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1.2 Representations of Motion

13 min read • december 22, 2022

Daniella Garcia-Loos

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Understanding Representations of Motion

In AP Physics 1, we study different representations of motion to understand and analyze the movement of objects. Some common types of representations include:

Graphical representations: These include position-time graphs , velocity-time graphs , and acceleration-time graphs . These graphs can be used to represent the motion of an object and to understand its characteristics, such as its speed and acceleration .

Numerical representations: These include tables or lists of numerical data that describe the motion of an object. These data may include position, velocity , and acceleration at different times.

Analytical representations: These include mathematical equations that describe the motion of an object. These equations may involve variables such as position, velocity , and acceleration , and can be used to make predictions about the motion of an object.

Diagrammatic representations: These include sketches or diagrams that show the position, velocity , and acceleration of an object at different times. These representations can help visualize and understand the motion of an object.

Overall, different representations of motion can be used to help understand the characteristics and behavior of moving objects, and can be useful tools for predicting and analyzing the motion of objects in different situations.

Center of Mass

The acceleration of the center of mass of a system is related to the net force exerted on the system, where a = F/m.

Key Vocabulary: Center of mass - a point on an object or system that is the mean position of the matter.

⟶ A force may be applied to this point to cause a linear acceleration without angular acceleration occurring.

4.A.1 Essential Knowledge

The linear motion of a system can be described by the displacement , velocity , and acceleration of its center of mass .

4.A.2 Essential Knowledge

The acceleration is equal to the rate of change of velocity with time, and velocity is equal to the rate of change of position with time.

Graphical Representations of Motion

As we covered in 1.1 Position, Velocity and Acceleration there are ways to represent all three quantities graphically. We also reviewed how to interpret these graphs,  but it is imperative to understand the relationships each graph has to one another. 

Image Courtesy of geogebra

From the image above, we can use position to find velocity for any given period of time by looking at the slope of the Position vs. Time Graph. Working from velocity to position we can look at the area underneath the curve to find displacement , but it is not possible to determine how far from the detector the object is located from a Velocity vs. Time Graph.  

When working from velocity to acceleration we look at the slope of the Velocity vs. Time Graph. Similarly, when working from acceleration to velocity we look at the area under the curve to find velocity . 

Linearization

Many graphs in physics will not be perfectly straight lines, but we can turn the curve into a straight sloped line! We accomplish this by squaring the x-axis value. This is called Linearization. This is a concept that will be used throughout the course, so get comfortable with it now!

Image Courtesy of x-engineer.org

Linearization of a graph is a method of approximating the behavior of a nonlinear function with a straight line. This can be useful for making predictions or understanding the general trend of the function. Here are some key points to consider when linearizing a graph:

A linear function is defined by a straight line, which can be represented by the equation y = mx + b, where m is the slope and b is the y-intercept.

A nonlinear function is any function that is not a straight line. Nonlinear functions can take many different shapes, including curves, loops, and jumps.

To linearize a nonlinear function, we need to find a straight line that is a good approximation of the function. This can be done by finding two points on the nonlinear function and drawing a straight line through them.

The equation of the tangent line can be found using the point-slope form of a linear equation: y - y1 = m(x - x1), where (x1, y1) is a point on the function and m is the slope of the tangent line at that point.

The accuracy of the linear approximation depends on how closely the straight line follows the shape of the nonlinear function. The closer the line is to the function, the more accurate the approximation will be.

Linearization can be useful in physics because many physical systems can be modeled with nonlinear functions. By linearizing these functions, we can make predictions and understand the general trend of the system's behavior.

Example Problem

A ball is dropped from a height of 10 meters, and we want to find the time it takes to hit the ground. We know that the equation for the position of the ball as a function of time is:

y = 10 - 4.9t^2

However, this is a nonlinear equation and it is difficult to solve for t. Instead, we can use linearization to find an approximate solution.

