematic equations are simplifications of object motion. Three of the equations assume constant acceleration (equations 1, 2, and 4), and the other equation assumes zero acceleration and constant velocity (equation 3).

When an object motion problem falls into these categories, we may use the kinematic equations to solve it. For example, we can use them to figure out how far a projectile flies as long as the projectile experiences constant acceleration during its flight. Another example of a simple and effective use of the kinematic equations is when a car driving at constant velocity. Since it has zero acceleration, we may use the equation that has no acceleration terms (equation 3).

How to Choose a Kinematic Equation

We must decide which kinematic equation is best for what we are solving. Generally, kinematics problems involve solving for some unknown. The unknown could be initial displacement, final displacement, change in displacement, initial velocity, final velocity, acceleration, or time.

We choose an equation based on what is known and what is not known. In some cases, we must use several equations sequentially to find the value of our unknown. The example problems below give more insight for the process of choosing an equation.

Kinematic Equations Example Problems

Kinematic equation 1 example.

A projectile is fired from the chamber of a cannon and accelerates at 1500 m/s 2 for 0.75 seconds before leaving the barrel. What is the projectile’s velocity as it leaves the barrel of the cannon?

• Since the projectile is not moving before it is fired, our initial velocity will be zero. We are given the values of acceleration (1500 m/s 2 ) and time (0.75 seconds), so the equation we will use is v = v 0 + at .
• Since we know the values of all variables except final velocity, we may plug in our known values to find v, this gives us v = (0 m/s) + (1500 m/s 2 )(0.75 seconds) = 1125 m/s .
• The projectile's velocity is 1125 m/s as it leaves the barrel.

Kinematic Equation 2 Example

A sprinter is moving at 5 m/s when they reach 20 meters into their race. They maintain a constant acceleration of 2 m/s 2 through 40 meters into the race. What is the sprinter’s velocity 40 meters into the race?

• We know the values of initial velocity (5 m/s), acceleration (2 m/s 2 ), and change in displacement (40 – 20 = 20 meters). The equation we will use is v 2 = v 0 2 + 2a(Δx) .
• Since we know the values of all variables except final velocity, we may plug in our known values and find v. v 2 = (5 m/s) 2 + 2(2 m/s 2 )(20 meters) v 2 = 105 v = 10.25 m/s
• The sprinter's velocity is 10.25 m/s at 40 meters into the race.

Kinematic Equation 3 Example

A car is 200 meters away from a building and starts driving away from the building at 20 m/s. How far away is the car from the building after driving for 6 seconds?

• We know the values of initial displacement (200 meters), initial velocity (20 m/s), and time in motion (6 seconds). We must find final displacement. The kinematic equation we will use is x = x 0 + v 0 t .
• Since we know the values of all variables but one, we may plug in our known values to find the unknown value of x. x = (200 meters) + (20 m/s)(6 seconds) = 320 meters
• After driving for 6 seconds, the car is 320 meters away from the building.

Kinematic Equation 4 Example

A person is standing 6 meters behind you. They throw a ball over your head with a horizontal velocity of 20 m/s. The ball experiences a constant 4 m/s 2 of horizontal deceleration while in flight and takes 2 seconds to land. How far in front of you does the ball land?

• We know the values of initial displacement (-6 meters), initial velocity (20 m/s), acceleration (-4 m/s 2 ), and time (2 seconds). The equation we will use is x = x 0 + v 0 t + 1 / 2 at 2 .
• Since we know the values of all variables except final displacement, we may plug in our known values and find x. x = (-6 meters) + (20 m/s)(2 seconds) + 1 / 2 (-4 m/s 2 )(2 seconds) 2 x = -6 + 40 - 8 x = 26 meters
• The ball lands 26 meters in front of you.

Bonus Kinematics Lesson

What is projectile motion.

Earlier we mentioned that projectile motion problems can be solved by using the kinematic equations. While projectile motion is just one of the many types of problems we come across in kinematics, it is very useful to understand exactly what it is.

