ap physics homework representing motion

Now that you have a good grasp of vectors and scalars, it is time to dive into understanding and representing motion, which is what Kinematics is all about. In this lesson we learn about the first part in representing motion. We will learn about , , and . Distance is a scalar quantity, whereas displacement is a vector. Let's see what you can do to the following problems with that information.

A rabbit hops 100 meters north to eat a piece of lettuce. It then hops 400 meters east to eat a carrot. What is the rabbit's distance from the starting point, in physics terms?

A rabbit hops 100 meters north to eat a piece of lettuce. It then hops 400 meters east to eat a carrot. What is the rabbit's displacement from the starting point?

Image a number line marked from 0 to 10. A person is at the mark 0. The person walks 7 units, then -5 units, and finally 6 units. What is the person's position?

By now you should have done all the example problems. Let's take a look at the correct way to do them, and what we learn from these problems.

A rabbit hops 100 meters, then 400 meters. Therefore 100+400=500, and the distance is 500 meters .◊

So distance is the total distance the object moves. In other words, you add up the distance (whether in meters, miles, kilometers, etc.), no matter if the objects turns or not, to get the distance. Distance is a scalar quantity.

We notice it is the same problem, but it is asking for displacement instead for distance. We also know displacement is a vector quantity. So therefore, by the Pythagorean Theorem, the answer is 412.313 meters with the direction shown on the image .◊

So we have a number line. The person is at 0. If he walks +7 (right 7) units, he will be at mark 7. Then he walks -5 (left 5) units, he will be at mark 2. Finally, if he walks +6 (right 6) units, he will be at mark 8 .◊

So here is what we have learned. Distance is how far you have traveled. Displacement is the length and direction of a straight line from the starting point to your end point. And finally, position is a number(s) designating your position (In 1D, position is a number; in 2D, it is a ordered pair).

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3.4 Projectile Motion

Learning objectives.

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

• Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory.
• Determine the location and velocity of a projectile at different points in its trajectory.
• Apply the principle of independence of motion to solve projectile motion problems.

The information presented in this section supports the following AP® learning objectives:

• 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2)
• 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1)

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile , and its path is called its trajectory . The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional Kinematics , is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible .

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction , where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x -axis and the vertical axis the y -axis. Figure 3.36 illustrates the notation for displacement, where s s size 12{s} {} is defined to be the total displacement and x x size 12{x} {} and y y size 12{y} {} are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s , x , and y . (Note that in the last section we used the notation A A size 12{A} {} to represent a vector with components A x A x size 12{A rSub { size 8{x} } } {} and A y A y size 12{A rSub { size 8{y} } } {} . If we continued this format, we would call displacement s s size 12{s} {} with components s x s x size 12{s rSub { size 8{x} } } {} and s y s y size 12{s rSub { size 8{y} } } {} . However, to simplify the notation, we will simply represent the component vectors as x x size 12{x} {} and y y size 12{y} {} .)

Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x - and y -axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: a y = – g = – 9.80 m /s 2 a y = – g = – 9.80 m /s 2 size 12{a rSub { size 8{y} } ="-g"="-9.80" "m/s" rSup { size 8{2} } } {} . (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, a x = 0 a x = 0 size 12{a rSub { size 8{x} } } {} . Both accelerations are constant, so the kinematic equations can be used.

Review of Kinematic Equations (constant a a )

Given these assumptions, the following steps are then used to analyze projectile motion:

Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so A x = A cos θ A x = A cos θ size 12{A rSub { size 8{x} } =A"cos"θ} {} and A y = A sin θ A y = A sin θ size 12{A rSub { size 8{y} } =A"sin"θ} {} are used. The magnitude of the components of displacement s s size 12{s} {} along these axes are x x size 12{x} {} and y. y. size 12{y} {} The magnitudes of the components of the velocity v v size 12{v} {} are v x = v cos θ v x = v cos θ size 12{v rSub { size 8{x} } =v"cos"θ} {} and v y = v sin θ, v y = v sin θ, size 12{v rSub { size 8{y} } =v"sin"θ} {} where v v size 12{v} {} is the magnitude of the velocity and θ θ size 12{θ} {} is its direction, as shown in Figure 3.37 . Initial values are denoted with a subscript 0, as usual.

Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time t t size 12{t} {} . The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below.

Step 4. Recombine the two motions to find the total displacement s s size 12{s} {} and velocity v v size 12{v} {} . Because the x - and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing A = A x 2 + A y 2 A = A x 2 + A y 2 size 12{v= sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } } {} and θ = tan − 1 ( A y / A x ) θ = tan − 1 ( A y / A x ) size 12{θ="tan" rSup { size 8{ - 1} } \( A rSub { size 8{y} } /A rSub { size 8{x} } \) } {} in the following form, where θ θ size 12{θ} {} is the direction of the displacement s s size 12{s} {} and θ v θ v size 12{θ rSub { size 8{v} } } {} is the direction of the velocity v v size 12{v} {} :

Total displacement and velocity

Example 3.4

A fireworks projectile explodes high and away.

