Chapter 3: Accelerated Motion

Chapter 3: Accelerated Motion

PHYSICS Principles and

Problems

Acceleration is the rate of change in an object’s velocity.

CHAPTER

3 Accelerated Motion

BIG IDEA

Section 3.1 Acceleration

Section 3.2 Motion with Constant Acceleration

Section 3.3 Free Fall

Table Of ContentsCHAPTER

3

Click a hyperlink to view the corresponding slides. Exit

MAIN IDEAYou can use motion diagrams to show how an object’s position changes over time.

• What is acceleration?

• How is acceleration different from velocity?

• What information can you learn from velocity-time graphs?

Essential Questions

SECTION3.1

Acceleration

Review Vocabulary• Vector a quantity that has magnitude and direction.

New Vocabulary• Acceleration

• Velocity-time graph

• Average acceleration

• Instantaneous acceleration

SECTION3.1

Acceleration

• An object in uniform motion moves along a straight line with an unchanging velocity.

• Few objects move in uniform motion all the time.

Nonuniform Motion Diagrams

SECTION3.1

Acceleration

• Object’s, more commonly, move in nonuniform motion, in which velocity is changing.– Nonuniform motion can be in a straight line: speeding

up, applying brakes, falling objects.

– Nonuniform motion can be along a circular path: hammer throw.

– Nonuniform motion can be the motion of thrown objects: baseball, football.

SECTION3.1

Acceleration

Nonuniform Motion Diagrams (cont.)

• You can feel a difference between uniform and nonuniform motion.

• When you move in a nonuniform motion, you feel pushed or pulled.

• In contrast, when you are in uniform motion and your eyes are closed, you feel as though you are not moving at all.

SECTION3.1

Acceleration


• Consider the particle-model motion diagram below showing the distance between successive positions.

a. The person is motionless b. Equally spaced images show her moving at a constant speed.

c. She is speeding up. d. She is slowing down.

SECTION3.1

Acceleration


• There are two major indicators of the change in velocity in this form of the motion diagram. The change in the spacing of the dots and the differences in the lengths of the velocity vectors indicate the changes in velocity.

SECTION3.1

Acceleration


• If an object speeds up, each subsequent velocity vector is longer.

• If the object slows down, each vector is shorter than the previous one.

SECTION3.1

Acceleration


• Both types of motion diagrams give an idea of how an object’s velocity is changing.


SECTION3.1

Acceleration

• For a motion diagram to give a full picture of an object’s movement, it should contain information about the object’s acceleration.

• The rate at which an object’s velocity changes is called the acceleration of the object. When the velocity of an object changes at a constant rate, it has a constant acceleration.

SECTION3.1

Acceleration


• You can use a particle motion diagram to draw an acceleration vector for the motion.


SECTION3.1

Acceleration

• You can determine the length and direction of an acceleration vector by subtracting two consecutive velocity vectors and dividing by the time interval.

________ _____________

SECTION3.1

Acceleration


• The time interval, Δt, is 1 s. This vector, (vf - vi)/1 s, shown in violet, is the average acceleration during that time interval.

• The velocities vi and vf refer to the velocities at the beginning and end of a chosen time interval.


SECTION3.1

Acceleration

• You need to know the direction of both the velocity and acceleration vectors in order to determine whether an object is speeding up or slowing down. – When the acceleration of an object is in the same

direction as its velocity, the object’s speed increases.

– When the acceleration of an object is in the opposite direction, the object’s speed decreases.

Direction of Acceleration

SECTION3.1

Acceleration

SECTION3.1

Acceleration

Direction of Acceleration (cont.)

• An object has a positive acceleration when the acceleration vector points in the positive direction and a negative acceleration when the acceleration vector points in the negative direction.

• The sign of the acceleration alone does not indicate whether the object is speeding up or slowing down.

SECTION3.1

Acceleration

Direction of Acceleration (cont.)

Click image to view the movie.

Velocity-Time Graphs• You can determine the acceleration from a velocity-time

graph by calculating the slope of the data.

SECTION3.1

Acceleration

• The average acceleration of an object is the change in velocity during some measurable time interval divided by that time interval.

Average and Instantaneous Acceleration

• A curved line on a velocity-time graph shows that the acceleration is changing.

• The slope indicates the average acceleration during a time interval that you choose.

SECTION3.1

Acceleration

• Average acceleration is measured in m/s2.

• The change in velocity at an instant of time is called instantaneous acceleration.

