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Ch 2 Kinematics in one dimension

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Goals for Chapter 2

• Become comfortable with displacement, velocity, and acceleration.

• Explore motions at constant acceleration.

• Be able to graph and interpret graphs as they describe motion.

• Be able to analyze multiple-concept problems.

• Examine the special case of freely falling bodies.

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Motion

Motion is divided into two areas of study:

DynamicsDynamics answers the “Why is this object moving?” question. Deals with the effect of forces.Will come in Chapter 4 and later.

KinematicsKinematics deals with the concepts which are needed to describe motion of objectsThis will be our focus in chapter 2.

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A One-Dimensional Coordinate System

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One-Dimensional Coordinates

Distance: total length of travel.SI unit: meter, m.

Displacement = change in position = final position – initial position.

Displacement = change in position = final position – initial position

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position initial ox

position final x

ntdisplaceme oxxx

2.1 Displacement

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x̂ m 0.2ox

x̂ m 0.7x

x̂ m 0.5x

x̂ m 0.5x̂ m 2.0x̂ m 7.0 oxxx

Displacement

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x̂ m 0.2x

x̂ m 0.7ox

x̂ m 0.5x

x̂ m 0.5x̂ m 7.0x̂ m 2.0 oxxx

Displacement

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x̂ m 0.2ox

x̂ m 0.7x

x̂ m 0.5x

x̂ m 0.7x̂ m .02x̂ m 5.0 oxxx

Displacement

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Average speed is the distance traveled divided by the timerequired to cover the distance.

timeElapsed

Distance speed Average

SI units for speed: meters per second (m/s)

2.2 Speed and velocity

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How far does a jogger run in 1.5 hours (5400 s) if his average speed is 2.22 m/s?

timeElapsed

Distance speed Average

m 12000s 5400sm 22.2

timeElapsedspeed Average Distance

Example: Average speed and distance

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Average velocity is the displacement divided by the elapsed time.

timeElapsed

ntDisplaceme velocityAverage

ttt o

o

xxx

v

Average velocity

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Andy Green in the car ThrustSSC set a world record of 341.1 m/s in 1997. To establish such a record, the driver makes two runs through the course, one in each direction, to nullify wind effects. From the data, determine the average velocity for each run.

Example: Average velocity

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sm5.339s 4.740

m 1609t

x

v

sm7.342s 4.695

m 1609t

x

v

Example: Average velocity

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Average velocity on a x vs. t graph

The slope of a line connecting two points on an x-versus-t plot is equal to the average velocity during that time interval.

tx,av

x

v

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The instantaneous velocity indicates how fast the car moves and the direction of motion at each instant of time.

tt

xv

0lim

Instantaneous velocity

• Instantaneous velocity is found with a tangent line to a position vs time graph.

• We need to be sure of sign conventions.

• What direction is the motion undergoing? Direction draws a distinction between scalars and vectors.

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Instantaneous velocity via graphing

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Instantaneous velocity: Tangent to x(t)

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t

xlimv

0t

Instantaneous velocity

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2.3 Acceleration

• The idea develops as it did with position vs time for velocity: now we plot velocity vs time. We need to consider sign conventions and “braking” as a new idea and application of vectors.

The notion of acceleration emerges when a change in velocity is combined with the time during which the change occurs.

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Cars Accelerating or Decelerating

Physical direction of a is the direction of change in vHere’s the connection:

a in same direction as v object is speeding upa in opposite dir. to v object is slowing down

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Definition of acceleration

Average Acceleration

SI unit: m/s2

Acceleration is a vector quantity.

ttt o

o

vvv

a

t

vlima

0t

Instantaneous

Acceleration

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Acceleration from a velocity vs time plot

t

vlima

0t

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Determine the average acceleration of the plane.

sm0ov

hkm260v

s 0ot s 29t

s

hkm0.9

s 0s 29

hkm0hkm260

o

o

tt

vva

Example: Acceleration and increasing velocity

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2sm0.5s 9s 12

sm28sm13

o

o

tt

vva

Example: Acceleration and decreasing velocity

Determine the average acceleration of the car

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Increasing velocity (‘speed up’)

Decreasing velocity (‘brake’)

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What is acceleration at A, B, C ?

t

vlima

0t

ss

m

05

010lim

0t

B

A

C

2s

m2

ss

m

515

1010lim

0t

2s

m0

ss

m

1525

105lim

0t

2s

m

2

1

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It is customary to dispense with the use of boldface symbols overdrawn with arrows for the displacement, velocity, and acceleration vectors (in 1-dim). We will, however, continue to convey the directions with a plus or minus sign.

o

o

tt

vv

a

o

o

tt

xx

v

o

o

tt

xxv

o

o

tt

vva

2.4 Equations of kinematics for constant acceleration

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o

o

tt

vva

t

vva o

ovvta

)4.2(atvv o

2.4 Equations of kinematics for constant acceleration

a = const. and to = 0

Velocity increases / decreases linearly with time

v

t

a

t

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Example: Passing speed

• A car initially traveling along a straight stretch of highway at 15 m/s accelerates with a constant acceleration of 2.0 m/s2 in order to pass a truck. What is the velocity of the car after 5.0 s ?

v = vo + at

= 15 + 2·(5.0) = 25.0 m/s

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Five kinematic variables:

1. displacement, x

2. acceleration (constant), a

3. final velocity (at time t), v

4. initial velocity, vo

5. elapsed time, t

Equations of kinematics for constant acceleration

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Equations of Kinematics for Constant Acceleration

Problem solving hint: Each equation contains 4 variables. If we know three variables, with the right equation we can solve the problem.