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0° 75.0° above the horizontal, as illustrated in Figure 3.38 . The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which a x = 0 a x = 0 size 12{ a rSub { size 8{x} } =0} {} and a y = – g a y = – g size 12{ a rSub { size 8{y} } =-g} {} . We can then define x 0 x 0 size 12{x rSub { size 8{0} } } {} and y 0 y 0 size 12{y rSub { size 8{0} } } {} to be zero and solve for the desired quantities.

Solution for (a)

By “height” we mean the altitude or vertical position y y size 12{y} {} above the starting point. The highest point in any trajectory, called the apex, is reached when v y = 0 v y = 0 size 12{ v rSub { size 8{y} } =0} {} . Since we know the initial and final velocities as well as the initial position, we use the following equation to find y y size 12{y} {} :

Because y 0 y 0 size 12{y rSub { size 8{0} } } {} and v y v y size 12{v rSub { size 8{y} } } {} are both zero, the equation simplifies to

Solving for y y size 12{y} {} gives

Now we must find v 0 y v 0 y size 12{v rSub { size 8{0y} } } {} , the component of the initial velocity in the y -direction. It is given by v 0 y = v 0 sin θ v 0 y = v 0 sin θ size 12{v rSub { size 8{0y rSup} =v rSub {0 rSup size 12{"sin"θ}} {} , where v 0 y v 0 y is the initial velocity of 70.0 m/s, and θ 0 = 75.0° θ 0 = 75.0° size 12{θ rSub { size 8{0} } } {} is the initial angle. Thus,

and y y size 12{y} {} is

Discussion for (a)

Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height.

Solution for (b)

As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use y = y 0 + 1 2 ( v 0 y + v y ) t y = y 0 + 1 2 ( v 0 y + v y ) t size 12{y=y rSub { size 8{0} } + { {1} over {2} } \( v rSub { size 8{0y} } +v rSub { size 8{y} } \) t} {} . Because y 0 y 0 size 12{y rSub { size 8{0} } } {} is zero, this equation reduces to simply

Note that the final vertical velocity, v y v y size 12{v rSub { size 8{y} } } {} , at the highest point is zero. Thus,

Discussion for (b)

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using y = y 0 + v 0 y t − 1 2 gt 2 y = y 0 + v 0 y t − 1 2 gt 2 size 12{y=y rSub { size 8{0} } +v rSub { size 8{0y} } t - { {1} over {2} } ital "gt" rSup { size 8{2} } } {} , and solving the quadratic equation for t t size 12{t} {} .)

Solution for (c)

Because air resistance is negligible, a x = 0 a x = 0 size 12{a rSub { size 8{x} } =0} {} and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by x = x 0 + v x t x = x 0 + v x t size 12{x=x rSub { size 8{0} } +v rSub { size 8{x} } t} {} , where x 0 x 0 size 12{x rSub { size 8{0} } } {} is equal to zero:

where v x v x size 12{v rSub { size 8{x} } } {} is the x -component of the velocity, which is given by v x = v 0 cos θ 0 . v x = v 0 cos θ 0 . size 12{v rSub { size 8{x} } =v rSub { size 8{0} } "cos"θ rSub { size 8{0} } "." } {} Now,

The time t t size 12{t} {} for both motions is the same, and so x x size 12{t} {} is

Discussion for (c)

The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below.

In solving part (a) of the preceding example, the expression we found for y y size 12{y} {} is valid for any projectile motion where air resistance is negligible. Call the maximum height y = h y = h size 12{y=h} {} ; then,

This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity.

Defining a Coordinate System

It is important to set up a coordinate system when analyzing projectile motion. One part of defining the coordinate system is to define an origin for the x x size 12{x} {} and y y size 12{y} {} positions. Often, it is convenient to choose the initial position of the object as the origin such that x 0 = 0 x 0 = 0 size 12{x rSub { size 8{0} } =0} {} and y 0 = 0 y 0 = 0 size 12{y rSub { size 8{0} } =0} {} . It is also important to define the positive and negative directions in the x x size 12{x} {} and y y size 12{y} {} directions. Typically, we define the positive vertical direction as upwards, and the positive horizontal direction is usually the direction of the object's motion. When this is the case, the vertical acceleration, g g size 12{g} {} , takes a negative value (since it is directed downwards towards the Earth). However, it is occasionally useful to define the coordinates differently. For example, if you are analyzing the motion of a ball thrown downwards from the top of a cliff, it may make sense to define the positive direction downwards since the motion of the ball is solely in the downwards direction. If this is the case, g g size 12{g} {} takes a positive value.

Example 3.5

Calculating projectile motion: hot rock projectile.

Kilauea in Hawaii is the world's most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an angle 35.0° 35.0° size 12{"35"°} {} above the horizontal, as shown in Figure 3.39 . The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock's velocity at impact?

Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for t t size 12{t} {} first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain v v size 12{v} {} and θ v θ v size 12{θ rSub { size 8{v} } } {} at the final time t t size 12{t} {} determined in the first part of the example.

