ch 2 1 chapter 2 kinematics in one dimension giancoli, physics,6/e © 2004. electronically...

Ch 2 1

Chapter 2

Kinematics in One Dimension

Giancoli, PHYSICS,6/E © 2004. Electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey

Ch 2 2

Module 2

Displacement, Velocity and Acceleration

Giancoli, Sec 2-1, 2, 3, 4, 8

AP Physics C Lesson 1


The following is an excellent lecture on this material.

Ch 2 3

Reference Frames• Any measurement of position, displacement, velocity or

acceleration must be made with respect to a defined reference frame—this is first step in problem solution.

Possible reference frames:

•Window with up = + or –

•Un-stretched net with up = + or -

•Stretched net with up = + or –

•Ground = not sufficient information

Ch 2 4

Coordinate Axis• We will use a set of coordinate axis where x is

horizontal and y is vertical

+y

-y

+x-xx1 x2

•Many problems will be motion in one dimension so we will plot x vs. time.

Ch 2 5

Displacement

Displacement: change in position

12 xxx

+y

-y

+x-xx1 x2

Displacement is a vector, so it has magnitude and direction. In one dimension we use + or minus sign to indicate direction.

Ch 2 6

Don’t Confuse Displacement and Distance

A person walks 70 m East and 30 m West.

Distance traveled =

Displacement =

100 m

40 m East or + 40 m

Ch 2 7

Negative Displacement

In the figure below the displacement is negative.

A negative displacement may indicate motion toward the West or something else depending on the situation and the coordinate system chosen.

12 xxx

Ch 2 8

Average Speed and Velocity

elapsed time

traveleddistance speed average

elapsed time

nt displaceme( velocity average )v

t

x

tt

xxv

12

12

Average velocity is a vector, so it has magnitude and direction. In one dimension we use + or minus sign to indicate direction.

Ch 2 9

t

x

t

xv

21ttt

hkm

km

hkm

kmt

990

2800

790

3100

Example 1. An airplane travels 3100 km at a speed of 790 km/h, and then encounters a tailwind that boosts its speed to 990 km/h for the next 2800 km. What was the total time for the trip? (Assume three significant figures)

hhht 75.683.292.3

2

2

1

1

v

x

v

xt

Ch 2 10

Example 1 (continued) An airplane travels 3100 km at a speed of 790 km/h, and then encounters a tailwind that boosts its speed to 990 km/h for the next 2800 km. What was the total time for the trip? (Assume three significant figures)

What was the average speed of the plane for this trip?

ht 75.6

t

xv

h

kmkm

75.6

28003100 h

km874

Note: A simple average of v1 and v2 gives 890 km/h and is not correct

Ch 2 11

Graphical Analysis of Linear Motion

graphtimevs. of slope isVelocity xt

xv

Ch 2 12

Acceleration

Average Acceleration: change in velocity divided by the time taken to make this change.

t

v

tt

vva

12

12

2 UNITS

s

m

ss

m

Ch 2 13

smvst

smvt

0.50.5

0.150

22

11

12

12

tt

vva

sss

ms

m

00.5

0.150.5

Example 2. A car traveling at 15.0 m/s slows down to 5.0 m/s in 5.0 seconds. Calculate the car’s acceleration.

Coordinate System: + is to the right

20.2s

m )left theto(

Ch 2 14

Acceleration

Instantaneous Acceleration: same definition as before but over a very short t.

t

va

0t

lim

Ch 2 15

Instantaneous Velocity

instantaneous velocity is defined as the average velocity over an infinitesimally short time interval.

t

xv

0t

lim

Ch 2 16


graphtimevs. of slope ison Accelerati vt

va

Ch 2 17

Example 6: Calculate the acceleration between points A and B and B and C.

t

va

0t

lim

) linestraight (t

va

20.00.150.20

0.15sm15.0

sm

sss

m

12

12

tt

vva BA

20.20.200.25

0.15sm5.0

sm

sss

m

12

12

tt

vvaBC

Ch 2 18

Module 3

Motion with Constant Acceleration

Giancoli, Sec 2-1, 2, 3, 4, 8


Ch 2 19

Motion with Constant Acceleration

t (s) v ( m/s) a ( m/ s2 )

0 0 15

1 15 15

2 30 15

3 45 15

4 60 15

5 75 15

Consider the case of a car that accelerates from rest with a constant acceleration of 15 m / s2.

t

v

tt

vva

12

12

)( 1212 ttavv )0(150 22 ts

m

We can make a table

Ch 2 20

Derivations•In the next 4 slides we will combine several known equations under the assumption that the acceleration is constant.

•This process is called a derivation.

•In general you will need to know the initial assumptions, the resultant equations and how to apply them.

•You do not need to memorize derivations

•But, I could ask you to derive an equation for a specific problem. This is very similar to an ordinary problem without a numeric answer.

Ch 2 21

Motion at Constant Acceleration - Derivation

•Consider the special case acceleration equals a constant: a = constant

• Use the subscript “0” to refer to the initial conditions

• Thus t0 refers to the initial time and we will set t0 = 0.

• At this time v0 is the initial velocity and x0 is the initial displacement.

