problem solving with kinematics equation 2

Kinematic Equations: Explanation, Review, and Examples

The Albert Team
Last Updated On: April 29, 2022

problem solving with kinematics equation 2

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*}

On the following page you can find over 40+ questions related to applying kinematics equations in velocity and acceleration:

Velocity and acceleration problems

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 uniformly accelerated kinematics problem

(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 moving car with two acceleration in kinematics problems

(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).

A driver notices a roadblock and apply the brakes.

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?

A car uniformly accelerate during two distinct phases

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?

Finding the speed of the last car of train moving uniformly

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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Physics library

Course: physics library > unit 1.

Average velocity for constant acceleration
Acceleration of aircraft carrier take-off
Airbus A380 take-off distance
Deriving displacement as a function of time, acceleration, and initial velocity
Plotting projectile displacement, acceleration, and velocity
Projectile height given time
Deriving max projectile displacement given time
Impact velocity from given height
Viewing g as the value of Earth's gravitational field near the surface
What are the kinematic formulas?

Choosing kinematic equations

Setting up problems with constant acceleration
Kinematic formulas in one-dimension

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Kinematics (Description of Motion) Problems

Also known as motion problems, these problems ask you to describe motion. Time is a key variable that tells you to work with the kinematic equations. If you are only asked for positions and velocities, you may also be able to work the problem using Conservation of Energy.

Example Problems

1-d kinematics problems.

Accelerating Car
Ball Thrown Straight Up
Accelerating Merry-Go-Round

2-D Kinematics Problems

2-D Kinematics Problem: Range of a Baseball
2-D Kinematics Problem: Height of an Arrow

Changing Acceleration Problems

Height With Changing Acceleration
X-T, V-T, and A-T Graphs

How to Solve Kinematic Problems

1. identify the problem.

2. Draw a Picture:

Because kinematic problems focus on describing motion, your picture should be a picture of the motion or path of the object of interest. Make sure to mark key points on the picture (any point about which you are given information, any point about which you are asked for information, and any point such as the top of the motion where you know information (like v y = 0 at the top of the arc) without being explicitly told. Be careful: you do NOT know that v = 0 when an object reaches the ground. As soon as it touches the ground, there is a new force on it and the problem changes. The kinematic equations you will use are valid only up until the point where the object almost touches down.

In addition to marking key points, you should pick an origin. In other words, pick a location where x = 0 and y = 0. All other position values are then measured from that location.

Finally, kinematic variables (x, v, and a) are vectors, and so you will only be able to work with the equations one direction at a time. It is very easy to forget to work with the correct component while you are in the middle of the problem, so taking a few minutes to divide all vectors into components and organize them in a table will save a lot of mistakes and time in the long run.

3.Select the Relation

There are three key kinematic equations. If you carefully select the equation which most directly describes the situation in your problem, you will not only solve the problem in fewer steps but also understand it better. The three equations, written for motion in the x-direction, are:

x = x 0 + v 0 Δt + ½ a(Δt) 2 (relates position and time)
v = v 0 + aΔt (relates velocity and time)
v 2 = v 0 2 + 2a(Δx) (relates velocity and position)

Note that some books give more than three equations. For example, they might reproduce these equations with –g instead of a for motion in the y-direction, or combine them together into a range equation. All of those are special case equations—they are partially solved versions of these three basic equations and only work in certain situations. (The range equation, for example, is only valid if the object lands at the same height from which it was thrown.) It is far better to always start with one of the three basic equations. They always apply and are always used in the same way.

The greatest challenge in kinematic problems is picking the best equation to use to solve your problem. The way to do this is to think about which answer you want. If the problem asks you for position, which position is it? If you want position at a given time, pick equation 1. If you want position at a given velocity, pick equation 2. v 0 and a occur in all equations, and so if you are asked to solve for either of those variables you will need to identify two known quantities and select the equation that relates them.

4. Solve the Problem

The kinematic equations are vector equations and so must be solved in the x- and y-directions separately. For many problems, you will only need to work in one dimension. If you ever need to relate the two (for example if your question asks for an x-position when the object has a given y-position) you need to go through time. Use what you know to solve for time in that direction, then use time in the other direction for your desired information.