SECTION3.1

Acceleration

Average and Instantaneous Acceleration (cont.)

• The instantaneous acceleration of an object can be found by drawing a tangent line on the velocity-time graph at the point of time in which you are interested. The slope of this line is equal to the instantaneous acceleration.

SECTION3.1

Acceleration

Average and Instantaneous Acceleration (cont.)

Velocity and Acceleration

How would you describe the sprinter’s velocity and acceleration as shown on the graph?

SECTION3.1

Acceleration

Step 1: Analyze and Sketch the Problem

From the graph, note that the sprinter’s velocity starts at zero, increases rapidly for the first few seconds, and then, after reaching about 10.0 m/s, remains almost constant.

SECTION3.1

Acceleration

Velocity and Acceleration (cont.)

Identify the known and unknown variables.

Known:

v = varies

Unknown:

a = ?

SECTION3.1

Acceleration


Step 2: Solve for the Unknown

SECTION3.1

Acceleration


Draw a tangent to the curve at t = 1.0 s and t = 5.0 s.

SECTION3.1

Acceleration


Solve for acceleration at 1.0 s:

SECTION3.1

Acceleration


The slope of the line at 1.0 s is equal to the acceleration at that time.

SECTION3.1

Acceleration


Solve for acceleration at 5.0 s:

SECTION3.1

Acceleration


The slope of the line at 5.0 s is equal to the acceleration at that time.

SECTION3.1

Acceleration


The acceleration is not constant because it changes from 2.9 m/(s2) at 1.0 s to 0.03 m/(s2) at 5.0 s.

The acceleration is in the direction chosen to be positive because both values are positive.

SECTION3.1

Acceleration


Step 3: Evaluate the Answer

SECTION3.1

Acceleration


Are the units correct?

SECTION3.1

Acceleration


Acceleration is measured in m/s2.

The steps covered were:Step 1: Analyze and Sketch the Problem

Step 2 Solve for the Unknown

Draw a tangent to the curve at t = 1.0 s and t = 5.0 s.

Solve for acceleration at 1.0 s.

Solve for acceleration at 5.0 s.

SECTION3.1

Acceleration


The steps covered were:


SECTION3.1

Acceleration


• Remember that acceleration of an object is the slope of that object’s velocity v. time graph.

• On a velocity v. time graph, slope equals:

Calculating Acceleration

SECTION3.1

Acceleration

• Acceleration can occur when speed is constant.

• Consider a satellite orbiting around Earth at a constant speed.

– Even though speed is constant, direction is changing so the velocity is changing.

– If velocity is changing, then there is acceleration.

Acceleration with Constant Speed

SECTION3.1

Acceleration

Which of the following statements correctly defines acceleration?

A. Acceleration is the rate of change of displacement of an object.

B. Acceleration is the rate of change of velocity of an object.

C. Acceleration is the amount of distance covered in unit time.

D. Acceleration is the rate of change of speed of an object.

SECTION3.1

Section Check

Answer

Reason: The rate at which an object’s velocity changes is called acceleration of the object.

SECTION3.1

Section Check

What happens when the velocity vector and the acceleration vector of an object in motion are in the same direction?

A. The acceleration of the object increases.B. The speed of the object increases.C. The object comes to rest.D. The speed of the object decreases.

SECTION3.1

Section Check

Reason: When the velocity vector and the acceleration vector of an object in motion are in the same direction, the speed of the object increases.

SECTION3.1

Section Check

Answer

On the basis of the velocity-time graph of a car moving up a hill, as shown on the right, determine the average acceleration of the car?

A. 0.5 m/s2

B. -0.5 m/s2

C. 2 m/s2

D. -2 m/s2

SECTION3.1

Section Check

Reason: Average acceleration of an object is the slope of the velocity-time graph.

Answer

SECTION3.1

Section Check

MAIN IDEAFor an object with constant acceleration, the relationships among position, velocity, acceleration and time can be described by graphs and equations.

• What do a position-time graph and a velocity-time graph look like for motion with constant acceleration?

• How can you determine the displacement of a moving object from its velocity, acceleration and time?

• What are the relationships among position, velocity, acceleration and time?

SECTION3.2

Motion with Constant Acceleration

Essential Questions

Review Vocabulary• Displacement a change in position having both

magnitude and direction: it is equal to the final position minus the initial position.

SECTION3.2


• The position data at different time intervals for a car with constant acceleration are shown in the table.

• The data from the table are graphed as shown on the next slide.