By definition, projectile motion is the motion of an object or particle that is only affected by the force of gravity. A projectile follows a parabolic trajectory similar to when a ball is thrown. This parabolic trajectory is also called a ballistic trajectory.

How to Solve Projectile Motion

Since we neglect the force of air resistance in projectile motion, we can figure out how far a projectile will fly by decomposing its velocity into vertical and horizontal components. The horizontal component of velocity will be constant during the flight. The vertical component of velocity will affect the projectile's time of flight, or hang time.

Here's the equations for a projectile's horizontal and vertical motion. They are derived from the kinematic equations.

• v x = v 0 cos(α)
• x = v 0 cos(α)t + x 0
• v y = v 0 sin(α)
• t = 2v y / g

Projectile Motion Example

A ball is thrown from the surface of Earth at 25 m/s on an angle of 30° above the horizontal. Neglecting air resistance, how far does the ball travel horizontally before hitting the ground?

• First, let's set our known variables. v 0 = 25 m/s α = 30° g = 9.81 m/s 2 .
• Let's find the velocity components. v x = v 0 cos(α) = 25cos(30°) = 21.65 m/s v y = v 0 sin(α) = 25sin(30°) = 12.5 m/s
• Now we can find the hang time, and use the hang time to calculate horizontal distance traveled. t = 2v y / g = 2(12.5) / 9.81 = 2.55 x = v 0 cos(α)t + x 0 = 25cos(30°)(2.55) + 0 = (21.65)(2.55) = 55.21
• The ball travels horizontally for 55.21 meters before hitting the ground.

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18 Motion With Constant Acceleration

Exercise 18.1: Motion Subject to a Constant Acceleration

These are the kinematic equations. Notice that the only assumption that goes into these equations is “constant acceleration.”  Therefore,

Every problem with constant acceleration can be solved using the kinematic equations. 

Conversely,

If an object does  NOT have constant acceleration, you should not use the kinetic equations.

Let’s solve a problem with constant acceleration to see how we can use the kinematic equations.

Exercise 18.2: Tesla Joy Ride

A Tesla roadster goes from 0 mph to 60 mph in 1.9 seconds (60 mph is about 26.8 m/s). This is quite a high acceleration, as should be clear from the following video:

Assuming the Tesla moves with constant acceleration, find the distance the vehicle has covered by the time it reaches 60 mph.

We’ll do this problem together.  Before we start with step 1 of the three steps of problem solving, I want you to take the time to summarize the story of the problem.  This story has a clear beginning and end, so make sure you identify these end points.

Step 1: D raw a picture. The picture should include all  of the information in the problem!  If you’re stuck, look at my solution, and fix anything that needs fixing so you can move on to step 2.

Step 2: Find relations.   Given the conditions of the problem, what are the relations that hold between the variables included in your picture?

If you’re stuck, you can look at my solution, and fix anything that needs fixing before moving on.

Step 3: Solve unknowns. Use the equations you found above to solve for the variables you were not given.  Try it, and then compare to my answer.

The problem above should help you see just how important it is to have a good understanding of the story behind the problem.  This is so much so, that from now on, I will officially add a “step 0” to the rules of problem solving.  The  four steps of problem solving are:

• Step 0: What’s the story?
• Step 1:  Draw a picture.
• Step 2:  Find Relations. 
• Step 3:   Solve Unknowns. 

I also want you to notice that when solving the problem, we ended up using the kinematic equations to relate the beginning of the problem to the end of the problem.  I find that this is the most useful way of thinking about the kinematic equations.  Consequently, from this point on I will write the kinematic equations as follows:

Throughout this book, you will occasionally run into tablets like the ones below.

These tablets contain equations that you absolutely should know by the end of this class/book.  I hope you will find that the number of equations you need to remember is, in fact, very small.

Exercise 18.3: Stopping a Bullet

Let’s start by looking at the following video (original video here ).

How long does it take for the bullet to stop as it travels through the sand?

Try solving the problem, starting with step 0: What is the story? 

Now step 1:  Draw a picture.  