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

If we take the initial position y 0 y 0 size 12{y rSub { size 8{0} } } {} to be zero, then the final position is y = − 20 .0 m . y = − 20 .0 m . size 12{y= - "20" "." 0" m" "." } {} Now the initial vertical velocity is the vertical component of the initial velocity, found from v 0 y = v 0 sin θ 0 v 0 y = v 0 sin θ 0 size 12{v rSub { size 8{0y} } =v rSub { size 8{0} } "sin"θ rSub { size 8{0} } } {} = ( 25 . 0 m/s 25 . 0 m/s size 12{"25" "." "0 m/s"} {} )( sin 35.0° sin 35.0° size 12{"sin 35"°} {} ) = 14 . 3 m/s 14 . 3 m/s size 12{"14" "." "3 m/s"} {} . Substituting known values yields

Rearranging terms gives a quadratic equation in t t size 12{t} {} :

This expression is a quadratic equation of the form at 2 + bt + c = 0 at 2 + bt + c = 0 size 12{ ital "at" rSup { size 8{2} } + ital "bt"+c=0} {} , where the constants are a = 4.90 a = 4.90 , b = – 14.3 b = – 14.3 , and c = – 20.0. c = – 20.0. Its solutions are given by the quadratic formula:

This equation yields two solutions: t = 3.96 t = 3.96 size 12{t=3 "." "96"} {} and t = – 1.03 t = – 1.03 size 12{t=3 "." "96"} {} . (It is left as an exercise for the reader to verify these solutions.) The time is t = 3.96 s t = 3.96 s size 12{t=3 "." "96""s"} {} or – 1.03 s – 1.03 s size 12{-1 "." "03""s"} {} . The negative value of time implies an event before the start of motion, and so we discard it. Thus,

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

From the information now in hand, we can find the final horizontal and vertical velocities v x v x size 12{v rSub { size 8{x} } } {} and v y v y size 12{v rSub { size 8{y} } } {} and combine them to find the total velocity v v size 12{v} {} and the angle θ 0 θ 0 size 12{θ rSub { size 8{0} } } {} it makes with the horizontal. Of course, v x v x size 12{v rSub { size 8{x} } } {} is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore:

The final vertical velocity is given by the following equation:

where v 0y v 0y size 12{v rSub { size 8{0y} } } {} was found in part (a) to be 14 . 3 m/s 14 . 3 m/s size 12{"14" "." "3 m/s"} {} . Thus,

To find the magnitude of the final velocity v v size 12{v} {} we combine its perpendicular components, using the following equation:

which gives

The direction θ v θ v size 12{θ rSub { size 8{v} } } {} is found from the equation:

The negative angle means that the velocity is 50 . 1° 50 . 1° size 12{"50" "." 1°} {} below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure 3.39 .)

One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile. On level ground, we define range to be the horizontal distance R R size 12{R} {} traveled by a projectile. Galileo and many others were interested in the range of projectiles primarily for military purposes—such as aiming cannons. However, investigating the range of projectiles can shed light on other interesting phenomena, such as the orbits of satellites around the Earth. Let us consider projectile range further.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed v 0 v 0 size 12{v rSub { size 8{0} } } {} , the greater the range, as shown in Figure 3.40 (a). The initial angle θ 0 θ 0 size 12{θ rSub { size 8{0} } } {} also has a dramatic effect on the range, as illustrated in Figure 3.40 (b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with θ 0 = 45° θ 0 = 45° size 12{θ rSub { size 8{0} }  = "45°"} {} . This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately 38° 38° size 12{"38°"} {} . Interestingly, for every initial angle except 45° 45° size 12{"45°"} {} , there are two angles that give the same range—the sum of those angles is 90° 90° size 12{"90°"} {} . The range also depends on the value of the acceleration of gravity g g size 12{g} {} . The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range R R size 12{R} {} of a projectile on level ground for which air resistance is negligible is given by

where v 0 v 0 size 12{v rSub { size 8{0} } } {} is the initial speed and θ 0 θ 0 size 12{θ rSub { size 8{0} } } {} is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described.

When we speak of the range of a projectile on level ground, we assume that R R size 12{R} {} is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure 3.41 .) If the initial speed is great enough, the projectile goes into orbit. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text.

Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of Velocities , we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic.

PhET Explorations

Projectile motion.

Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. Make a game out of this simulation by trying to hit a target.

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• Authors: Gregg Wolfe, Erika Gasper, John Stoke, Julie Kretchman, David Anderson, Nathan Czuba, Sudhi Oberoi, Liza Pujji, Irina Lyublinskaya, Douglas Ingram
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• Book title: College Physics for AP® Courses
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Unit 5: Simple harmonic motion and rotational motion

About this unit.

Let's swing, buzz and rotate into the study of simple harmonic and rotational motion! Learn about the period and energy associated with a simple harmonic oscillator and the specific kinematic features of rotational motion.

Period of simple harmonic oscillators

• Period of a Pendulum (Opens a modal)
• Definition of amplitude and period (Opens a modal)
• Period dependence for mass on spring (Opens a modal)
• Equation for simple harmonic oscillators (Opens a modal)
• Period of simple harmonic oscillators Get 3 of 4 questions to level up!