SECTION3.2


Position with Constant Acceleration

• The graph shows that the car’s motion is not uniform: the displacements for equal time intervals on the graph get larger and larger.

• The slope of a position-time graph of a car moving with a constant acceleration gets steeper as time goes on.

• For an object with constant acceleration, the position-time graph is a parabola.

SECTION3.2


Position with Constant Acceleration (cont.)

• The slopes from the position time graph can be used to create a velocity-time graph as shown on the right.

• Note that the slopes shown in the position-time graph are the same as the velocities graphed in the velocity-time graph.

SECTION3.2



• A unique position-time graph cannot be created using a velocity-time graph because it does not contain any information about the object’s position.

• However, the velocity-time graph does contain information about the object’s displacement.

• Recall that for an object moving at a constant velocity,

SECTION3.2



• On the graph shown on the right, v is the height of the plotted line above the t-axis, while Δt is the width of the shaded rectangle. The area of the rectangle, then, is vΔt, or Δx. Thus, the area under the v-t graph is equal to the object’s displacement.


SECTION3.2


• If an object’s average acceleration during a time interval is known, then it can be used to determine how much the velocity changed during that time.

• The definition of average acceleration:

Velocity with Average Acceleration

can be rewritten as follows:

SECTION3.2


• The equation for final velocity with average acceleration can be written as follows:

• The final velocity is equal to the initial velocity plus the product of the average acceleration and time interval.

Velocity with Average Acceleration (cont.)

SECTION3.2


• In cases in which the acceleration is constant, the average acceleration, ā, is the same as the instantaneous acceleration, a. The equation for final velocity can be rewritten to find the time at which an object with constant acceleration has a given velocity.

• It also can be used to calculate the initial velocity of an object when both the velocity and the time at which it occurred are given.


SECTION3.2


Finding the Displacement from a v-t Graph

The v-t graph shows the motion of an airplane. Find the displacement of the airplane at Δt = 1.0 s and at Δt = 2.0 s.

SECTION3.2


Step 1: Analyze and Sketch the Problem• The displacement is the area under the v-t graph.• The time intervals begin at t = 0.0.

Finding the Displacement from a v-t Graph (cont.)

SECTION3.2


Identify the known and unknown variables.

Known:

v = +75 m/s

Δt = 1.0 s

Δt = 2.0 s

Unknown:

Δx = ?

SECTION3.2



Step 2: Solve for the Unknown


SECTION3.2


Solve for displacement during Δt = 1.0 s.

SECTION3.2



Substitute v = +75 m/s, Δt = 1.0 s

SECTION3.2



Solve for displacement during Δt = 2.0 s.

SECTION3.2



Substitute v = +75 m/s, Δt = 2.0 s

SECTION3.2




SECTION3.2



Are the units correct?

Displacement is measured in meters.

Do the signs make sense?

The positive sign agrees with the graph.

Is the magnitude realistic?

Moving a distance of about one football field in 2 s is reasonable for an airplane.

SECTION3.2




Step 1: Analyze and Sketch the ProblemThe displacement is the area under the v-t graph.The time intervals begin at t = 0.0.

Step 2 Solve for the UnknownSolve for displacement during Δt = 1.0 s.Solve for displacement during Δt = 2.0 s.

SECTION3.2





SECTION3.2



• The graph describes constant acceleration that started with an initial nonzero velocity.

• To determine the displacement, you can divide the area under the graph into a rectangle and a triangle.


SECTION3.2


• The total area is then:

• Substituting for the change in velocity in the equation yields:

• When the initial or final position of the object is known the equation can be written as follows:


SECTION3.2


• If the initial time=0, the equation then becomes the following: Position with Average Acceleration


SECTION3.2


• Often, it is useful to relate position, velocity, and constant acceleration without including time.

• Rearrange the velocity with average equation:

An Alternative Expression

to solve for time:

• This equation can be solved for velocity (vf), at any position (xf).

Velocity with Constant Acceleration:

SECTION3.2


A position-time graph of a bike moving with constant acceleration is shown below. Which statement is correct regarding the displacement of the bike?

A. The displacement in equal time intervals is constant.B. The displacement in equal time intervals progressively increases.C. The displacement in equal time intervals progressively decreases.D. The displacement in equal time intervals first increases, then after

reaching a particular point, it decreases.

SECTION3.2

Section Check

Reason: You will see that the slope gets steeper as time progresses, which means that the displacement in equal time intervals progressively gets larger and larger.

