Chapter 2 One-Dimensional Kinematicsnsmn1.uh.edu/rbellwied/classes/spring2013/ch2_notes.pdf · • Freely falling objects: constant acceleration g = 9.81 m/s2. Title: ch2-notes.pdf

Copyright © 2010 Pearson Education, Inc.

Chapter 2

One-DimensionalKinematics


Units of Chapter 2• Position, Distance, and Displacement

• Average Speed and Velocity

• Instantaneous Velocity

• Acceleration

• Motion with Constant Acceleration

• Applications of the Equations of Motion

• Freely Falling Objects


2-1 Position, Distance, and Displacement

Before describing motion, you must set up acoordinate system – define an origin and apositive direction.



The distance is the total length of travel; if youdrive from your house to the grocery store andback, you have covered a distance of 8.6 mi.



Displacement is the change in position. If youdrive from your house to the grocery store andthen to your friend’s house, your displacementis 2.1 mi and the distance you have traveled is10.7 mi.


0 x (cm)21−2 −1

cm 4

cm 2 cm 2

!=

!!=

!=" if xxx

Example: A ball is initially at x = +2 cm and is moved to x =-2 cm. What is the displacement of the ball?


Example: At 3 PM a car is located 20 km south of its startingpoint. One hour later its is 96 km farther south. After twomore hours it is 12 km south of the original starting point.

(a) What is the displacement of the car between 3 PM and6 PM?

xi = –20 km and xf = –12 km

( ) km 8km 20 km 12 +=!!!=

!=" if xxx

Use a coordinatesystem wherenorth is positive.


(b) What is the displacement of the car from the startingpoint to the location at 4 pm?

(c) What is the displacement of the car from 4 PM to 6 PM?

Example continued

xi = 0 km and xf = –96 km

( ) km 96km 0 km 96 !=!!=

!=" if xxx

xi = –96 km and xf = –12 km

( ) km 84km 96 km 12 +=!!!=

!=" if xxx


2-2 Average Speed and Velocity

The average speed is defined as the distancetraveled divided by the time the trip took:

Average speed = distance / elapsed time

Is the average speed of the red car 40.0 mi/h,more than 40.0 mi/h, or less than 40.0 mi/h?



Average velocity = displacement / elapsed time

If you return to your starting point, youraverage velocity is zero.



Graphical Interpretation of Average Velocity

The same motion, plotted one-dimensionallyand as an x-t graph:


2-3 Instantaneous Velocity

Definition:

(2-4)

This means that we evaluate the averagevelocity over a shorter and shorter period oftime; as that time becomes infinitesimallysmall, we have the instantaneous velocity.


2-3 Instantaneous VelocityThis plot shows the average velocity beingmeasured over shorter and shorter intervals.The instantaneous velocity is tangent to thecurve.


2-3 Instantaneous Velocity

Graphical Interpretation of Average andInstantaneous Velocity


Example: Speedometer readings are obtained andgraphed as a car comes to a stop along a straight-linepath. How far does the car move between t = 0 and t = 16seconds?

Since there is not a reversal of direction, the areabetween the curve and the time axis will represent thedistance traveled.


Example continued:

The rectangular portion has an area of Lw = (20 m/s)(4 s)= 80 m.

The triangular portion has an area of ½bh = ½(8 s) (20 m/s)= 80 m.

Thus, the total area is 160 m. This is the distance traveledby the car.


2-4 Acceleration

Average acceleration:

(2-5)


2-4 AccelerationGraphical Interpretation of Average andInstantaneous Acceleration:


2-4 Acceleration

Acceleration (increasing speed) anddeceleration (decreasing speed) should not beconfused with the directions of velocity andacceleration:


Example: The graph shows speedometer readings as a carcomes to a stop. What is the magnitude of the accelerationat t = 7.0 s?

The slope of the graph at t = 7.0 sec is

( )( )

2

12

12

avm/s 5.2

s 412

m/s 200=

!

!=

!

!=

"

"=

tt

vv

t

va

x


2-5 Motion with Constant Acceleration

If the acceleration is constant, the velocitychanges linearly:

(2-7)

Average velocity:



Average velocity:

(2-9)

Position as a function of time:

(2-10)

(2-11)

Velocity as a function of position:

(2-12)



The relationship between position and timefollows a characteristic curve.