Energy of simple harmonic oscillator

• Energy graphs for simple harmonic motion (Opens a modal)
• Energy of a simple harmonic oscillator Get 3 of 4 questions to level up!

Rotational kinematics

• Angular motion variables (Opens a modal)
• Relating angular and regular motion variables (Opens a modal)
• Rotational kinematic formulas (Opens a modal)
• Rotational kinematics Get 3 of 4 questions to level up!

Torque and angular acceleration

• Torque Basics (Opens a modal)
• Finding torque for angled forces (Opens a modal)
• Torque and angular acceleration Get 3 of 4 questions to level up!

Angular momentum and torque

• Angular momentum (Opens a modal)
• Rotational version of Newton's second law (Opens a modal)
• Angular momentum of an extended object (Opens a modal)
• Angular momentum and torque Get 3 of 4 questions to level up!

Conservation of angular momentum

• Conservation of angular momentum (Opens a modal)
• Ball hits rod angular momentum example (Opens a modal)
• Conservation of angular momentum Get 3 of 4 questions to level up!

Find what you need to study

1.1 Position, Velocity, and Acceleration

8 min read • december 22, 2022

Daniella Garcia-Loos

Position , velocity , and acceleration are three fundamental concepts in physics that are related to the motion of an object. Together, these three concepts form the basis for understanding the motion of objects. In AP Physics 1, you will learn more about these concepts and how to use them to solve problems involving the motion of objects.

Frames of Reference 👨‍💻

All forces share certain common characteristics when considered by observers in inertial reference frames. (A frame of reference in which a body remains at rest or moves with constant linear velocity unless acted upon by forces)

Key Vocabulary: Frame of Reference - a point of view 👀

⟶ Motion involves the change in position of an object over a period of time, and it is measured in reference to another object. 

Essential Knowledge 3.A.1 🏘

An observer in a reference frame can describe the motion of an object using such quantities as position , displacement , distance , velocity , speed , and acceleration .

A frame of reference is a set of points or objects that are used as a reference for measuring positions and movements. In physics, frames of reference are used to describe the motion of objects and to assign values to physical quantities such as position , velocity , and acceleration . Here are some key points to remember about frames of reference in AP Physics 1:

• A frame of reference can be fixed, meaning it does not move, or it can be moving at a constant velocity .
• The choice of a frame of reference can affect the values of physical quantities such as position , velocity , and acceleration .
• When describing the motion of an object, it is important to specify the frame of reference in which the measurements are taken.
• In a non-accelerating frame of reference , an object's velocity is constant if and only if its acceleration is zero.
• In a frame of reference that is accelerating, an object's velocity will change even if its acceleration is zero. This is known as the " fictitious force " or the " acceleration due to the change in frame of reference ."
• When describing the motion of an object in a frame of reference that is accelerating, it is useful to use the concept of relative velocity , which takes into account the acceleration of the frame of reference .

Key Vocabulary: Position - a location relative to a fixed point 

⟶ You can represent position in a Position (m) vs. Time (s) Graph (pictured below)

Image courtesy of ck12.org

Interpreting the Graph

To determine which way an object is moving look at which way the Position vs. Time Graph is sloped

A front slash / indicates that an object is moving away from the detector

A black slash \ indicates that an object is moving towards the detector

The slope of a Position vs. Time Graph is equal to velocity  

When the slope is a straight line it has constant velocity

When the slope is a curved lived there is acceleration (a change in velocity )

When the slope is zero the object is at rest

The y-intercept is the initial displacement of an object

⟶ Still feeling a little confused on Position vs. Time Graphs? Don’t worry! Check out this video from Khan Academy for more practice! 

Scalar vs. Vector Quantities 💫

Key Vocabulary: Scalar - quantities that are described by magnitude (a numerical value) alone 

Example: She is five feet tall

Distance and Speed are scalar quantities 

Key Vocabulary: Vector - quantities that are described by a size (magnitude) and a direction (ex. East, Up, Right, etc.) 

Example: The gas station is five miles west from the car

Displacement, Velocity , and Acceleration are vector quantities

Vectors can also be represented by arrows, and the length of the arrow should represent the magnitude of the described quantity. From the image below you can see the 5m arrow is smaller in length than the 50m arrow to reflect the difference in magnitude of the two quantities. 

Here are some key points to remember about the difference between scalar and vector quantities in AP Physics 1:

• Scalar quantities can be added or subtracted using simple arithmetic, but vector quantities require the use of vector addition or subtraction.
• Vector quantities have a direction associated with them, while scalar quantities do not.
• Scalar quantities are described by a single number, while vector quantities are described by both a magnitude and a direction.
• Examples of scalar quantities include mass, volume, density, and temperature. Examples of vector quantities include displacement, velocity , acceleration , and force.
• Scalar quantities can be represented by a single point on a number line, while vector quantities are represented by an arrow with a magnitude and a direction.
• Scalar quantities are often denoted by lowercase letters, while vector quantities are often denoted by uppercase letters.