Answer

SECTION3.2

Section Check

A car is moving with an initial velocity of vi m/s. After reaching a highway, it moves with a constant acceleration of a m/s2, what will be the velocity (vf) of the car after traveling for t seconds?

A. vf = vi + at

B. vf = vi + 2at

C. vf2 = vi

2 + 2at

D. vf = vi – at

SECTION3.2

Section Check

Reason: Since a = Δv/Δtvf - vi = a (tf - ti)

Also since the car is starting from rest, ti = 0

Therefore vf = vi + at (where t is the total time)

Answer

SECTION3.2

Section Check

From the graph as shown on the right, if a car slowing down with a constant acceleration from initial velocity vi to the final velocity vf, calculate the total distance (Δx) traveled by the car.

SECTION3.2

Section Check

Reason: Acceleration is the area under the graph.

Solving for Δx, we get

Now since (vf – vi) = aΔt, then (vi – vf) = –aΔt

SECTION3.2

Section Check

Answer

MAIN IDEAThe acceleration of an object in free fall is due to gravity alone.

• What is free-fall acceleration?

• How do objects in free fall move?

SECTION3.3

Free Fall

Essential Questions

Review Vocabulary• Origin the point at which both variables in a coordinate

system have the value.

New Vocabulary• Free-fall• Free-fall acceleration

SECTION3.3

Free Fall

• Free-fall is the motion of an object when gravity is the only significant force acting on it.

• After a lot of observation, Galileo concluded that, neglecting the effect of the air, all objects in free fall had the same acceleration. It didn’t matter what they were made of, how much they weighed, what height they were dropped from, or whether they were dropped or thrown.

• The acceleration of an object due only to the effect of gravity is known as free-fall acceleration.

Galileo’s Discovery

SECTION3.3

Free Fall

• The acceleration of falling objects, given a special symbol, g, is equal to 9.80 m/s2 downward. (Near Earth’s surface)

• When analyzing free fall, whether you treat the acceleration as positive or negative depends on the coordinate system you use.

Galileo’s Discovery (cont.)

SECTION3.3

Free Fall

Click image to view movie.

Free-Fall Acceleration

SECTION3.3

Free Fall

• At the top of its flight, the ball’s velocity is 0 m/s. What would happen if its acceleration were also zero? Then, the ball’s velocity would not be changing and would remain at 0 m/s.

• If this were the case, the ball would not gain any downward velocity and would simply hover in the air at the top of its flight.

Free-Fall Acceleration (cont.)

SECTION3.3

Free Fall

• Because this is not the way objects tossed in the air behave on Earth, you know that the acceleration of an object at the top of its flight must not be zero. Further, because you know that the object will fall from that height, you know that the acceleration must be downward.


SECTION3.3

Free Fall

• Amusement parks use the concept of free fall to design rides that give the riders the sensation of free fall.

• These types of rides usually consist of three parts: the ride to the top, momentary suspension, and the plunge downward.

• When the cars are in free fall, the most massive rider and the least massive rider will have the same acceleration.


SECTION3.3

Free Fall

• When David Scott tested Galileo’s discovery on the surface moon in 1971, the object did fall with the same acceleration as Galileo had predicted but the acceleration was not 9.8m/s2.

• Factors that affect the value of free-fall acceleration: – Mass of the object. Ex. Earth or Moon– Distance of the object in free-fall from the object

responsible for the gravitational pull. Ex. Greater distance, less acceleration.

Variations in Free Fall

SECTION3.3

Free Fall

What is free fall?

Answer: Free fall is the motion of the body when air resistance is negligible and the action can be considered due to gravity alone.

SECTION3.3

Section Check

If a stone is thrown vertically upward with a velocity of 25 m/s, what will be the velocity of the stone after 1 second?

A. 9.8 m/s

B. 15.2 m/s

C. 25 m/s

D. 34.8 m/s

SECTION3.3

Section Check

Reason: Since the ball is thrown upward, the velocity and acceleration are in opposite directions. Therefore the speed of the ball decreases. After 1 s, the ball’s velocity is reduced by 9.8 m/s (as acceleration due to gravity is 9.8 m/s2 downward), so it is now traveling at 25 m/s – 9.8 m/s = 15.2 m/s.

SECTION3.3

Section Check

Answer

A 50-kg bag and a 100-kg bag are dropped from a height of 50 m. Which of the following statements is true about their acceleration? (Neglect air resistance.)