To linearize the equation, we need to find the equation of the tangent line at a specific point on the curve. Let's choose the point where t = 1 second. At this point, the position of the ball is:

y = 10 - 4.9(1^2) = 5.1

Now we need to find the equation of the tangent line at this point. We can do this by using the point-slope form of a linear equation: y - y1 = m(x - x1), where (x1, y1) is the point on the curve and m is the slope of the tangent line at that point.

Since the point on the curve is (1, 5.1), we can set y1 = 5.1 and x1 = 1. To find the slope of the tangent line, we can plot a second point on the curve that is close to (1, 5.1) and find the slope between the two points. For example, we can choose the point where t = 1.1 seconds:

y = 10 - 4.9(1.1^2) = 4.61

The slope between the two points is (4.61 - 5.1)/(1.1 - 1) = -0.49.

Now we can plug the values into the point-slope form of the linear equation to find the equation of the tangent line:

y - 5.1 = -0.49(x - 1)

y = -0.49x + 5.6

We can use this linear equation to approximate the position of the ball as it falls. For example, if we want to find the position of the ball after 0.5 seconds, we can substitute t = 0.5 into the linear equation to get:

y = -0.49*0.5 + 5.6 = 5.355

This means that the ball has fallen about 5.355 meters after 0.5 seconds.

This is just an approximate solution, but it is a much simpler calculation than solving the nonlinear equation for the position of the ball. Linearization can be useful for finding approximate solutions to problems involving nonlinear functions, especially when the nonlinear function is difficult to work with or when we only need an approximate solution.

⟶ Still feeling a little confused on Linearization? Don’t worry! Check out this video from AP Physics 1 Online for more practice!

Mathematical Representations of Motion

In Kinematics there are four major equations you must understand to begin calculations. They relate acceleration , displacement , initial and final velocity , and time together . 

Variable Interpretation: Δx is horizontal displacement in meters, Vf is final velocity in meters/second, Vo is initial velocity in meters/second, t is time in seconds, and a is acceleration in m/s/s. 

⟶ In order to solve for a variable without having all four other quantities known, we look at the ‘Variable Missing’ column to pick the equation that best suits our question. 

Key Vocabulary: Free Fall - an object only under the influence of gravity

Equation: velocity = force of gravity x time

Key Vocabulary: Acceleration due to Gravity - 9.8 m/s/s (it is acceptable to round up to 10 m/s/s on the AP Physics 1 exam)

Variable Interpretation: Δy is vertical displacement in meters, Vf is final velocity in meters/second, Vo is initial velocity in meters/second, t is time in seconds, and g is acceleration due to gravity in m/s/s. 

⟶ In free fall equations, we now replace Δx with Δy and a with g giving us a modified list of The Big Four as seen in the table above. 

Object Dropped (trip down)

Vo (initial velocity ) = 0 m/s

g works in the direction of motion

Object Tossed (trip up)

Vf (final velocity ) = 0 m/s

At maximum height of its trip, an object has a velocity of 0 m/s

g works against the direction of motion

Projectile Motion

Projectile motion is the motion of an object that is thrown, launched, or projected into the air and is then subject only to the force of gravity. Here are some key points to consider when solving projectile motion problems:

In projectile motion, the object moves in two dimensions: horizontally and vertically.

The horizontal motion of a projectile is uniform, which means it moves at a constant speed in a straight line. This is because there are no forces acting on the projectile in the horizontal direction.

The vertical motion of a projectile is affected by the force of gravity, which pulls the projectile downward. The force of gravity causes the projectile to accelerate downward at a rate of 9.8 m/s^2.

The path of a projectile is called its trajectory , and it is a parabolic curve.

The initial velocity of a projectile is a vector that specifies both the magnitude and direction of the projectile's motion. It can be broken down into its horizontal and vertical components.

The range of a projectile is the horizontal distance it travels from the starting point to where it lands.

The maximum height of a projectile is the highest point it reaches during its motion.

The time of flight of a projectile is the total time it takes to complete its motion from the starting point to the landing point.

The angle at which a projectile is launched has an effect on its range, maximum height, and time of flight.