The only tricky algebra in kinematics is if you need to solve equation 1 when v 0 is not equal to zero. In that case, you will need to use the quadratic equation .

5. Understand the Results

Once you have completed the problem, look at it again. Does your answer make sense? Did it give the behavior you intuitively expected to find? Can you now do steps that caused you problems earlier or do a better job identifying which equation is best to use? Can you explain in words what is happening? If you only recognized it as a kinematics problem because of the section heading in your text book, identify the information that you would use to recognize a problem like this on the final exam.

Help! I can’t find an example that looks like the problem I need to work!

Are you certain your problem is a kinematics problem.

A common mistake is to think too hard. If you are told distance and asked for average velocity, or vice versa, the two are directly related through just the definition of average velocity. Check Definition and Ratio problems to see if you can find a useful example.

It is also possible that your problem is better solved using energy and momentum. There is a lot of overlap between kinematics and energy problems, so check Energy and Momentum problems to see if you can find a useful example there.

Yes, my problem is definitely a kinematics problem.

In that case, think more broadly about what makes a useful example. Remember, you were given your assignment to practice the problem solving approach, not because the answers to your problems are particularly interesting. An example in which you merely substitute your numbers for those in the problem will give you practice entering numbers on your calculator but will teach you nothing about physics, and when you take your exam every problem on it will feel new and different to you. So think about your example as support for helping you to practice the problem solving approach.

And every single problem in this section uses the very same approach, so any problem is an appropriate example to help you approach your problem. It isn’t the way a problem looks that determines how you solve it, it is the type of problem (in this case, kinematics) that you need to consider. That said, different situations require you to do different side problems along the way. If you are asked for time you might need to solve the quadratic equation, and if the problem is two dimensional you might need to go between the x- and y-directions. So if your problem has any of these features, you may find it useful to pick an example that does as well. But don’t worry, you don’t need (or want) an example to look exactly like your problem!

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).

The Kinematic Equations

Kinematic Equations Introduction
Solving Problems with Kinematic Equations
Kinematic Equations and Free Fall
Sample Problems and Solutions
Kinematic Equations and Kinematic Graphs

There are a variety of quantities associated with the motion of objects - displacement (and distance), velocity (and speed), acceleration , and time. Knowledge of each of these quantities provides descriptive information about an object's motion. For example, if a car is known to move with a constant velocity of 22.0 m/s, North for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car is fully described. And if a second car is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s 2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an eastward displacement of 96 meters, then the motion of this car is fully described. These two statements provide a complete description of the motion of an object. However, such completeness is not always known. It is often the case that only a few parameters of an object's motion are known, while the rest are unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22 m/s, East and is capable of a skidding acceleration of 8.0 m/s 2 , West. However you do not know the displacement that your car would experience if you were to slam on your brakes and skid to a stop; and you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters can be determined using physics principles and mathematical equations (the kinematic equations).

The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. They can never be used over any time period during which the acceleration is changing. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car is known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object's motion.

The four kinematic equations that describe an object's motion are:

There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the velocity of the object; a subscript of i after the v (as in v i ) indicates that the velocity value is the initial velocity value and a subscript of f (as in v f ) indicates that the velocity value is the final velocity value.

Each of these four equations appropriately describes the mathematical relationship between the parameters of an object's motion. As such, they can be used to predict unknown information about an object's motion if other information is known. In the next part of Lesson 6 we will investigate the process of doing this.

Introduction to Two-Dimensional Kinematics

Chapter outline.

The arc of a basketball, the orbit of a satellite, a bicycle rounding a curve, a swimmer diving into a pool, blood gushing out of a wound, and a puppy chasing its tail are but a few examples of motions along curved paths. In fact, most motions in nature follow curved paths rather than straight lines. 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. Both two- and three-dimensional kinematics are simple extensions of the one-dimensional kinematics developed for straight-line motion in the previous chapter. This simple extension will allow us to apply physics to many more situations, and it will also yield unexpected insights about nature.

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Is it made out of tin foil , paper 2024/576, on complexity of the problem of solving systems of tropical polynomial equations of degree two.

