chapter 4: accelerated motion in a straightline

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Chapter 4: Accelerated Motion in a StraightLine. 4.1 Acceleration 4.2 A Model for Accelerated Motion 4.3 Free Fall and the Acceleration due to Gravity. Chapter Objectives. Calculate acceleration from the change in speed and the change in time. - PowerPoint PPT Presentation

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Page 1: Chapter 4: Accelerated Motion in a StraightLine
Page 2: Chapter 4: Accelerated Motion in a StraightLine

Chapter 4: Accelerated Motion in a Straight Line

4.1 Acceleration

4.2 A Model for Accelerated Motion

4.3 Free Fall and the Acceleration due to

Gravity

Page 3: Chapter 4: Accelerated Motion in a StraightLine

Chapter Objectives Calculate acceleration from the change in speed and the

change in time.

Give an example of motion with constant acceleration.

Determine acceleration from the slope of the speed versus time graph.

Calculate time, distance, acceleration, or speed when given three of the four values.

Solve two-step accelerated motion problems.

Calculate height, speed, or time of flight in free fall problems.

Explain how air resistance makes objects of different masses fall with different accelerations.

Page 4: Chapter 4: Accelerated Motion in a StraightLine

Chapter Vocabulary acceleration

acceleration due to gravity (g)

air resistance

constant acceleration

delta (Δ)

free fall

initial speed

m/s2

term

terminal velocity

time of flight

uniform acceleration

Page 5: Chapter 4: Accelerated Motion in a StraightLine

Inv 4.1 Acceleration

Investigation Key Question:

How does acceleration relate to velocity?

Page 6: Chapter 4: Accelerated Motion in a StraightLine

4.1 Acceleration Acceleration is the rate of change in the

speed of an object.

Rate of change means the ratio of the amount of change divided by how much time the change takes.

Page 7: Chapter 4: Accelerated Motion in a StraightLine

4.1 Acceleration in metric units If a car’s speed

increases from 8.9 m/s to 27 m/s, the acceleration in metric units is 18.1 m/s divided by 4 seconds, or 4.5 meters per second per second.

Meters per second per second is usually written as meters per second squared (m/s2).

Page 8: Chapter 4: Accelerated Motion in a StraightLine

4.1 The difference between velocity and acceleration Velocity is fundamentally different from

acceleration.

Velocity can be positive or negative and is the rate at which an object’s position changes.

Acceleration is the rate at which velocity changes.

Page 9: Chapter 4: Accelerated Motion in a StraightLine

4.1 The difference between velocity and acceleration

The acceleration of an object can be in the same direction as its velocity or in the opposite direction.

Velocity increases when acceleration is in the same direction.

Page 10: Chapter 4: Accelerated Motion in a StraightLine

4.1 The difference between velocity and acceleration

When acceleration and velocity have the opposite sign the velocity decreases, such as when a ball is rolling uphill.

Page 11: Chapter 4: Accelerated Motion in a StraightLine

4.1 The difference between velocity and acceleration If both velocity and acceleration are negative the

speed increases but the motion is still in the negative direction.

Suppose a ball is rolling down a ramp sloped downhill to the left.

Motion to the left is defined to be negative so the velocity and acceleration are both negative.

The velocity of the ball gets LARGER in the negative direction, which means the ball moves faster to the left.

Page 12: Chapter 4: Accelerated Motion in a StraightLine

4.1 Calculating acceleration

Acceleration is the change in velocity divided by the change in time. The Greek letter delta (Δ) means “the change in.”

a = v t

Change in speed (m/sec)

Change in time (sec)

Acceleration (m/sec2)

Page 13: Chapter 4: Accelerated Motion in a StraightLine

4.1 Calculating acceleration The formula for acceleration can also be

written in a form that is convenient for experiments.

Page 14: Chapter 4: Accelerated Motion in a StraightLine

1. You are asked for acceleration.2. You are given times and speeds from an

experiment.3. Use the relationship a = (v2 – v1) ÷ (t2 – t1)4. Choose any two pairs of time and speed data since

the change in speed is constant. a = (6 m/s 4 m/s) ÷ (3 s – 4 s) = (2 m/s) ÷ (-1 s) a = −2 m/s

Calculating acceleration in m/s2

A student conducts an acceleration experiment by coasting a bicycle down a steep hill. A partner records the speed of the bicycle every second for five seconds. Calculate the acceleration of the bicycle.

Page 15: Chapter 4: Accelerated Motion in a StraightLine

4.1 Constant speed and constant acceleration

Constant acceleration is different from constant speed.

If an object is traveling at constant speed in one direction, its acceleration is zero.

Motion with zero acceleration appears as a straight horizontal line on a speed versus time graph.

Page 16: Chapter 4: Accelerated Motion in a StraightLine

4.1 Uniform acceleration Constant acceleration is

sometimes called uniform acceleration.

A ball rolling down a straight ramp has constant acceleration because its speed is increasing at the same rate.