⟶ Are you still feeling a little confused about Scalar vs. Vector Quantities? Don’t worry! Check out this video from Khan Academy for more practice! 

Displacement vs. Distance 🚴‍♀️

Key Vocabulary: Displacement - how far an object is from its original position  

Vector quantity

Express with a Sign (+ or -) or Direction (North, Down, Left, etc.)

SI Unit: Meter (m)

We use the symbol Δx to indicate displacement

⟶ Typical Displacement Question: How far are you from home? 

Key Vocabulary: Distance - how far an object has traveled 

Scalar quantity

Needs no frame of reference

⟶ Typical Distance Question: How far did you travel?

Displacement and distance are two important concepts in physics that are often confused with one another. Here are some key points to remember about the difference between displacement and distance in AP Physics 1:

• Displacement is a vector quantity that represents the change in the position of an object. It is defined as the final position of an object minus its initial position .
• Distance is a scalar quantity that represents the total length of the path traveled by an object. It does not take into account the direction of travel.
• Displacement has both a magnitude and a direction, while distance has only a magnitude.
• Displacement is a measure of the change in the position of an object, while distance is a measure of the total length of the path traveled by an object.

As you can see from the image below, distance takes into account the journey an object takes whereas displacement is concerned with the frame of reference of the original position .

Image Couretsy of thescienceclassroom.org

⟶ Still feeling a little confused about Distance vs. Displacement? Don’t worry! Check out this video from Khan Academy for more practice! 

Speed vs. Velocity 🏇

Key Vocabulary: Speed - describes how fast a particle is moving

SI Unit: Meters (m)/Seconds (s)

Equation: S = D/t

Key Vocabulary: Velocity - speed in a given direction

Equation: V = x/t

⟶ You can represent velocity in a Velocity (m/s) vs. Time (s) Graph (pictured below)

Image Courtesy of ck12.org

To determine which way the object is moving look at whether the Velocity vs. Time Graph is above or below the horizontal axis (x-axis)

An object is moving away from the detector if it’s above

An object is moving towards the detector if it’s below

The y-intercept is the initial velocity of an object

The slope of a velocity graph is equal to the acceleration  

When the slope is zero the object has constant velocity

When the slope is a straight line it has constant acceleration

When the slope is a curved line there is changing acceleration

The area under the curve is displacement

The object is stopped when y = 0

⟶ The table below is a type of motion cheat sheet. Memorizing this will help you ace any quizzes or tests with graph interpretation present!

Acceleration 🚀

Key Vocabulary: Acceleration - a change in velocity (magnitude or direction)

Equation: Aavg = V/t

⟶ You can represent acceleration in an Acceleration (m/s/s) vs. Time (s) Graph (pictured below)

Image courtesy of khanacademy.org

The y-intercept is the initial acceleration of an object

When the slope is zero the object has constant acceleration

The area under the curve is velocity

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Key Terms to Review ( 15 )

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AP Physics 1 Course Syllabus Important Physics 1 Dates   Test is still Tues 16 Feb in spite of 2 hour delay! --> Text:   College Physics - Explore and Apply (2nd Edition) by Etkina, Planinsic, and Van Heuvelen Text:   Open Stax, College Physics by Paul Peter Urone & Roger Hinrichs Physics 1 Overview from the College Board Physics 1 Course & Exam Description from the College Board Physics 1 Home multiple resources and information from the College Board AP Exam Info and Equation Sheet for Physics 1 2021 AP Exam Day Information --> AP Physics 1 Released Free Response - by year, includes scoring guidelines * Review Sessions May 18 - 20, 3:40 - 4:45 pm, Mr. M's room O102 * --> Astronomy and Physics Tutorials --> Mr. M's favorite internet Links for physics Lab Information and Homework Requirements

Measurement Review  - Units, Uncertainty, Graphical Analysis     Science Skills Assessment - Analyze your skills and consult this table --> solution guide     The following seven links lead to optional material that can be very useful and worthwhile review and reference:     Objectives & Homework     (see also Homework Requirements )     In-Class Practice     Lab Report - the Pendulum     (see also Lab Information )     Graphing - directions, examples, free graph paper     Notes - uncertainty, significant digits, and units     Functions Overview - summary of important types of equations     Review Problems

Kinematics  - a Mathematical Description of Motion     Objectives & Homework - Kinematics     (see also Homework Requirements )         Answer Key     In-Class Practice - Vector Magnitude & Direction     In-Class Practice - Word Problems     In-Class Practice - Position Graphs     Desmos graphs: Bolt Olympics ,   Bolt WR ,   Car & Stoplight     In-Class Practice - Velocity Graphs     Desmos graphs: Fan Cart ,   Block on Ramp     Interactive Desmos Graph:   Kinematics Model - constant acceleration functions, vectors, & animation     Mini-Lab Graph Matching - walk a graph!     Mini-Lab VW Bus - kinematics of a spring powered toy     Mini-Lab Jug Toss - kinematics of an airborne milk jug     Review Problems     Lab Report - Freefall Acceleration     (see also Lab Information )     Class Notes:   Vectors ,   Position, Distance, & Displacement ,   Speed & Velocity ,   Acceleration ,   Formulas ,   Freefall