The three key equations for Ch.2:


2-6 Applications of the Equations of MotionHit the Brakes!


Example: A trolley car in New Orleans starts from rest at theSt. Charles Street stop and has a constant acceleration of1.20 m/s2 for 12.0 seconds.

0

2

4

6

8

10

12

14

16

0 2 4 6 8 10 12 14

t (sec)

v (

m/s

ec)

(a) Draw a graph of vx versus t.


(b) How far has the train traveled at the end of the 12.0seconds?

The area between the curve and the time axisrepresents the distance traveled.

( )

( )( ) m 4.86s 12m/s 4.142

1

tsec 12t2

1

==

!"==! vx

(c) What is the speed of the train at the end of the 12.0 s?

This can be read directly from the graph, vx = 14.4 m/s.

Example continued:


Example: A train of mass 55,200 kg is traveling along astraight, level track at 26.8 m/s. Suddenly the engineer seesa truck stalled on the tracks 184 m ahead. If the maximumpossible braking acceleration has magnitude of 1.52 m/s2,can the train be stopped in time?

( )( )

m 236m/s 52.12

m/s 8.26

2

02

2

22

22

=!

!=

!="

="+=

x

ix

xixfx

a

vx

xavv

Know: ax = −1.52 m/s2, vix = 26.8 m/s, vfx = 0

Using the given acceleration, compute the distancetraveled by the train before it comes to rest.

The train cannot be stopped in time.


2-7 Freely Falling Objects

Free fall is the motion of an object subjectonly to the influence of gravity. Theacceleration due to gravity is a constant, g.



An object falling in air is subject to airresistance (and therefore is not freely falling).



Free fall from rest:


Example: A penny is dropped from the observation deck ofthe Empire State Building 369 m above the ground. Withwhat velocity does it strike the ground? Ignore airresistance.

369 m

x

yGiven: viy = 0 m/s, ay = −9.8 m/s2,Δy = −369 m

Unknown: vyf

Use:

yav

ya

yavv

yyf

y

yiyfy

!=

!=

!+=

2

2

222

ay


( )( ) m/s 0.85m 369m/s 8.9222 =!!="= yav yyf

How long does it take for the penny to strike the ground?

sec 7.82

2

1

2

1 22

=!

=!

!=!+!=!

y

yyiy

a

yt

tatatvy

(downward)

Example continued:

Given: viy = 0 m/s, ay = −9.8 m/s2, Δy = −369 mUnknown: Δt


Example: You throw a ball into the air with speed 15.0 m/s;how high does the ball rise?

Given: viy = +15.0 m/s; ay = −9.8 m/s2

2

2

1tatvy

yiy!+!=!

x

yviy

ay

tavv yiyfy !+=

To calculate the final height, weneed to know the time of flight.

Time of flight from:


sec 53.1m/s 9.8

m/s 0.15

0

2=

!!=!="

="+=

y

iy

yiyfy

a

vt

tavvThe ball risesuntil vfy = 0.

( )( ) ( )( )

m 5.11

s 53.1m/s 8.92

1s 53.1m/s 0.15

2

1

22

2

=

!+=

"+"=" tatvyyiy

The height:

Example continued:


2-7 Freely Falling ObjectsTrajectory of a projectile:


Summary of Chapter 2

• Distance: total length of travel

• Displacement: change in position

• Average speed: distance / time

• Average velocity: displacement / time

• Instantaneous velocity: average velocitymeasured over an infinitesimally small time


Summary of Chapter 2• Instantaneous acceleration: averageacceleration measured over an infinitesimallysmall time

• Average acceleration: change in velocitydivided by change in time

• Deceleration: velocity and acceleration haveopposite signs

• Constant acceleration: equations of motionrelate position, velocity, acceleration, and time

• Freely falling objects: constant accelerationg = 9.81 m/s2

Documents

Chapter 2 One-Dimensional Kinematicsnsmn1.uh.edu/rbellwied/classes/spring2013/ch2_notes.pdf · • Freely falling objects: constant acceleration g = 9.81 m/s2. Title: ch2-notes.pdf