Now bring it home: find relations and solve unknowns.

D. The barrel of a desert eagle 50 caliber gun is 6 in long.  Which acceleration is larger in magnitude, the acceleration of the bullet as it shoots out of the gun, or the bullet’s acceleration as it is stopped by the sand? Explain you answer in English: you don’t need to do any algebra.

Key Takeaways

The motion of objects with constant acceleration is described by the  kinematic equations.   Write down these equations  from memory .

Practice Problems

PP 18.1: A car is driving at 60 mph down the highway.  At that speed, the typical stopping distance of a car is 55 m. How long does it take for a car to break this distance?

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Kinematic Equations for Constant Acceleration Calculator

This kinematics calculator will help you to solve constant acceleration problems using kinematic equations

As you may know, there are two main kinematic equations of motion for uniform, or constant, acceleration.

Thus, we have five parameters of motion: initial velocity V₀ , final velocity V , acceleration a , time t , and displacement, or distance, S , and two equations. Therefore, to use these equations, we need three known parameters and two unknown parameters. Also, as Combinatorics – combinations, arrangements and permutations tells us, the number of combinations of 3 from 5 is 10, so there are ten types of kinematic equations problems at all; each has a different set of known parameters.

This calculator allows you to enter any three known parameters and clear the parameters that should be found, and it kindly finds them. Kinematic equations for each set of parameters are listed below the calculator. BTW, by default, the acceleration has the value of gravity constant g , making a free-fall problem.

Kinematic Equations Calculator

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• m/s² meters per second per second
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Kinematic equations

Below are ten types of problems along with the solution formulas.

Case 1. Find the unknowns given initial velocity, acceleration, and time

Example problem: An airplane accelerates down a runway at n m/s² for m seconds until it finally lifts off the ground. Determine the final velocity and the distance traveled before takeoff. Solution: Although the problem specifies only the acceleration and the time, the third parameter is known implicitly. The problem assumes that an airplane starts from the rest, hence, its initial velocity is zero. Thus, we can use our kinematic equations as is

Case 2. Find the unknowns given initial velocity, final velocity, and time

Example problem: An airplane accelerates down a runway for n seconds until it finally lifts off the ground having a speed of m m/s. Determine the acceleration and the distance traveled before takeoff. Solution: Again we know that the initial velocity is zero. To solve the problem, we need to rearrange our kinematic equations like so:

Case 3. Find the unknowns given initial velocity, final velocity, and acceleration

Example problem: An airplane accelerates down a runway at n m/s² until it finally lifts off the ground having a speed of m m/s. Determine the time and the distance traveled before takeoff. Solution: The same case with the initial velocity being zero. So, our kinematic equations will be:

Case 4. Find the unknowns given initial velocity, final velocity, and distance

Example problem: An airplane accelerates down a runway until it finally lifts off the ground having a speed of n m/s after traveling the distance of m meters. Determine the time and the acceleration. Solution: As usual we know that the initial velocity is zero. And we use the kinematic equations like so:

Here it is easier to find the time first, then plug the time into the second equation.

Case 5. Find the unknowns given initial velocity, time, and distance

Example problem: An airplane accelerates down a runway for n seconds until it finally lifts off the ground after traveling the distance of m meters. Determine the liftoff speed and the acceleration. Solution: In this problem, the initial velocity is zero as well. The kinematic equations for this case are:

Note that here it is easier to find the final velocity first, then plug it into the second equation.

Case 6. Find the unknowns given initial velocity, acceleration, and distance

Example problem: An airplane accelerates down a runway at n m/s² until it finally lifts off the ground after traveling the distance of m meters. Determine the liftoff speed and the time. Solution: Yet again the initial velocity is zero. However, this case is quite complicated, because the only way to find the time is to solve the quadratic equation :

Of course, to find the time you need to pick up the positive root. After that, you can plug the time into the next equation to find the final velocity:

Case 7. Find the unknowns given final velocity, time, and distance

Example problem: An car decelerates for n seconds until it finally stops after traveling the distance of m meters. Determine the initial speed and the deceleration. Solution: Now the initial velocity is unknown and the final velocity is zero. Also, we will get a negative value for acceleration, meaning that the car decelerates. To solve the problem, we need to use the following form of the kinematic equations:

Here it is easier to find the initial velocity first, then plug it into the second equation.