Vectors  - Two-Dimensional Kinematics     Objectives & Homework - Vectors         Answer Key     Vector Addition PhET - Exploration and inquiry using PhET interactive     In-Class Practice     Boat in River - analysis of the timeless classic...     Projectile Exploration - inquiry with PhET Projectile Motion     Graphing Projectiles - Parametric equations on TI Calculators     Mini-Lab Projectile - tennis ball through photogate     Lab Report     Review Problems Lab Report - Projectile Motion Mini Launcher -->     Optional Lab - Pick one: Projectile Video or Strobe Photograph or Mini Launcher     Lab Report - Projectile Motion Video Analysis-->             Videos:   Nerf Football   4-Square Ball   Superball     Class Notes:   Addition/Graphical ,   Components ,   Addition/Numerical ,   Relative Motion ,   Projectiles

Forces  - Newton's 3 Laws of Motion (Dynamics)     Objectives & Homework - Forces         Answer Key     In-Class Practice     net Force - virtual mini-lab using PhET simulation     MiniLab - Net Force - using Vernier GoDirect Force and Acceleration sensor     MiniLab - Friction - using Vernier GoDirect Motion Detector     Review Problems     Lab Report - Newton's 2nd Law     Lab tips and pics     Class Notes:   Laws 1 & 2 ,   Net Force ,   Weight & Mass ,   3rd Law & Normal Force ,   Friction ,   Components & Inclines

Periodic Motion & Gravity  - SHM, Orbits, & Force Fields     Objectives & Homework - Periodic Motion         Answer Key     Solar System Data - Sun, planets, moons     In-Class Practice     Rotating Frames Desmos Graphs:   Space Station Ball Toss     SHM Summary - a concise overview (more complete notes at link below)     Review Problems     Mini-Lab Mass on Spring - interactive exploration of Hooke's Law and simple harmonic motion using PhET simulation     Lab Report - the Conical Pendulum     Class Notes:   Circular Motioin ,   Gravitation ,   Simple Harmonic Motion

Conservation of Energy & Momentum  - an Alternate Form of Mechanics     Objectives & Homework - Energy         Answer Key     Objectives & Homework - Momentum         Answer Key     In-Class Practice - Energy     In-Class Practice - Momentum     Energy Form Examples     Review Problems     Effect of Speed     Mini-Lab: Rubberband Energy     Mini-Lab: Ping Blaster!         Solution Summary     Mini-Lab Spring Energy - interactive exploration of conservation of energy using PhET simulation     Virtual Lab: Momentum & Center of Mass     Lab Report - Mass on a Spring       Lab Setup and Tips             Logger Pro file download:     SHM Energy.cmbl             Doing the report Using Graphical Analysis App on Chromebook (instead of Logger Pro)-->     Class Notes:   Energy & Conservation ,   Work ,   Springs ,   Power & Machines     Class Notes:   Momentum, Force, & Impulse ,   Conservation of Momentum, Elasticity

Rotation  - Mechanics of Rotating Bodies     Objectives & Homework - Rotation         Answer Key     In-Class Practice     Mini-Lab Cardboard Tube - rotational kinematics - using GoDirect gyro     Mini-Lab Torque - using GoDirect Force and a meter stick     Mini-Lab Gyro1 - rotational kinematics of a gyroscope     Mini-Lab Gyro2 - rotational dynamics of a gyroscope and falling mass     Mini-Lab 2nd Law - 2nd Law for rotation, using Rotary Motion sensor     Mini-Lab Ramp Rollers - the effect of rotational inertia on a ramp     Review Problems     Class Notes:   Rotational Kinematics ,   Torque & Rotational Inertia ,   Rotational Work & Energy, Angular Momentum

Electricity  - Coulomb's Law & DC Circuits     Objectives & Homework - Electricity Answer Key -->     In-Class Practice     The Digital Multimeter - a Primer     Review Problems     Some Notes     Lab Report - Virtual Circuits     Lab Report - Ohm's Law     Class Notes:   Electric Charge & Force ,   Electric Field & Potential     Class Notes:   Current & Power ,   Resistance ,   Series & Parallel ,   Combinations & Kirchoff's Laws

Waves & Interference  - Mechanical Waves and Sound     Objectives & Homework - Waves Answer Key -->     Wave Graphs - Homework Problem 21     Lab Report - Virtual Waves on a String     In-Class Practice     Class Notes:   Wave Parameters ,   Sound & Wave Graphs ,   Interference & Standing Waves     Review Problems

Final Exam!     About the Final Exam . . .     Early Dismissal Form - Turn in ONLY the ONE form that best applies to you     FHS Attendance and AP Incentives     Review Questions and Problems - Multiple Choice     Practice Problems based on DAWN     Practice Problems using Interactive Simulations     Free Response Practice - Actual AP exam questions from the College Board

AP® Physics 1 & 2​

Rotational motion: ap® physics 1 & 2 crash course review.

