# Kinematics Practice Problems

On this page, several problems related to kinematics are given. The solutions to the problems are initially hidden, and can be shown in gray boxes or hidden again by clicking “Show/hide solution.” It is advised that students attempt to solve each problem before viewing the answer, then use the solution to determine if their answer is correct and, if not, why. Remember to include units on all final answers.

Content that appears in a box similar to this is content that applies only to the AP C curriculum. AP B and AICE students can skip any content contained in these boxes.

### Displacement

1. A rock is thrown straight upward off the edge of a balcony that is 5 m above the ground. The rock rises 10 m, then falls all the way down to the ground below the balcony. What is the rock’s displacement?

• A child walks 5 m east, then 3 m north, then 1 m east.

a.) What is the magnitude of the child’s displacement?

b.) What is the direction of the child’s displacement?

As you can see in this diagram, the displacement is equivalent to the hypotenuse of a right triangle whose legs are 6 m and 3 m long. So, we can calculate the magnitude of the displacement using the Pythagorean theorem:

• An athlete runs exactly once around a circular track with a total length of 500 m. Find the runner’s displacement for the race.

### Speed and Velocity

1. If the child from problem 2 completes his journey in 20 seconds, what is the magnitude of his average velocity?

• If the runner from problem 3 runs the lap in 1 minute 18 seconds, find his/her average speed and the magnitude of his/her average velocity.

• a.) Is it possible to move with constant speed but not constant velocity?

b.) Is it possible to move with constant velocity but not constant speed?

### Acceleration

1. A car drives in a straight line at a constant speed of 60 miles per hour for 5 seconds. Find its acceleration.

• A remote control car is driven along a straight track at 2 m/s. The child controlling the car then activates the toy’s turbo mode so that, 3 seconds later, the car’s speed is 3.2 m/s. Find its average acceleration.

• Shortly after, the remote control car in the previous example exits turbo mode, slowing from 3.2 m/s back to 2 m/s in 2 seconds. Find the car’s average acceleration over this interval.

### Uniform Acceleration w/ the Big Five

1. A particle moves along the x-axis with an initial velocity of 4 m/s and constant acceleration. After 3 seconds, its velocity is 14 m/s. How far did it travel during this interval

• A car is initially moving at 10 m/s and accelerates at a constant rate of 2 m/s 2 for 4 seconds, in a straight line. How far did the car travel during this time?

• A rock is dropped from a cliff that is 80 m above the ground. If the rock hits the ground with a velocity of 40 m/s, what acceleration did it undergo?

a = (v 2 )/(2 x) = (40 2 )/(2 * 80) = 1600/160 = 10 m/s 2 , toward the ground.

### Free Fall

1. A rock is dropped 80 meters from a cliff. How long does it take to reach the ground?

Since we are not given nor asked for v, we can use the Big 5 number 3, adapted to the y direction:

• A ball is thrown straight up with an initial speed of 20 m/s. How high will the ball travel?

20 2 /(-2 * -10) = 20 2 /20 = 20 m

• One second after being thrown straight down, a rock is falling at 20 m/s. How fast will it be falling 2 seconds later?

-10 * 2 – 20 = -20 – 20 = -40 m/s

• An object is thrown straight upward with an initial speed of 8 m/s and strikes the ground 3 seconds later. What height was the object thrown from?

-0.5 * -10 * 3 2 – 8 * 3 = 5 * 9 – 24 = 45 – 24 = 21 m

### Projectile Motion

1. A tennis ball is thrown horizontally with an initial speed of 10 m/s. If it hits the ground after 4 seconds, how far did it drop before hitting the ground?

0.5 * -10 * 4 2 = -5 * 16 = -80 m

• A ball is thrown horizontally from a height of 100 m with an initial speed of 15 m/s. How far does it travel horizontally in the first 2 seconds?

• A javelin travels in a parabolic arc for 6 seconds before hitting the ground. Compare its horizontal velocity 1 second after being thrown to its horizontal velocity 4 seconds after being thrown.

• A circus performer is launched out of a cannon with a launch angle of 30 o and an initial velocity of 40 m/s. How long after launch will it take for the performer to reach the top of his trajectory, and how high is this point?