Case 8. Find the unknowns given final velocity, time, and acceleration

Example problem: An car accelerates at n m/s² for t seconds until it reaches m m/s. Determine the initial speed and the distance traveled. Solution: Again we do not know the initial velocity. We need to use the kinematic equations like so:

Case 9. Find the unknowns given final velocity, acceleration, and distance

Example problem: An car accelerates at n m/s² for m meters until it reaches m m/s. Determine the initial speed and the time. Solution: It is also a complicated case, there we again need to solve the quadratic equation to find the time. Our equation will be:

Then time is found, find the initial velocity:

Case 10. Find the unknowns given time, acceleration, and distance

Example problem: An car accelerates at n m/s² for m seconds and it traveled s meters. Determine the initial speed and the final speed. Solution: Use the following kinematic equations:

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How to use the Kinematic Equations (w/ Derivations)

The kinematics equations describe the motion of an object undergoing constant acceleration. These equations relate the variables of time, position, velocity and acceleration of a moving object, allowing any of these variables to be solved for if the others are known.

Below is a depiction of an object undergoing constant acceleration motion in one dimension. The variable ‌ t ‌ is for time, position is ‌ x, ‌ velocity ‌ v ‌ and acceleration ‌ a ‌. The subscripts ‌ i ‌ and ‌ f ‌ stand for "initial" and "final" respectively.

Kinematic Equations

There are three primary kinematic equations of motion listed below which apply when working in one dimension with constant acceleration. These equations are:

where ‌ a ‌ is the acceleration, ‌ v_i ‌ is the starting velocity, ‌ v_f ‌ is the final velocity of the object after time interval ‌ Δt ‌, and ‌ Δx ‌ is the distance traveled. It can be seen above that this set of equations allows for an exclusion of one quantity from the calculations. The first kinematic equation does not use position (‌ Δx) ‌ at all, the second equation does not have final velocity, and the third equation is timeless, so it does not utilize ‌ Δt. ‌ You can solve for each of these equations and other equivalent relationships by substituting them into one another and simplifying; they act as a system of equations.

Notes About the Kinematic Equations

• These kinematic formulas only work with a constant acceleration (which may be zero in the case of constant velocity).
• Depending upon which source you read, the final quantities may not have a subscript ‌ f ‌, and/or might be represented in function notation as ‌ x(t) ‌ – read “‌ x ‌ as a function of time” or “‌ x ‌ at time ‌ t ‌” – and ‌ v(t) ‌. Note that ‌ x(t) ‌ does NOT mean ‌ x ‌ multiplied by ‌ t ‌
• Sometimes the quantity ‌ x f - x i ‌ is written ‌ Δx ‌, meaning “the change in ‌ x ‌,” or even simply as ‌ d ‌, meaning displacement. All are equivalent.
• Position, velocity and acceleration are vector quantities, meaning they have direction associated with them. In one dimension, direction is typically indicated by signs – positive quantities are in the positive direction and negative quantities are in the negative direction.
• Subscripts: "0" might be used for initial position and velocity instead of ‌ i ‌. This "0" means "at ‌ t ‌ = 0," and ‌ x o ‌ and ‌ v o ‌ are typically pronounced "x-naught" and "v-naught."
• Only one of the equations does not include time. When writing out givens and determining what equation to use, this is key!
• The distance traveled is a a total distance traveled to the final position, not the displacement (i.e. it is not necessarily a straight line from the starting position to the final location).

A Special Case: Free Fall

Free-fall motion is the motion of an object accelerating due to gravity alone in the absence of air resistance. The same kinematic equations apply; however, the acceleration value near the Earth’s surface is known. The magnitude of this acceleration is often represented by ‌ g ‌, where g = 9.8 m/s 2 . The direction of this acceleration is downward, towards the Earth’s surface. (Note that some sources may approximate ‌ g ‌ as 10 m/s 2 , and others may use a value that is accurate to more than two decimal places.)