Falling objects also undergo uniform acceleration.

Page 17: Chapter 4: Accelerated Motion in a StraightLine

4.1 Constant negative acceleration Consider a ball rolling up a

ramp.

As the ball slows down, eventually its speed becomes zero and at that moment the ball is at rest.

However, the ball is still accelerating because its velocity continues to change.

Page 18: Chapter 4: Accelerated Motion in a StraightLine

4.1 The speed vs. time graph for accelerated motion

In this experiment, velocity and acceleration are in the same direction.

No negative quantities appear, and the analysis simply uses speed instead of velocity.

Page 19: Chapter 4: Accelerated Motion in a StraightLine
Page 20: Chapter 4: Accelerated Motion in a StraightLine

4.1 Slope and Acceleration Use slope to recognize

acceleration on speed vs. time graphs. Level sections (A) on the

graph show an acceleration of zero.

The highest acceleration (B) is the steepest slope on the graph.

Sections that slope down (C) show negative acceleration (slowing down).

Page 21: Chapter 4: Accelerated Motion in a StraightLine

1. You are asked for maximum acceleration.2. You are given a graph of speeds vs. time.3. Use the relationship a = slope of graph4. The steepest slope is between 60 and 70

seconds, when the speed goes from 2 to 9 m/s. a = (9 m/s – 2 m/s) ÷ (10 s) a = 0.7 m/s2

Calculating acceleration

The graph shows the speed of a bicyclist going over a hill. Calculate the maximum acceleration of the cyclist and calculate when in the trip it occurred.

Page 22: Chapter 4: Accelerated Motion in a StraightLine

Chapter 4: Accelerated Motion in a Straight Line

4.1 Acceleration

4.2 A Model for Accelerated Motion

4.3 Free Fall and the Acceleration due to

Gravity

Page 23: Chapter 4: Accelerated Motion in a StraightLine

Inv 4.2 Accelerated Motion

Investigation Key Question:

How does acceleration relate to velocity?

Page 24: Chapter 4: Accelerated Motion in a StraightLine

4.2 A Model for Accelerated Motion To get a formula for solve for the speed of an accelerating object, we can rearrange the experimental formula we had for acceleration.

Page 25: Chapter 4: Accelerated Motion in a StraightLine

4.2 The speed of an accelerating object In physics, a piece of an equation is called a term.

One term of the formula is the object’s starting speed, or its initial velocity (v0)

The other term is the amount the velocity changes due to acceleration.

Page 26: Chapter 4: Accelerated Motion in a StraightLine

1. You are asked for speed.2. You are given initial speed, acceleration and time.3. Use the relationship v = v0 + at4. Substitute values

v = 2 m/s + (0.75 m/s2)(10 s) v = 9.5 m/s2

Calculating speed

A ball rolls at 2 m/s off a level surface and down a ramp. The ramp creates an acceleration of 0.75 m/s2. Calculate the speed of the ball 10 s after it rolls down the ramp.

Page 27: Chapter 4: Accelerated Motion in a StraightLine

4.2 Distance traveled in accelerated motion

The distance traveled by an accelerating object can be found by looking at the speed versus time graph.

The graph shows a ball that started with an initial speed of 1 m/s and after one second its speed has increased.

Page 28: Chapter 4: Accelerated Motion in a StraightLine

4.2 Distance traveled in accelerated motion

The area of the shaded rectangle is the initial speed v0 multiplied by the time t, or v0t.

The second term is the area of the shaded triangle.

Page 29: Chapter 4: Accelerated Motion in a StraightLine

4.2 A Model for Accelerated Motion It is possible that a moving object may not start at the origin.

Let x0 be the starting position.

The distance an object moves is equal to its change in position (x – x0).

Page 30: Chapter 4: Accelerated Motion in a StraightLine
Page 31: Chapter 4: Accelerated Motion in a StraightLine

1. You are asked for distance.2. You are given initial speed and acceleration. Assume an initial

position of 0 and a final speed of 0.3. Use the relationship v = v0 + at and x = x0 + v0t + 1/2at2

4. At the highest point the speed of the ball must be zero. Substitute values to solve for time, then use time to calculate distance.

0 = 2 m/s + (- 0.5 m/s2)(t) = - 2 m/s = - 0.5 m/s2 (t) t = 4 s x = (0) + (2 m/s) ( 4 s) + (0.5) (-0.5 m/s2) (4 s)2 = 4 meters

Calculating position from speedand acceleration

The angle of the ramp creates an acceleration of - 0.5 m/s2. What distance up the ramp does the ball travel before it turns around and rolls back?

A ball traveling at 2 m/s rolls up a ramp.

Page 32: Chapter 4: Accelerated Motion in a StraightLine

4.2 Solving motion problems with

acceleration Many practical problems involving accelerated motion have more than one step. List variables Cancel terms that are zero.

Speed is zero when it starts from rest. Speed is zero when it reaches highest point

Use another formula to find the missing piece of information.