0.5 * -10 * 2 2 + 20 * 2 = -5 * 4 + 40 = -20 + 40 = 20 m

• A cannonball is fired from the ground at a 30 o angle with an initial speed of 60 m/s. If the cannonball lands back at the same height it was launched from:

a.) For how long will the cannonball be in the air?

b.) How far will it travel horizontally?

60cos(30 o )6 = 312 m

### Kinematics w/ Graphs

1. An object’s position during a 10 second time interval is shown by the graph below:

a.) Determine the object’s total distance traveled and displacement.

b.) What is the object’s velocity at the following times: t = 1, t = 3, and t = 6.

c.) Determine the object’s average velocity and average speed from t = 0 to t = 10.

d.) What is the object’s acceleration at t = 5?

The displacement of the object is simply the final position minus the initial position, or -2 – 0 = -2 m.

• An object’s velocity during a 10 second time interval is shown by the graph below:

a.) Determine the object’s total distance traveled and displacement.

b.) At t = 0, the object’s position is x = 2 m. Find the object’s position at t = 2, t = 4, t = 7, and t = 10.

c.) What is the object’s acceleration at the following times: t = 1, t = 3, and t = 6.

d.) Sketch the corresponding acceleration vs. time graph from t = 0 to t = 10.

The total distance traveled by the object is simply the sum of all these areas: 3 + 6 + 4.5 + 2 + 2 = 17.5 m

The displacement is found in a similar fashion, except areas below the x-axis are considered negative: 3 + 6 + 4.5 – 2 – 2 = 9.5 m

x(10) = 2 + 3 + 6 + 4.5 – 2 – 2 = 11.5 m

• An object’s position during a given time interval is shown by the graph below:

a.) At which of the marked points is the object’s velocity the greatest? The least?

b.) Is the object’s acceleration positive or negative between points A and B?

d.) Using the function from part c, determine the object’s maximum and minimum positions and velocities within the interval from t = 1 to t = 6.

v(1) = 3 * 1 2 – 19 * 1 + 23 = 3 – 19 + 23 = 7 m/s

v(3) = 3 * 3 2 – 19 * 3 + 23 = 27 – 57 + 23 = -7 m/s

v(5) = 3 * 5 2 – 19 * 5 + 23 = 75 – 95 + 23 = 3 m/s

a(3) = 6 * 3 – 19 = 18 – 19 = -1 m/s 2

a(5) = 6 * 5 – 19 = 30 – 19 = 11 m/s 2

t = 1.63008 s or t = 4.70326 s

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# Holt Physics Holt Physics

ISBN: 9780030735486 / 0030735483

Author: Jerry S. Faughn Serway

### Chapter 1

The Science Of Physics

### Chapter 2

Motion In One Dimension

### Chapter 3

Two-Dimensional Motion And Vectors

### Chapter 4

Forces And The Laws Of Motion

### Chapter 5

Work And Energy

### Chapter 6

Momentum And Collisions

### Chapter 7

Circular Motion And Gravitation

### Chapter 11

Vibrations And Waves

### Chapter 13

Light And Reflection

### Chapter 15

Interference And Diffraction

### Chapter 16

Electric Forces And Fields

### Chapter 17

Electrical Energy And Current

### Chapter 18

Circuits And Circuit Elements

# CHEAT SHEET

## Physics Q A

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# NCERT Exemplar Problems class 12 Physics

NCERT Exemplar Problems class 12 Physics in PDF format are available to download. NCERT books and solutions are also available to download along with the answers given at the end of the book. Buy NCERT Books online , If you are having any suggestion for the improvement, your are welcome. The improvement of the website and its contents are based on your suggestion and feedback. NCERT exemplar questions are very good in concept developing and revision. These questions also gives excellent practice for JEE Mains exam as well as the other competitive examination. After completing the syllabus of 2017 2018 , students are advise to to these questions.

### Chapter 1: Electric Charges and Fields

Quantization of Electric Charge: The magnitude of all charges found in nature are in integral multiple of a fundamental charge. Q = ne, where, e is the fundamental unit of charge.

### Chapter 2: Electrostatic Potential and Capacitance

Equipotential Surface: A surface on which electric potential is equal at all the points is called an equipotential surface. The direction of electric field is normal to the equipotential surface.