In this paper, we investigate the computational complexity of the problem of solving a one-sided system of equations of degree two of a special form over the max-plus algebra. Also, we consider the asymptotic density of solvable systems of this form. Such systems have appeared during the analysis of some tropical cryptography protocols that were recently suggested. We show how this problem is related to the integer linear programming problem and prove that this problem is NP-complete. We show that the asymptotic density of solvable systems of this form with some restrictions on the coefficients, the number of variables, and the number of equations is 0. As a corollary, we prove that this problem (with some restrictions on the coefficients, the number of variables, and the number of equations) is decidable generically in polynomial time.

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2.7: Problem-Solving Basics for One-Dimensional Kinematics

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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 Basics for One-Dimensional Kinematics

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.

Close-up photo of a hand writing in a notebook. On top of the notebook is a graphing calculator.

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 $0.40 m/s^2$ 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,

\[v=v0+at=0+(0.40m/s2)(100s)=40m/s.\]

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/s2, 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/s2 for 100 s (almost two minutes).

The six basic problem solving steps for physics are:

Step 1. Examine the situation to determine which physical principles are involved.

Step 2. Make a list of what is given or can be inferred from the problem as stated (identify the knowns).

Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns).

Step 4. Find an equation or set of equations that can help you solve the problem.

Step 5. Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete with units.

Step 6. Check the answer to see if it is reasonable: Does it make sense?

IMAGES

Problem Solving in Kinematics Example
Kinematics Formula
PPT
PPT
Solving Problems Using Kinematic Equations
Kinematic Equations problem set 2

VIDEO

Problem 2.35
Intro to Motion Problem Solving Techniques
Engineering Mechanics 2
Solving 2d kinematics problems
short questions physics chapter 2
Paper 2 Answering Techniques

COMMENTS

Kinematic Equations: Sample Problems and Solutions
A useful problem-solving strategy was presented for use with these equations and two examples were given that illustrated the use of the strategy. Then, the application of the kinematic equations and the problem-solving strategy to free-fall motion was discussed and illustrated. In this part of Lesson 6, several sample problems will be presented.
Kinematic Equations: Explanation, Review, and Examples
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.
Kinematics Practice Problems with Answers
On the following page you can find over 40+ questions related to applying kinematics equations in velocity and acceleration: Velocity and acceleration problems. Problem (2): A moving object slows down from 12\, {\rm m/s} 12m/s to rest at a distance of 20 m. Find the acceleration of the object (assumed constant).
What are the kinematic formulas? (article)
At this point, we solve the quadratic equation for time t ‍ . The solutions of a quadratic equation in the form of a t 2 + b t + c = 0 ‍ are found by using the quadratic formula t = − b ± b 2 − 4 a c 2 a ‍ . For our kinematic equation a = 1 2 (− 9.81 m s 2) ‍ , b = 18.3 m/s ‍ , and c = − 12.2 m ‍ .
2.4: Problem-Solving for Basic Kinematics
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 ...
2.6 Problem-Solving Basics for One-Dimensional Kinematics
Step 4. 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 ...
8.2: Kinematics
Equation 8.2.2 becomes: vf = vo + at (8.2.4) We can follow a similar procedure to determine how position varies with time for the case of constant acceleration. Starting with definition of velocity: v = dx dt (8.2.5) and integrating to solve for change in position over time: Δx = xf −xo = ∫t 0 vdt (8.2.6)
Choosing kinematic equations (video)
Choosing kinematic equations. Kinematic equations help solve for an unknown in a problem when an object has either a constant velocity or constant acceleration. This video will help you choose which kinematic equations you should use, given the type of problem you're working through.
Solving Problems with Kinematic Equations
Solving problems with kinematic equations can be broken down into simple steps; Step 1: Write out the variables that are given in the question, both known and unknown, and use the context of the question to deduce any quantities that aren't explicitly given e.g. for vertical motion a = ± 9.81 m s -2, an object which starts or finishes at rest will have u = 0 or v = 0
2.6: Problem-Solving Basics for One-Dimensional Kinematics
The six basic problem solving steps for physics are: Step 1. Examine the situation to determine which physical principles are involved. Step 2. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns).
PDF Chapter 5 Two Dimensional Kinematics
There are several important things to notice about Figures 5.5 and 5.6. The first point is that the abscissa axes are different in both figures. The second thing to notice is that at t = 0 , the slope of the graph in Figure 5.5 is equal to dy ⎛ dy / dt ⎞ v. y,0. = ⎜ ⎟ = = tanθ dx ⎝ dx / dt ⎠ v.
Using the Kinematic Equations to Solve Problems
This video tutorial lesson is the third of three lessons on the Kinematic Equations. The purpose of this video is to demonstrate through three examples an ef...
2.1.7 Solving Problems with Kinematic Equations
Step 1: Write out the variables that are given in the question, both known and unknown, and use the context of the question to deduce any quantities that aren't explicitly given e.g. for vertical motion a = ± 9.81 m s -2, an object which starts or finishes at rest will have u = 0 or v = 0; Step 2: Choose the equation which contains the quantities you have listed
10.2 Kinematics of Rotational Motion
The equations given above in Table 10.2 can be used to solve any rotational or translational kinematics problem in which a a and α α are constant. Problem-Solving Strategy for Rotational Kinematics Examine the situation to determine that rotational kinematics (rotational motion) is involved .
Kinematics (Description of Motion) Problems
There are three key kinematic equations. If you carefully select the equation which most directly describes the situation in your problem, you will not only solve the problem in fewer steps but also understand it better. The three equations, written for motion in the x-direction, are: x = x 0 + v 0 Δt + ½ a (Δt) 2 (relates position and time)
CC
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 f2 = v i2 + 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.
Kinematic Equations
The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration 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. The arc of a basketball, the orbit of a satellite, a bicycle rounding a curve, a swimmer diving into a pool, blood gushing out of a ...
Kinematics Calculator
Download Kinematics Calculator App for Your Mobile, So you can calculate your values in your hand. 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 ...
PDF On Complexity of The Problem of Solving Systems of Tropical Polynomial
tem of tropical polynomial equations. A finite set of two-sided tropical polynomial equations is called a two-sided system of tropical polynomial equations. Using the matrix notation, we can write any one-sided system of tropical linear equations as A⊗X= B, (2) and any two-sided system of tropical linear equations as A⊗X+ B= C⊗X+ D. (3)
On complexity of the problem of solving systems of tropical polynomial
In this paper, we investigate the computational complexity of the problem of solving a one-sided system of equations of degree two of a special form over the max-plus algebra. Also, we consider the asymptotic density of solvable systems of this form. Such systems have appeared during the analysis of some tropical cryptography protocols that were recently suggested. We show how this problem is ...
2.7: Problem-Solving Basics for One-Dimensional Kinematics
The six basic problem solving steps for physics are: Step 1. Examine the situation to determine which physical principles are involved. Step 2. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns).