Problem Solving Strategy for Kinematics Problems in One Dimension:

Sketch a diagram of the situation and choose an appropriate coordinate system. (Recall that ‌ x ‌, ‌ v ‌ and ‌ a ‌ are all vector quantities, so by assigning a clear positive direction, it will be easier to keep track of signs.)

Write a list of known quantities. (Beware that sometimes the known values are not always obvious. Look for phrases like “starts from rest,” meaning that ‌ v i ‌ = 0, or “hits the ground,” meaning that ‌ x f ‌ = 0, and so on.)

Determine which quantity the question wants you to find. What is the unknown you will be solving for?

Choose the appropriate kinematic equation. This will be the equation that contains your unknown quantity together with known quantities.

Solve the equation for the unknown quantity, then plug in known values and compute the final answer. (Be careful about units! Sometimes you will need to convert units before computing.)

One-Dimensional Kinematics Examples

‌ Example 1: ‌ An advertisement claims that a sports car can go from 0 to 60 mph in 2.7 seconds. What is the acceleration of this car in m/s 2 ? How far does it travel during these 2.7 seconds?

‌ Solution: ‌

Known and unknown quantities:

The first part of the question requires solving for the unknown acceleration. Here we can use equation #1:

Before we plug in numbers, however, we need to convert 60 mph to m/s:

So the acceleration is then:

In order to find how far it goes in that time, we can use equation #2:

‌ Example 2: ‌ A ball is thrown up at a speed of 15 m/s from a height of 1.5 m. How fast is it going when it hits the ground? How long does it take to hit the ground?

To solve the first part, we can use equation #3:

Everything is already in consistent units, so we can plug in values:

There are two solutions here. Which one is correct? From our diagram, we can see that the final velocity should be negative. So the answer is:

To solve for time, we can use either equation #1 or equation #2. Since equation #1 is simpler to work with, we will use that one:

‌ Note that the answer to the first part of the this question was not 0 m/s. ‌ While it is true that after the ball lands, it will have 0 velocity, this question wants to know how fast it is going in that split second before impact. ‌ Once the ball makes contact with the ground, our kinematic equations no longer apply because acceleration will not be constant. ‌

Kinematic Equations for Projectile Motion (Two Dimensions)

A projectile is an object moving in two dimensions under the influence of Earth's gravity. Its path is a parabola because the only acceleration is due to gravity. The kinematic equations for projectile motion take a slightly different form from the kinematic equations listed above. We use the fact that motion components that are perpendicular to each other – such as the horizontal ‌ x ‌ direction and the vertical ‌ y ‌ direction – are independent.

Problem Solving Strategy for Projectile Motion Kinematics Problems:

‌ Example 1: ‌ A projectile is launched horizontally from a cliff of height 20 m with an initial velocity of 50 m/s. How long does it take to hit the ground? How far from the base of the cliff does it land?

We can find the time it takes to hit the ground by using the second vertical motion equation:

Then to find where it lands, ‌ x f ‌, we can use the horizontal motion equation:

‌ Example 2: ‌ A ball is launched at 100 m/s from ground level at an angle of 30 degrees with the horizontal. Where does it land? When is its velocity the smallest? What is its location at this time?