Page 33: Chapter 4: Accelerated Motion in a StraightLine

1. You are asked to find the length of the ramp.2. You are given v0 = 0, v = 2 m/s at t = 1 s, t = 3 s at the

bottom of the ramp, and you may assume x0 = 0.3. After canceling terms with zeros, v = at and x = ½ at2

4. This is a two-step problem. First, calculate acceleration, then you can use the position formula to find the length of the ramp.

a = v ÷ t = (2 m/s ) ÷ (1 s ) = 2 m/s2

x = ½ at2 = (0.5)(2 m/s )(3 s )2 = 9 meters

Calculating position from timeand speed

After one second, the speed of the ball is 2 m/s. How long does the ramp need to be so that the ball can roll for 3 seconds before reaching the end?

A ball starts to roll down a ramp with zero initial speed.

Page 34: Chapter 4: Accelerated Motion in a StraightLine

1. You are asked to find the time and speed.2. You are given v0 = 0, x = 440 m, and a = 6 m/s2; assume

x0 = 0.3. Use v = v0 + at and x = x0 + v0t + ½ at2

4. Since x0 and v0 = 0, the equation reduces to x = ½at2 440 m = (0.5)(6 m/s2) (t)2

t2 = 440 ÷ 3 = 146.7 s t = 12.1 s

Calculating time from distanceand acceleration

A car at rest accelerates at 6 m/s2. How long does it take to travel 440 meters, or about a quarter-mile, and how fast is the car going at the end?

Page 35: Chapter 4: Accelerated Motion in a StraightLine

Chapter 4: Accelerated Motion in a Straight Line

4.1 Acceleration

4.2 A Model for Accelerated Motion

4.3 Free Fall and the Acceleration due to

Gravity

Page 36: Chapter 4: Accelerated Motion in a StraightLine

Inv 4.3 Free Fall

Investigation Key Question:

What kind of motion is falling?

Page 37: Chapter 4: Accelerated Motion in a StraightLine

4.3 Free Fall and the Acceleration due to Gravity

An object is in free fall if it is moving under the sole influence of gravity.

Free-falling objects speed up, or accelerate, as they fall.

The acceleration of 9.8 m/s2 is given its own name and symbol—acceleration due to gravity (g).

Page 38: Chapter 4: Accelerated Motion in a StraightLine

4.3 Free fall with initial velocity The motion of an object in free fall is described by the

equations for speed and position with constant acceleration.

The acceleration (a) is replaced by the acceleration due to gravity (g) and the variable (x) is replaced by (y).

Page 39: Chapter 4: Accelerated Motion in a StraightLine
Page 40: Chapter 4: Accelerated Motion in a StraightLine

4.3 Free fall with initial velocity When the initial speed

is upward, at first the acceleration due to gravity causes the speed to decrease.

After reaching the highest point, its speed increases exactly as if it were dropped from the highest point with zero initial speed.

Page 41: Chapter 4: Accelerated Motion in a StraightLine
Page 42: Chapter 4: Accelerated Motion in a StraightLine

4.3 Solving problems with free fall Most free-fall problems ask you to find

either the height or the speed.

Height problems often make use of the knowledge that the speed becomes zero at the highest point of an object’s motion.

If a problem asks for the time of flight, remember that an object takes the same time going up as it takes coming down.

Page 43: Chapter 4: Accelerated Motion in a StraightLine

1. You are asked for distance.2. You are given an initial speed and time of flight.3. Use v = v0 - gt and y = y0 + v0t - ½ gt2

4. Since y0 and v0 = 0, the equation reduces to x = -½ gt2

y = - (0.5) (9.8 m/s2) (1.6s)2

y = -12.5 m (The negative sign indicates the height is lower than the initial height)

Calculating height from the time of falling

A stone is dropped down a well and it takes 1.6 seconds to reach the bottom. How deep is the well? You may assume the initial speed of the stone is zero.

Page 44: Chapter 4: Accelerated Motion in a StraightLine

4.3 Air Resistance and Mass

The acceleration due to gravity does not depend on the mass of the object which is falling.

Air creates friction that resists the motion of objects moving through it.

All of the formulas and examples discussed in this section assume a vacuum (no air).

Page 45: Chapter 4: Accelerated Motion in a StraightLine

4.3 Terminal Speed You may safely assume that a = g = 9.8 m/sec2 for

speeds up to several meters per second.

The air resistance from friction increases as a falling object’s speed increases.

Eventually, the rate of acceleration is reduced to zero and the object falls with constant speed.

The maximum speed at which an object falls when limited by air friction is called the terminal velocity.

Page 46: Chapter 4: Accelerated Motion in a StraightLine

Antilock braking systems (ABS) are standard on most new cars and trucks.

If brakes are applied too hard or too fast, a rolling wheel locks up, which means it stops turning and the car skids.

With the help of constant computer monitoring, these systems give the driver more control when stopping quickly.

Anti-lock Brakes