### Chapter 3: Current Electricity

Potetiometer: It is a device in which one can obtain a continuously varying potential difference between any two points which can be measured simultaneously. The potential difference between any two points of a petentiometer wire is directly proportional to the distance between that two points.

### Chapter 4: Moving Charges and Magnetism

Ampere s Circuital Law: The line integral of magnetic field on a closed curve in a magnetic field, is equal to the product of the algebraic sum of the electric currents enclosed by that closed curve and the permeability of vacuum.

### Chapter 5: Magnetism and Matter

The magnetic field lines do not intersect at a point. It forms continuous closed loops. It emerges out from the magnetic north pole, reach the magnetic south pole and then passing through the magnet reach to the north pole to complete the loop.

### Chapter 6: Electromagnetic Induction

Self-Induction: When a current flowing through the coil is changed, the magnetic flux linked with the coil itself changes. In such circumstances an emf is induced in the coil. Such emf is called self-induced emf and this phenomenon is called self-induction.

# NCERT Exemplar Problems class 12 Physics

NCERT Exemplar Problems class 12 Physics in PDF format are available to download. NCERT books and solutions are also available to download along with the answers given at the end of the book. Buy NCERT Books online

### Chapter 1: Electric Charges and Fields

Quantization of Electric Charge: The magnitude of all charges found in nature are in integral multiple of a fundamental charge. Q = ne, where, e is the fundamental unit of charge.

### Chapter 2: Electrostatic Potential and Capacitance

Equipotential Surface: A surface on which electric potential is equal at all the points is called an equipotential surface. The direction of electric field is normal to the equipotential surface.

### Chapter 3: Current Electricity

Potetiometer: It is a device in which one can obtain a continuously varying potential difference between any two points which can be measured simultaneously. The potential difference between any two points of a petentiometer wire is directly proportional to the distance between that two points.

### Chapter 4: Moving Charges and Magnetism

Ampere’s Circuital Law: The line integral of magnetic field on a closed curve in a magnetic field, is equal to the product of the algebraic sum of the electric currents enclosed by that closed curve and the permeability of vacuum.

### Chapter 5: Magnetism and Matter

The magnetic field lines do not intersect at a point. It forms continuous closed loops. It emerges out from the magnetic north pole, reach the magnetic south pole and then passing through the magnet reach to the north pole to complete the loop.

### Chapter 6: Electromagnetic Induction

Self-Induction: When a current flowing through the coil is changed, the magnetic flux linked with the coil itself changes. In such circumstances an emf is induced in the coil. Such emf is called self-induced emf and this phenomenon is called self-induction.

# answers to physics problems

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# Free Mathematics Tutorials

• Each side of the square pyramid shown below measures 10 inches. The slant height, H, of this pyramid measures 12 inches.

.

1. What is the area, in square inches, of the base of the pyramid?

• The parallelogram shown in the figure below has a perimeter of 44 cm and an area of 64 cm 2 . Find angle T in degrees.

.

• Find the area of the quadrilateral shown in the figure.(NOTE: figure not drawn to scale)

.

• In the figure below triangle OAB has an area of 72 and triangle ODC has an area of 288. Find x and y.

.

• Find the dimensions of the rectangle that has a length 3 meters more that its width and a perimeter equal in value to its area?

• Find the circumference of a circular disk whose area is 100 pi square centimeters.

• The semicircle of area 1250 pi centimeters is inscribed inside a rectangle. The diameter of the semicircle coincides with the length of the rectangle. Find the area of the rectangle.

Answers to the Above Questions

a) 100 inches squared

b) 100 + 4*(1/2)*12*10 = 340 inches squared

c) h = sqrt(12 2 – 5 2 ) = sqrt(119)

d) Volume = (1/3)*100*sqrt(119)

= 363.6 inches cubed (approximated to 4 decimal digits)

• .

height = area / base

sin(T) = (32/7) / 8 = 32/56 = 4/7, T = arcsin(4/7) = 34.8 o

.