First we need to break the velocity vector into components:

First we need to find the time the ball is in flight. We can do this with the second vertical equation‌ . ‌ Note that we use symmetry of the parabola to determine that the final ‌ y ‌ velocity is the negative of the initial:

Then we determine how far it moves in the ‌ x ‌ direction in this time:

Using the symmetry of the parabolic path, we can determine that the velocity is smallest at ‌ 5.1 s ‌, when the projectile is at the peak of its motion and the vertical component of velocity is 0. The x- and y-components of its motion at this time are:

‌ Equation #1: ‌ If the acceleration is constant, then:

Solving for the velocity, we have:

‌ Equation #2: ‌ The average velocity can be written in two ways:

If we replace ‌ v f ‌with the expression from equation #1, we get:

Solving for ‌ x f ‌ gives:

Equation #3: Start by solving for ‌ t ‌ in equation #1

Plug this expression in for ‌ t ‌ in the average velocity relationship:

Rearranging this expression gives:

Sketch a diagram of the situation. Just as with one-dimensional motion, it is helpful to sketch the scenario and indicate the coordinate system. Instead of using the labels ‌ x ‌, ‌ v ‌ and ‌ a ‌ for position, velocity and acceleration, we need a way of labeling the motion in each dimension separately.

For the horizontal direction, it’s most common to use ‌ x ‌ for position and ‌ v x ‌ for the x-component of velocity (note that acceleration is 0 in this direction, so we don’t need a variable for it.) In the ‌ y ‌ direction, it is most common to use ‌ y ‌ for position and ‌ v y ‌ for the y-component of velocity. Acceleration can either be labeled ‌ a y ‌ or we can use the fact that we know the acceleration due to gravity is ‌ g ‌ in the negative y-direction, and just use that instead.

Write a list of known and unknown quantities by splitting the problem into two sections: vertical and horizontal motion. Use trigonometry to find the x- and y-components of any vector quantities that do not lie along an axis. It can be helpful to list this out in two columns:

Note: If velocity is given as a magnitude together with an angle, ‌ Ѳ ‌, above the horizontal, then use vector decomposition:

We can consider our three kinematic equations from before and adapt them to the x and y directions respectively.

X direction:

Y direction:

‌ Note that the acceleration in the ‌ y ‌ direction is -g if we assume up is positive. ‌ A common misconception is that g = -9.8 m/s 2 , but this is incorrect; ‌ g ‌ itself is simply the magnitude of the acceleration: g = 9.8 m/s 2 , so we need to specify that the acceleration is negative.

Solve for one unknown in one of those dimensions, and then plug in what is common across both directions. While the motion in the two dimensions is independent, it happens on the same time scale, so the time variable is the same in both dimensions. (The time it takes the ball to undergo its vertical motion is the same as the amount of time it takes to undergo its horizontal motion.)

Rotational Kinematics

These kinematics equations are also equally valid to describe rotational motion, albeit with angular acceleration, tangential acceleration, and other rotational quantities instead of their conventional counterparts. This is beyond the scope of this article, but there are many great resources for analyzing the kinematics of rotational motion .).

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• Isaac Physics: Equations of Motion
• Georgia State University: HyperPhysics: Motion

About the Author

Gayle Towell is a freelance writer and editor living in Oregon. She earned masters degrees in both mathematics and physics from the University of Oregon after completing a double major at Smith College, and has spent over a decade teaching these subjects to college students. Also a prolific writer of fiction, and founder of Microfiction Monday Magazine, you can learn more about Gayle at gtowell.com.

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Kinematics Calculator

Select the parameters and write the required ones against them. The calculator will readily calculate results by employing kinematics equations.

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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.

What is Kinematics?

Kinematics is referred to as a subfield of physics that developed in classical mechanics. In physics, it represents the motion of points, bodies, as well as the system of bodies without considering the forces, which cause them to move. More specifically, kinematics indicated as the study of the objects in motion, there velocity, acceleration and momentum. Example: Train moving, Flowing water in a river. It doesn't matter whether you are dealing with the motion of points, or an object, this kinematics calculator helps you to determine the kinematics.

What are Kinematics Formulas?