Air resistance is ignored

When dealing with horizontal projectile launches, launches that have an entirely horizontal initial velocity (ex. a ball rolling off a table), we break it up into two sets of equations: vertical and horizontal. 

Variable Interpretation: Δy is vertical displacement in meters, Δx is horizontal displacement in meters, Vfy is vertical final velocity in meters/second, Voy is vertical initial velocity in meters/second, Vx is horizontal velocity in m/s, t is time in seconds, and g is acceleration due to gravity in m/s/s. 

As shown in the chart above, there are three equations we use for the vertical component of launches and one for the horizontal. You cannot put an x-component into a formula without a y-component!

⟶ Still feeling a little confused on Projectile Launches? Don’t worry! Check out this live stream from Fiveable for more practice!

Angled Motion

Key Vocabulary: Angled Launches - launches at an angle that includes both a horizontal and vertical component of initial velocity

Key Vocabulary: Vector Components - the horizontal and vertical parts of a vector

Angled launches require you to find the Vox and Voy, or vector components , based on the initial velocity Vo and the angle Ө. Now you can solve for the following: Vo, Vox, Voy, Ө, t, X, Ymax, and Vf. 

If an object is shooting upward, Voy is (+)

If an object is shooting downward, Voy is (-)

Vertical velocity is 0 m/s at the top

Flight is symmetric if the projectile starts and ends at the same height

Variable Interpretation: Voy is vertical initial velocity in meters/second, Vox is initial horizontal velocity in m/s, t is time in seconds, t is time in seconds, and g is acceleration due to gravity in m/s/s. 

⟶ Still feeling a little confused on Angled Launches ? Don’t worry! Check out this video from Khan Academy for more practice!  Want more practice - Check out the Fiveable Live streams on this topic:

🎥 Watch AP Physics 1 - 2D Motion & Freefall

🎥 Watch AP Physics 1 - Horizontal Launch Problems

🎥 Watch AP Physics 1 - Angle Launch Problems

Key Terms to Review ( 15 )

Acceleration

Acceleration-Time Graphs

Angled Launches

Center of mass

Cosine Function (cos)

Displacement

Initial Velocity Vector Components

Numerical Data

Position-Time Graphs

Sine Function (sin)

Vector Components

Velocity-Time Graphs

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AP Physics 1/Kinematics/Representing Motion Part 1

Solution [ edit | edit source ].

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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AP Physics 1 Practice Test 29: Simple Harmonic Motion

• AP Physics 1 Practice Tests

Questions 1-4 refer to the following information.

Questions below are based on the following figure of a mass-spring system. Assume the mass is pulled back to position +A and released, and it slides back and forth without friction.

1. When the mass reaches position -A, what can be said about its speed?

2. When the mass reaches position 0, what can be said about its speed?

3. At what position does the mass have the greatest acceleration?

4. The mass is released from the -A position at time t = 0, and it oscillates with period T , measured in seconds. Which equation best represents the displacement?

5. A mass is suspended from a spring and allowed to oscillate freely. When the amplitude of vibration is doubled, what happens to the frequency of vibration?

6. The Moon has a gravitational field strength that is approximately one-sixth of the gravitational field on the Earth. What is the ratio between the period of a pendulum on the Moon and the period of an identical pendulum on the Earth?

7. Which of the following are the best examples of simple harmonic motion?

8. Which of the following best represents periodic motion?

9. Which of the following significantly affects the period of a simple pendulum?

10. A mass is suspended from a vertical spring attached to a support. Which of the following significantly affects the period of oscillation of this system?

11. A mass oscillates from the end of a vertical spring. What may be done to increase the frequency of oscillation?

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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)
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• 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)
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Rotational kinematics

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Torque and angular acceleration

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Angular momentum and torque

• Angular momentum (Opens a modal)
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Conservation of angular momentum

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AP® Physics 1

Describing motion from an equation, suggested time.

By upgrading a subject, you'll have access to the rest of the  Prompt, a Sample Response, and an Explanation.

Projectile Motion

Projectile Motion - Local Version

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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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