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o

o

tt

xxv

0xo 0to

tvx

Let the object be at the origin when the clock starts.

t

xv

Equations of kinematics for constant acceleration

For constant acceleration, the velocity increases at a linear rate. Then average velocity is midway between initial velocity vo

and final velocity v

v )8.2(2

1 tvvx o

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atvv o

tatvvtvvx oo21

o21

)8.2(2

21 attvx o

Equations of kinematics for constant acceleration

x

t

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m 110

s 0.8sm0.2s 0.8sm0.6

attvx22

21

221

o

Equations of kinematics for constant acceleration

Find the position of the accelerating speedboat at t = 8.0 s.

vo = 6.0 m/s

a = +2.0 m/s2

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A jet is taking off from a deck of an aircraft carrier. Starting from rest, the jet is catapulted with constant acceleration 31 m/s2 along a straight line and reaches a velocity of 62 m/s.

Find its displacement.

sm0vo

??x

2sm31a

sm62v

Example: Catapulting a jet

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a

vvvvtvvx o

o21

o21

t

vva o

a

vvt o

)9.2(

22

2a

vvx o

Equation of kinematics for constant acceleration

Need relation between acceleration and distance. Eliminate time t.

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m 62

sm312

sm0sm62

a2

vvx

2

222o

2

Equation of kinematics for constant acceleration

Example 6

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Summary: Constant acceleration equations

atvv o v

t

a

t

x

t

a = const.

2oo at

2

1tvxx

ax2vv 2o

2 afs

Reasoning Strategy

1. Make a drawing.

2. Decide which directions are to be called positive (+) and

negative (-).

3. Write down the values that are given for any of the five

kinematic variables.

4. Verify that the information contains values for at least three

of the five kinematic variables. Select the appropriate equation.

5. When the motion is divided into segments, remember that

the final velocity of one segment is the initial velocity for the next.

6. Keep in mind that there may be two possible answers to a kinematics problem.

2.5 Applications of the Equation of kinematics

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Example 9 A motocycle ride

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Example 9 A motocycle ride

x a Vf Vo t

Segment #1 120 2.6 ? 0

Segment #2 ? -1.5 12 V1

Segment #1

smV

axVV

f

of

/0.25

6241206.22222

Segment #2

mx

aVVx of

1600.3/]481[

)5.1(2/])25()12[(2/)( 2222

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2.6 Freely Falling bodies

In the absence of air resistance, it is found that all bodiesat the same location above the Earth fall vertically with the same acceleration. If the distance of the fall is smallcompared to the radius of the Earth, then the accelerationremains essentially constant throughout the descent.

This idealized motion is called free-fall and the acceleration of a freely falling body is called the acceleration due to gravity.

22 sft2.32or sm80.9g

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Freely Falling bodies

2sm80.9g

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Freely Falling bodies

Example 10 A Falling Stone

A stone is dropped from the top of a tall building. After 3.00sof free fall, what is the displacement y of the stone?

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Freely Falling bodies

y a v vo t

? -9.80 m/s2 0 m/s 3.00 s

my

attvy o

1.44

)3)(8.9()5.0(02

1 22

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Example: Motion in one dimension• When throwing a ball straight up, which of the following is true about its velocity v and its acceleration a at the highest point in its path?

y

(a) v = 0 and a = 0.

(b) v 0, but a = 0.

(c) v = 0, but a 0.

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The referee tosses the coin up with an initial speed of 5.00m/s.In the absence if air resistance, how high does the coin go above its point of release?

Example 12: How high does it go ?

Given:a -9.80 m/s2

vo + 5.00 m/sv 0 m/s

Find y

ay2vv 2o

2

a2

vvy

2o

2

m 28.1

sm80.92

sm00.5sm0y

2

22

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There are three parts to the motion of the coin. On the way up, the coin has a vector velocity that is directed upward and has decreasing magnitude. At the top of its path, the coin momentarily has zero velocity. On the way down, the coin has downward-pointing velocity with an increasing magnitude.

In the absence of air resistance, does the acceleration of the coin, like the velocity, change from one part to another?

Conceptual Example: Acceleration versus velocity

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Symmetry of a freely falling body

The motion of an object that is thrown upward and eventually falls back to earth has a symmetry that is useful to keep in mind.

In example 12, the initial velocity is 5.0 m/s, what is the velocity of the coin when it is at a height of 1.04 m?

smv

sm

ayvv o

/15.2

)/(62.4

)04.1)(8.9(2522

222

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sm4s 2

m 8 Slope

t

x

2.7 Graphical analysis: Velocity

tt

xv

0lim

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2sm6s 2

sm 12

t

v Slope

2.7 Graphical analysis: Acceleration

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Velocity vs. time plots

• Find the acceleration in each of the following plots

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Summary

atvv 0

200 at

2

1tvxx

consta

For constant acceleration we find:

)xx(a2vv 02

02

Velocity

t

xlimv

0t

Acceleration

t

vlima

0t

x

a

vt

t

t

Position