• The Albert Team
• Last Updated On: March 1, 2022

Introduction to Rotational Motion

Speaking of rotational motion, you may think of merry-go-round, a fan or even our rotating earth. Rotational motion is all about taking an object and spinning it. The object follows many rules while spinning just like the way they do when moving on a straight line. You may think it’s easy to get confused by those rules, but being analogous to physics of linear motion actually makes the rules of rotational motion very easy to memorize.

In this article, we are going to discuss some basic terms and a set of equations in rotational motion. AP® Physics 1 & 2 Crash Course Review will go through all you need to know about rotation and talk about rotation-related questions in AP® Physics and how those equations apply to solve real test questions.

Basic Elements in Rotation

Center of mass.

To tackle rotation-related questions, we should first expand the moving object from a point to a real object with shape and scale. In linear motion, we usually treat objects as if they are single particles, and we assume that all the forces are exerted only on a single point that represents the object. But how can we be sure? Why does that point summarize and represent the whole object?

If you have ever balanced a pen on your finger, you should be very familiar that you could always find a specific point to keep the pen from falling. The position of the point varies from pen to pen and are not always in the middle. That certain point is called center of mass. In other words, the center of mass is the point where all the mass of the object could be considered to be concentrated.

Pivot Point

A pivot point is a center of rotation. It could be the center of mass or any point that you select. Force exerted on this point will not contribute to the rotation.

Torque is the measure of a force’s effectiveness as making an object accelerate rotationally. Still don’t know what a torque is? To understand torque, first, let’s recall a time that you use a wrench to loosen a bolt.

When exerted on the same spot, a greater force is usually more effective than a small force. It is easier to turn the wrench when we push it by a greater force.

If exerted on a different spot, the same force may have different effects. You will find it more effective pushing the wrench at the end than in the middle. The closer the spot is to the rotation axis, the harder you are likely to turn the wrench.

What about changing the angle of the force exerted on the wrench? As showed in the figure below, F1 is always perpendicular to the wrench while F2 has an angle and F3 is parallel to the wrench. F3 certainly will not contribute to turning the wrench and you may find F2 less efficient than F1 when doing the job.

To summarize what we discussed earlier, we can find that the effectiveness of turning the wrench is actually related to 3 factors: Force(in example 1), distance from the pivot point (in example 2) and the angle(example 3). So then people think of combining these factors together and…Boom, we get the torque equation:

Now let’s see what we are really calculating. Torque is technically a vector, but for AP® Physics 1&2, you only need to be able to calculate the magnitude of torque and the direction of rotation (clockwise or counterclockwise). The term Fsin { \theta } equals the magnitude of force perpendicular to the lever arm. We are therefore getting a variable that is the product of force and a distance, so the unit for torque is Newton∙meter (Nm). Using this equation, if θ is between 0 to 90 degree, τ is getting smaller and smaller as θ decreases. The change of radius and force also affect torque in the same way.

Angular Velocity and Angular Acceleration

If the object is already spinning, what are variables to characterize its movement? Just like velocity and acceleration in linear motion, angular velocity and angular acceleration depict the movement of an object or of a particular point on the object. Angular velocity depicts the rate an object is spinning. It equals to the angle turned divided by time, notated as \omega =\dfrac { \Delta \Phi }{ \Delta t } . The direction of angular velocity is perpendicular to the rotational plane, determined by the right-hand rule. Angular acceleration is the change rate of ω, denoted as α.

Moment of Inertia

Also known as rotational inertia, the moment of inertia is used to characterize the tendency of an object to continue rotating before a torque is exerted on it. To better understand inertia, let’s compare inertia to mass. Mass is the tendency of an object to resist changes in its motion. To make an object move, we add a force on it. To make an object rotate, we exert a torque. While F = ma, torque equals to moment of inertia times angular acceleration. Rotational inertia is denoted as I.

So what determine the moment of inertia? Let’s first look at an example in figure skating. If the athlete wants to speed up while spinning, they draw in their arms and legs. Why will this do? Because inertia is related to the distribution of mass. The further an object is from the pivot point or the axis, the harder it is to make it spin. So when the athletes fold their arms, they are reducing their rotational inertia, therefore reduce the torque required to make them spin and elevates the angular velocity.

About Rotational Equilibrium

After getting familiar with those terms in rotation, we can see how those terms apply in solving problems. First, we would like to determine when an object can reach the equilibrium. The translational equilibrium is reached when the sum of force acting on the object is zero. Similarly, the rotational equilibrium is reached when the sum of torque acting on the object is zero. Be careful, being in a rotational equilibrium state does not necessarily mean the object is not rotating, it could also be rotating around its center of mass at a constant speed. If the object is totally at rest, it is said to be in a static equilibrium.

Tackling Rotation-related Questions

The first type of questions you might meet on test date is finding center of mass. The position of the center is related with distribution of mass in the object. In general, we use the equation below to compute the position of the center of mass.

For objects with uniform density, homogenous objects, the center of mass located at the geometric center. For example for a uniformed stick, the center of mass is the mid point of the stick. Always remember that center of mass can be located outside the body of object, such as a ring’s center of mass locate at its center. For other systems, simply use the equation above.