The kinematic formulas are referred to as a set of formulas that use the five kinematic variables given below:

\(s\) = Displacement

\(t\) = Time taken

\(u\) = Initial velocity

\(v\) = Final velocity

\(a\) = Constant acceleration

If you know any three of these five kinematic variables \((s, t, u, v, a)\) for an object under constant acceleration, then you can use a kinematic formula. Typically, the kinematic formulas are written as the given four equations. Even our kinematic equations calculator uses the following four equations to find the unknown variables: $$ v = u + at $$ $$ s = ut + \frac {1}{2}at^2 $$ $$ v^2 = u^2 + 2as $$ $$ s = (\frac {v + u}{2}) t $$ Remember that these formulas are only accurate if the acceleration is constant during the time taken considered, so, you should be careful not to use them when the acceleration is changing. Also, the variables of the kinematic equations are referring to the same direction: horizontal x, vertical y. For ease, you can also use our online acceleration calculator for the calculations of the acceleration of the moving object from the different calculation formulas.

Few Manual Examples Where You Can Use the Kinematics Calculator:

The calculations for the moving object become very easy with this tool. Here we have some manual examples. After reading, you will be able to completely solve different kinematics equations: Read on!

An object start with the velocity of \(2ms^{-1}\) , after \(8\) seconds attain the velocity of \(30ms^{-1}\). Determine the acceleration and the distance covered by the object?

\(u = 2 ms^{-1}\)

\(v = 30 ms^{-1}\)

So, By first equation of motion:

\(v = u + at\)

\(30 = 2 + a(8)\)

\(30-2 = 8a\)

\(28 = 8a\)

\(a = 28/8\)

\(a = 3.5ms^{-12}\)

Now, using second equation of motion:

\(S = ut + ½ at^{2}\)

\(S = (2)(8) + ½ (3.5)(8)^{2}\)

\(S = 16 + ½ (3.5)(64)\)

\(S = 16 + ½ (224)\)

\(S = 16 +112\)

\(S = 128m\)

A body moves with an acceleration of \(4ms^{-2}\) in \(14 s\) and covers a displacement of \(40 m\). Find the initial & final velocity of the body?

\(a = 4ms^{-2}\)

\(t = 14s\)

\(S = 40m\)

From the second equation of motion:

\(u = S - ½ at^{2} / t \)

\(u = 40 - ½ (4)(14)^{2}/ 14\)

\(u = 40 - ½ (4)(196) / 14\)

\(u = 40 – (0.5) (4)(196) / 14\)

\(u = 40 – 392 / 14\)

\(u =– 352 / 14\)

\(u = -25.14ms^{-1}\)

Now, using first equation:

\(v = -25.14 + 4*14\)

\(v = -25.14 + 56\)

\(v = 30.85ms^{-1}\)

No doubt, it is very hard to remember these formulas for kinematics, but, thanks to the kinematic equations calculator that helps you to solve equations of motion problems accurtaely.

How to Use This Online Kinematics Calculator?

Follow the below-mentioned steps to use this kinematics solver.

• First of all, choose which two variables you want to find from the dropdown menu.
• Then, enter the data in all the fields according to the selected option.
• Lastly hit the calculate button.

Outputs: Once you fill all the fields, the calculator shows:

• Initial Velocity.
• Final Velocity.
• Displacement
• Acceleration
• Formulas that are used.

Note: No matter, what’s your input is, the kinematics calculator shows you the results according to your selected kinematic equations.

Frequently Ask Questions (FAQ's):

How do i find average acceleration in kinematics.

The acceleration is the rate of change of velocity of a moving object. Simply, dividing the velocity with the time taken by an object gives the acceleration of the object.

Is Time a Kinematic Variable?

Kinematic variables including position, velocity & acceleration of the body can be used to describe the state of rest or motion of the body. They are used in many real-life fields like mechanical engineering, biomechanics, and robotics to describe the motion of the engine, the skeleton of the human body, or the robot. So, to solve the kinematics formulas for any of the variables, you can try this online kinematics calculator that helps you to do the calculations for the state of the moving object accurately.

References:

From the source of Wikipedia : General overview of Kinematics From the site of Khanacademy : Kinematic formulas & Equation of motion From the source of physicsclassroom : How to use the kinematic equations  

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Intro to Kinematic Equations Video Tutorial

• What are the four kinematic equations and what do the symbols of the equations mean?
• How are the kinematic equations used to solve problems?

View on YouTube