A. The 100-kg bag will fall with a greater acceleration.

B. The 50-kg bag will fall with a greater acceleration.

C. Both will fall at the same and constant acceleration.

D. Both will fall at the same acceleration, which changes equally as time progresses.

SECTION3.3

Section Check

Reason: Any body falling freely toward Earth, falls with a same and constant acceleration of 9.8 m/s2. It doesn’t matter how much it weighs or what height it was dropped from.

Answer

SECTION3.3

Section Check

Physics Online

Study Guide

Chapter Assessment Questions

Standardized Test Practice

CHAPTER

3Resources

Accelerated Motion

• Acceleration is the rate at which an object’s velocity changes.

• Velocity and acceleration are not the same thing. An object moving with constant velocity has zero acceleration. When the velocity and the acceleration of an object are in the same direction, the object speeds up; when the are in opposite directions, the object slows down.

Study Guide

SECTION3.1

Acceleration

• You can use a velocity-time graph to find the velocity and the acceleration of an object. The average acceleration of an object is the slope of its velocity-time graph.

SECTION3.1

AccelerationStudy Guide

• If an object is moving with constant acceleration, its position-time graph is a parabola, and its velocity-time graph is a straight line.

• The are under an object’s velocity-time graph is its displacement.

SECTION3.2

Motion with Constant AccelerationStudy Guide

• In motion with constant acceleration, position, velocity, acceleration and time are related.

SECTION3.2

Motion with Constant AccelerationStudy Guide

• Free-fall acceleration on Earth is about 9.8m/s2 downward. The sign associated with free-fall acceleration in equations depends on the choice of the coordinate system.

• When an object is in free-fall, gravity is the only force acting on it. Equations for motion with constant acceleration can be used to solve problems involving objects in free fall.

SECTION3.3

Free FallStudy Guide

A. Acceleration and velocity are the same thing.

B. Acceleration is the rate of change in velocity.

C. Acceleration is velocity with direction.

D. Velocity is the rate of change in acceleration.

Chapter Assessment

CHAPTER


Which correctly describes the relationship between acceleration and velocity?

Reason: Acceleration is the rate at which velocity changes with time.

Chapter Assessment

CHAPTER


Chapter Assessment

CHAPTER


What will be the change in displacement (Δd) of a train starting from rest and moving with a constant acceleration (a), if the motion lasts fort seconds?

Reason: From the equation of position with

constant acceleration, we can write:

∆x = vi(t) + at2 (where vi is the initial

velocity). In this case, the train is

starting from rest vi = 0. Hence,

∆x = at2.

Chapter Assessment

CHAPTER


A car starting from rest moves with a constant acceleration of a m/s2. What will be the velocity (vf) of the car after traveling d meters?

Chapter Assessment

CHAPTER


Reason: From the equation of velocity with constant acceleration, we can write vf 2 = vi

2 + 2ax (where d is the change in displacement). But since the car is starting from rest vi = 0. Therefore,

Chapter Assessment

CHAPTER


The acceleration due to gravity on the Moon is 6 times less than the gravity on Earth. If a ball on the Moon is dropped from a height of 20 m, what is the velocity of the ball after 1 s?

A.

B.

C.

D.

Chapter Assessment

CHAPTER


Reason: The acceleration due to gravity on the

Moon is m/s2. The rate of change of

velocity of the ball on the Moon is

m/s2. Hence, after 1 s, the ball will fall with a

velocity of m/s.

Chapter Assessment

CHAPTER


A ball is thrown vertically upward. Which of the following statements is true when the ball is at its maximum height?

A. The acceleration of the ball is zero.

B. The velocity of the ball is at its maximum.

C. The velocity of the ball is zero.

D. The rate of change of velocity is zero.

Chapter Assessment

CHAPTER


Reason: When the ball reaches its maximum height, it stops, and starts falling down. So, when the ball is at its maximum height, the velocity of the ball is zero.

Chapter Assessment

CHAPTER


A ball rolls down a hill with a constant acceleration of 2.0 m/s2. The ball starts at rest and travels for 4.0 s before it stops. How far did the ball travel before it stopped?

A. 8.0 m

B. 12 m

C. 16 m

D. 20 m

CHAPTER

3 Accelerated MotionStandardized Test Practice

A ball rolls down a hill with a constant acceleration of 2.0 m/s2. The ball starts at rest and travels for 4.0 s before it stops. What was the ball’s velocity just before it stopped?