Now we have a uniformed 2 kg bar with the length of L. M1 = 40kg, M2 = 60kg, M3 = 70kg, M2 is attached to a quarter point of the bar. If we attached it to the ceiling, where should the string attached in order to reach a equilibrium?

This is actually asking you to find the center of mass. First chose a point as the reference point. You can chose wherever you like as long as it’s easy to calculate. Then figure out the distance between the object and reference point. At last, make use of the formula. For example let’s chose the mid point to calculate.

The result shows that the center of mass lies exactly on the mid point.

The second type of questions might be about how to reach equilibrium.

A unified 2 kg bar is attached to the wall at the mid point. The thread has a 30 degree angle with the wall. M = 40kg. The system is at equilibrium. Find tension in the thread.

You may wonder why normal force exerted by the wall is not contributing to rotation. It’s because normal force goes right through the pivot, so the torque it exerts is zero.

To sum up, rotational motion is very similar to linear motion by all means. This chart below will help you better memorize the relationship between linear motion and rotational motion.

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AP Physics C: Mechanics : Motion

Study concepts, example questions & explanations for ap physics c: mechanics, all ap physics c: mechanics resources, example questions, example question #1 : mechanics exam.

Which of the following is not a vector quantity?

Acceleration

Speed is defined as the magnitude of velocity.   In other words, speed visually represents the size of a velocity vector, but is indepedent of its direction.

Velocity, acceleration, and force are all vector quantities. Each is defined by the magnitude of the measurement as well as its direction of action.

A dog is initially standing near a fire hydrant. The dog moves 3 meters to the right. Then it runs 7 meters to the left. What is the dog's final displacement from its original position?

This question tests your understanding of displacement as a vector quantity. The best way to solve it is to create a number line to track the motion of the dog.

Let the initial position of the dog be 0m on our number line:

Then the dog moves three meters to the right, which we can represent as follows:

Then the dog moves seven meters to the left. Note that it is moving seven meters to the left from the position it is at now (3m):

Remember that displacement is the change in position, and not the total distance traveled. We only care about the initial position and the final position, regardless of the path traveled. Positive and negative signs indicate the direction of motion.

Example Question #1 : Ap Physics C

You drive your car from your house all the way to your school which is 50km away. After you are done with classes you drive back through the same route and park exactly where you had your car at the beginning of the day. By the end of the day, what were the distance and displacement of your motion?

This question tests your conceptual understanding of distance as a scalar quantity vs your understanding of displacement as a vector quantity. 

Distance measures the total length that was traveled in a given motion, and does not care about the direction since it is a scalar value. In your day, you traveled 50km on your way to school and 50km on your way back home. In total you traveled 100km, so that is your distance.

Displacement is a vector quantity that measures the change in position. It cares about your final and initial positions, taking into account the direction of the change in position. In this scenario you started and ended your motion exactly at the same position, so overall at the end of the day your car did not change position at all. Therefore your displacement was 0m.

This is a simple question that tests your conceptual understanding of speed as a scalar quantity and velocity as a vector quantity. The motion of the object is quite simple, so you need only to be mindful of the fact that velocity, as a vector, must tell you both the magnitude and direction (how fast it is going and where), while speed only tells you the magnitude (how fast it is going).

Since the acceleration is constant in this problem, we can apply the kinematics given equation to calculate the distance:

First, we need to calculate the acceleration.

Plug in our velocity and time values to find the acceleration.

Now we can return to the kinematics equation and solve for the distance traveled:

First, find the displacement equations for both cars. Car 1 will be the car that is initially stationary; car 2 will be the car traveling with constant velocity.

The accelerating car will catch and pass the car traveling at constant velocity after 4 seconds.

Take the derivative of the displacement equation.

Set the velocity equal to zero.

Solve for the time.

Set the acceleration fuction equal to zero and solve for the time.

A projectile is launched out of a cannon or launch tube with zero air resistance or friction. At what angle (in degrees) should the projectile be launched to maximize the distance it travels?

The projectile travels the farthest when the vertical component of its velocity matches the sum of its horizontal component and whatever the wind/friction adds or subtracts. If the wind has no effect, then a 45-degree angle will be the best because the horizontal and vertical components of the velocity will create a right isosceles triangle (remember special triangles).

The velocity (in meters per second) of a moving particle is given by the following function:

If the particle's initial position is 0m, what is the position of the particle after two seconds?

To solve this problem you need to obtain a function of position with respect to time. For that, you need to understand velocity as the rate of change of displacement with respect to time. In other words, velocity is "how fast" (i.e. how much time it takes) an object changes position (remember displacement is the change in position). This means that velocity is the derivative of displacement with respect to time.

Therefore, to obtain a function of position with respect to time you need to take the antiderivative of the velocity function, so we integrate:

Therefore, after two seconds have passed, we have t = 2s and

Note: we know that position is given in meters since the question specified that velocity is measured in meters per second. 

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Projectile Motion

Projectile Motion - Local Version