Kinematic Equations: Explanation, Review, and Examples

The Kinematic Equations

The First Kinematic Equation

The Second Kinematic Equation

The Third Kinematic Equation

The Fourth Kinematic Equation

How to Approach a Kinematics Problem

Step 1: Identify What You Know

Step 2: Identify the Goal

Step 3: Gather Your Tools

Step 4: Put it all Together

Kinematic Equation 1: Review and Examples

Step 4: Put it All Together

Kinematic Equation 2: Review and Examples

Kinematic Equation 3: Review and Examples

Kinematic Equation 4: Review and Examples

Problem-Solving Strategies

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Kinematics (Description of Motion) Problems

Example Problems

2-D Kinematics Problems

Changing Acceleration Problems

How to Solve Kinematic Problems

2. Draw a Picture:

3.Select the Relation

4. Solve the Problem

5. Understand the Results

Help! I can’t find an example that looks like the problem I need to work!

Yes, my problem is definitely a kinematics problem.

How to solve kinematics problems, part 2

Kinematics Equations for Constant Acceleration

1-Dimensional Problem Solving Steps

Avoiding Common Mistakes: Hidden Quantities

Avoiding Common Mistakes: Top of Flight

Avoiding Common Mistakes: Positive and Negative Numbers

Example: A Woman and Her Ball

Part A: How high does the ball reach?

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

Part C: Put it all together.

The Kinematic Equations

Introduction to Two-Dimensional Kinematics

What a lovely hat

2.7: Problem-Solving Basics for One-Dimensional Kinematics

Problem-Solving Basics for One-Dimensional Kinematics

Problem-Solving Steps

Unreasonable Results

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VIDEO

COMMENTS

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