# NCERT Exemplar Problems class 12 Physics

NCERT Exemplar Problems class 12 Physics in PDF format are available to download. NCERT books and solutions are also available to download along with the answers given at the end of the book. Buy NCERT Books online , If you are having any suggestion for the improvement, your are welcome. The improvement of the website and its contents are based on your suggestion and feedback. NCERT exemplar questions are very good in concept developing and revision. These questions also gives excellent practice for JEE Mains exam as well as the other competitive examination. After completing the syllabus of 2017 2018 , students are advise to to these questions.

### Chapter 1: Electric Charges and Fields

Quantization of Electric Charge: The magnitude of all charges found in nature are in integral multiple of a fundamental charge. Q = ne, where, e is the fundamental unit of charge.

### Chapter 2: Electrostatic Potential and Capacitance

Equipotential Surface: A surface on which electric potential is equal at all the points is called an equipotential surface. The direction of electric field is normal to the equipotential surface.

### Chapter 3: Current Electricity

Potetiometer: It is a device in which one can obtain a continuously varying potential difference between any two points which can be measured simultaneously. The potential difference between any two points of a petentiometer wire is directly proportional to the distance between that two points.

### Chapter 4: Moving Charges and Magnetism

Ampere s Circuital Law: The line integral of magnetic field on a closed curve in a magnetic field, is equal to the product of the algebraic sum of the electric currents enclosed by that closed curve and the permeability of vacuum.

### Chapter 5: Magnetism and Matter

The magnetic field lines do not intersect at a point. It forms continuous closed loops. It emerges out from the magnetic north pole, reach the magnetic south pole and then passing through the magnet reach to the north pole to complete the loop.

### Chapter 6: Electromagnetic Induction

Self-Induction: When a current flowing through the coil is changed, the magnetic flux linked with the coil itself changes. In such circumstances an emf is induced in the coil. Such emf is called self-induced emf and this phenomenon is called self-induction.

# University of Wisconsin Green Bay

### Dynamics (Force or NewtonвЂ™s 2nd Law) Problems

Dynamics (Force) problems ask you to relate motion to the forces causing it. Note that the word вЂњforceвЂќ isnвЂ™t always used explicitly in the statement of the problem. You know many forces such as gravity, tension, and normal force that are present even if not listed in the problem.

#### How to Solve Force Problems

##### 1.Identify the Problem

Any problem that asks you to relate force and motion is a NewtonвЂ™s Second Law problem, no matter what was given or requested in the problem. In some cases, NewtonвЂ™s Second Law is easy to identifyвЂ”for example, a problem might ask you for the value of a particular force.

In other cases, it is harder to recognize 2nd Law problems. You may know forces that are present without having force ever mentioned in the problem, and you may be asked for things like how fast an object moves in a circle or how much of an object is underwater. Recognizing how to approach these problems comes from experienceвЂ”taking the time to answer the question вЂњhow did I know this was a 2nd Law problemвЂќ for every single force example you work will help to build that intuition which you will need on the final exam.

##### 2. Draw a Picture

When you approach force problems, a free body diagram will allow you to both picture what is happening and directly map the picture into the equation. All the physics is done in the process of drawing the diagramвЂ”only algebra remains after this step.

The first thing that you will need to do is to identify what object(s) will be the focus of the diagramвЂ”in other words, what system do you need to consider in order to answer the question. In some cases, you will need to consider several objects as separate systems. In those cases, you will draw free body diagrams and set up equations for each object separately. You will very likely use NewtonвЂ™s Third Law to relate forces on the two objects.

Once you have identified your system, think about all forces on that system, discard any that are too small to matter. Only forces acting on the object should be shown, since you are trying to understand what causes the motion of the object. The free body diagram maps directly into the left side of в€‘F=ma. Acceleration is the result not the causeвЂ”if you wish to sketch the acceleration, make sure that you do so off to the side and not on the sketch of forces.

Because NewtonвЂ™s Second Law is a vector equation, you will need to divide all forces into their x- and y- components in order to work with the equation. Math is always easiest if you pick one axis to be along the direction of acceleration. That way, one component of a will be zero and you will have fewer linked equations.

##### 3. Select the Relation

All force problems begin with the relation в€‘F=ma. This true no matter what you are asked to find. If any additional information is needed, it will become apparent as you work through the problem.