A. 2.0 m/s

B. 8.0 m/s

C. 12 m/s

D. 16 m/s

CHAPTER


A driver of a car enters a new 110-km/h speed zone on the highway. The driver begins to accelerate immediately and reaches 110 km/h after driving 500 m. If the original speed was 80 km/h, what was his rate of acceleration?

A. 0.44 m/s2

B. 0.60 m/s2

C. 8.4 m/s2

D. 9.80 m/s2

CHAPTER


A flowerpot falls off the balcony of a penthouse suite 85 m above the street. How long does it take to hit the ground?

A. 4.2 s

B. 8.3 s

C. 8.7 s

D. 17 s

CHAPTER


A rock climber’s shoe loosens a rock, and her climbing buddy at the bottom of the cliff notices that the rock takes 3.20 s to fall to the ground. How high up the cliff is the rock climber?

A. 15.0 m

B. 31.0 m

C. 50.0 m

D. 1.00×102 m

CHAPTER


Tables

If a test question involves a table, skim the table before reading the question. Read the title, column heads, and row heads. Then read the question and interpret the information in the table.

CHAPTER


Test-Taking Tip

Motion Diagram of a Jogger

Chapter Resources

CHAPTER


Particle-model Version of a Jogger’s Motion Diagram

Chapter Resources

CHAPTER


Slope of a Velocity-Time Graph

Chapter Resources

CHAPTER


Velocity Vectors

Chapter Resources

CHAPTER


Graph of a Sprinter’s Velocity and Acceleration

Chapter Resources

CHAPTER


Motion Diagrams of Four Possible Ways to Move

Chapter Resources

CHAPTER


Graph of a Car in Non-uniform Motion

Chapter Resources

CHAPTER


Velocity-Time Graph for Constant Acceleration

Chapter Resources

CHAPTER


Graph Showing an Object’s Displacement

Chapter Resources

CHAPTER


Graph of an Object Moving With Constant Acceleration

Chapter Resources

CHAPTER


Graphs for a Ball Thrown UpwardChapter Resources

CHAPTER


Suppose you run wind sprints back and forth across the gym. You first run at 4.0 m/s toward the wall. Then, 10.0 s later, you run at 4.0 m/s away from the wall. What is your average acceleration if the positive direction is toward the wall?

Calculating Acceleration

Chapter Resources

CHAPTER


Chapter Resources

CHAPTER


Calculating Acceleration (cont.)

The negative sign indicates that the direction of acceleration is away from the wall. The velocity changes when the direction of motion changes, because velocity includes the direction of motion. A change in velocity results in acceleration. Thus, acceleration is also associated with a change in the direction of motion.

Chapter Resources

CHAPTER



There are several parallels between acceleration and velocity. Both are rates of change: acceleration is the time rate of change of velocity, and velocity is the time rate of change of position. Both acceleration and velocity have average and instantaneous forms.

Chapter Resources

CHAPTER



On the v-t graph shown on the right, for an object moving with constant acceleration that started with an initial velocity of vi, derive the object’s displacement.

Position with Constant AccelerationChapter Resources

CHAPTER


Chapter Resources

CHAPTER



Solving for the total area under the graph results in the following:

When the initial or final position of the object is known, the equation can be written as follows:

Chapter Resources

CHAPTER



If the initial time, ti = 0, the equation then becomes the following:

An object’s position at a later time is equal to the sum of its initial position, the product of the initial velocity and the time, and half the product of the acceleration and the square of the time.

Chapter Resources

CHAPTER



Suppose the free-fall ride at an amusement park starts at rest and is in free fall for 1.5 s. What would be its velocity at the end of 1.5 s?

Choose a coordinate system with a positive axis upward and the origin at the initial position of the ride. Because the ride starts at rest, vi would be equal to 0.00 m/s.

Acceleration Due to GravityChapter Resources

CHAPTER


To calculate the final velocity, use the equation for velocity with constant acceleration.

Chapter Resources

CHAPTER


Acceleration Due to Gravity

How far does the ride fall? Use the equation for displacement when time and constant acceleration are known.

Chapter Resources

CHAPTER


Acceleration Due to Gravity

Velocity and AccelerationHow would you describe the sprinter’s velocity and acceleration as shown on the graph?

Chapter Resources

CHAPTER


Finding the Displacement from a v-t Graph

The v-t graph shows the motion of an airplane. Find the displacement of the airplane at Δt = 1.0 s and at Δt = 2.0 s.

Chapter Resources

CHAPTER


Chapter 3: Accelerated Motion

Documents

Transcript of Chapter 3: Accelerated Motion