##### 4. Solve the Problem

в€‘F=ma is a vector equation and so is set up separately in the x- and y- directions. Think carefully about the direction of each force and include the appropriate sign. Once you have filled your forces into these equations, you have only algebra left and can solve the equations in any way that works. In general, it is easiest to solve the equation in which a=0 first as you will sometimes need those values in the other expression.

##### 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? Can you explain in words what is happening? If you only recognized it as a 2nd Law 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 canвЂ™t find an example that looks like the problem I need to work!

##### Are you certain your problem is a NewtonвЂ™s Second Law problem?

One of the most common mistakes is to think too hard. If you are told net force and asked for acceleration, or vice versa, you donвЂ™t need to go through all of the steps of a NewtonвЂ™s Second Law problem. 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 kinematics (description of motion) or energy and momentum. Can you clearly explain to yourself that your problem requires you to relate the cause of motion (forces) to the effect of those forces (acceleration or changing velocity?)

##### Yes, my problem is definitely a NewtonвЂ™s Second Law problem.

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 isnвЂ™t the way a problem looks that determines how you solve it, it is the type of interaction (in this case, force) that you need to consider. That said, different situations require you to do different side problems along the way. Circular motion requires you to know about centripetal acceleration, buoyant force requires you to know about density, an inclined plane required you to know how to divide vectors into components, and multiple object problems require you to understand NewtonвЂ™s Third Law. So if your problem has any of these features, you may find it useful to pick an example that does as well. But donвЂ™t worry, you donвЂ™t need (or want) an example to look exactly like your problem!

# NCERT Exemplar Problems class 11 Physics

NCERT Exemplar Problems class 11 Physics in PDF format are available to download. NCERT books, Syllabus for the academic year 2017 2018 and solutions are also available to download (as well as to Purchase Online ) along with the answers given at the end of the book. After going through your NCERT Books Solutions , you must see these Exemplar books to understand the concepts properly. These books contain MCQ questions, Short questions and Long answers questions for complete revision of the chapters. If you are having any suggestion for the improvement, your are welcome. The improvement of the website and its contents are based on your suggestion and feedback.

## NCERT Exemplar Problems class 11 Physics

### Chapter 15: Waves

Facts Must See

• The word भौतिक विज्ञान (used in Hindi/Gujarati) related to the science of physical world is derived from Sanskrit word Bautiki .
• English word Physics comes from a Greek word meaning Nature .
• Physics deals with the study of basic laws of nature related to matter, energy and their manifestation in different phenomena.
• The scope of the physics is extended on two basic domains Macroscopic to microscopic. It also deals with static and dynamic systems.
• The basic laws of physics are universal and apply in widely different contexts and conditions.
• The gravitational force, electromagnetic force, strong nuclear force and weak nuclear force are the four fundamental forces in nature. The attempts towards the unification of forces are going on.
• The physical quantity that remains unchanged with time is called conserved quantity.
• The laws of conservation of energy, charge, linear momentum and angular momentum are considered as the fundamental laws of physics.

# NCERT Exemplar Problems class 11 Physics

NCERT Exemplar Problems class 11 Physics in PDF format are available to download. NCERT books, Syllabus for the academic year 2017 – 2018

## NCERT Exemplar Problems class 11 Physics

### Chapter 15: Waves

Facts – Must See

• The word ‘भौतिक विज्ञान’ (used in Hindi/Gujarati) related to the science of physical world is derived from Sanskrit word ‘Bautiki’.
• English word ‘Physics’ comes from a Greek word meaning ‘Nature’.
• Physics deals with the study of basic laws of nature related to matter, energy and their manifestation in different phenomena.
• The scope of the physics is extended on two basic domains – Macroscopic to microscopic. It also deals with static and dynamic systems.
• The basic laws of physics are universal and apply in widely different contexts and conditions.
• The gravitational force, electromagnetic force, strong nuclear force and weak nuclear force are the four fundamental forces in nature. The attempts towards the unification of forces are going on.
• The physical quantity that remains unchanged with time is called conserved quantity.
• The laws of conservation of energy, charge, linear momentum and angular momentum are considered as the fundamental laws of physics.