•At a later time t, v is the velocity and x is the displacement

•In the equations t1t0 and t2 t

Ch 2 22

Motion at Constant Acceleration - Derivation

1)(Eqn.0

0

0

t

xx

tt

xxv

) 2 Eqn.(Constant0

t

vva

The average velocity during this time is:

The acceleration is assumed to be constant

From this we can write

) 3 Eqn. (0 tavv

t

v

ov

Ch 2 23

Motion at Constant Acceleration - DerivationBecause the velocity increases at a uniform rate, the average velocity is the average of the initial and final velocities

)4Eqn.(2

0 vvv

From the definition of average velocity

tatvv

xtvv

xtvxx )2

()2

( 000

000

And thus

)5Eqn.(2

1 200 tatvxx

t

v

ov

Ch 2 24

Motion at Constant Acceleration

• The book derives one more equation by eliminating time

• The 4 equations listed below only apply when a = constant

tavv 0

200 2

1tatvxx

)(2 020

2 xxavv

20vv

v

Ch 2 25

mxsmv

xv

0.155.11

0000

0

2

0

2 2 xxavv

)(20

2

0

2

xx

vva

Example 3. A world-class sprinter can burst out of the blocks to essentially top speed (of about 11.5 m/s) in the first 15.0 m of the race. What is the average acceleration of this sprinter, and how long does it take her to reach that speed? (Note: we have to assume a=constant)

)00.15(2

0)5.11( 22

ms

ma

241.4s

ma

atvv 0

a

vvt 0

241.4

05.11

sms

mt

st 61.2

Ch 2 26

2

00 2

1tatvxx

tttt

tsmx t )25(

200 2

1tatvxx cccc

Example 4. A truck going at a constant speed of 25 m/s passes a car at rest. The instant the truck passes the car, the car begins to accelerate at a constant 1.00 m / s2. How long does it take for the car to catch up with the truck.

How far has the car traveled when it catches the truck?

22 )0.1(

2

1t

smx c

When the car catches the truck:tc

xx

tsmts

m )25()0.1(2

1 22

smts

m 25)0.1(2

12

st 50

22 )50)(0.1(

2

1ss

mxc m1250

Ch 2 27


graphtimevs. of slope isVelocity xt

xv

graphtimevs. of slope ison Accelerati vt

va

Ch 2 28


t

xv

0t

lim

t

va

0t

lim

v is slope of position vs. time graph.

a is slope of velocity vs. time graph.

Ch 2 29

Example 6: Calculate the acceleration between points A and B and B and C.

t

va

0t

lim

C) B & B A for linestraight (

t

va

20.00.150.20

0.15sm15.0

sm

sss

m

12

12

tt

vva BA

20.20.200.25

0.15sm5.0

sm

sss

m

12

12

tt

vvaBC

Ch 2 30

Module 4

Falling Objects

Giancoli, Sec 2-1, 2, 3, 4, 8


Ch 2 31

Falling Objects

•Galileo showed that for object falling from rest with no air resistance

y t2

•Note that this is true when acceleration is constant

Ch 2 32

Falling ObjectsGalilei showed that

•near the surface of the earth

•at the same location

•in the absence of air resistance

all objects fall with the same constant

acceleration g, the acceleration

due to gravity

g = 9.8 m/s2

Note: g is a positive number. When you define your coordinate system, you can decide whether up or down is positive.

smv /8.91

smv /6.192

smv /4.293

Ch 2 33

Up and Down Motion

For object that is thrown upward and returns to starting position:

• assumes up is positive

•velocity changes sign (direction) but acceleration does not

•Velocity at top is zero

•time up = time down

•Velocity returning to starting position = velocity when it was released but opposite sign

2

00 2

1gttvyy

gtvv 0

Ch 2 34

Acceleration due to Gravity

Ch 2 35

Example 5

myys

masmv

700

8.90.12

0

20

2

00 2

1gttvyy

29.412070 tt

(2-47) A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 70.0 m high (Fig. 2–34). (a) How much later does it reach the bottom of the cliff? (b) What is its speed just before hitting? (c) What total distance did it travel?

(a) UP = POSITIVE

070129.4 2 tt

)9.4)(2(

)70)(9.4)(4()12()12( 2 t

sst 20.5,75.2

Use quadratic equation:

Answer = 5.20 s

Ch 2 36

Example 5

atvv 0

)2.5)(8.9(12 2 ssm

sm

2-47 A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 70.0 m high (Fig. 2–34). (a) How much later does it reach the bottom of the cliff? (b) What is its speed just before hitting? (c) What total distance did it travel?

(b) sm9.38

(c) Find maximum height, where 0v

)(20

2

0

2 yyavv

))(2(

2

0

2

0 a

vvyy

)8.9)(2(

)12(00

2

2

sms

my

m35.7

Total distance = mmm 0.7035.735.7 m7.84

)(2 020

2 yyavv

Ch 2 37

Example 5 (2-47) continued. A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 70.0 m high (Fig. 2–34). (a) How much later does it reach the bottom of the cliff? (b) What is its speed just before hitting? (c) What total distance did it travel? Excel Calculation—use the equation for displacement and velocity to get y and vy vs time.

Time Y (m) v y (m/s)(s)

0.00 0.00 12.000.25 2.69 9.550.50 4.78 7.100.75 6.24 4.651.00 7.10 2.201.25 7.34 -0.251.50 6.98 -2.701.75 5.99 -5.152.00 4.40 -7.602.25 2.19 -10.052.50 -0.63 -12.502.75 -4.06 -14.953.00 -8.10 -17.403.25 -12.76 -19.853.50 -18.03 -22.303.75 -23.91 -24.754.00 -30.40 -27.204.25 -37.51 -29.654.50 -45.23 -32.104.75 -53.56 -34.555.00 -62.50 -37.005.25 -72.06 -39.45

Ch 2 38

Problem Solving Tips1. Read the whole problem and make sure you

understand it. Then read it again.

2. Decide on the objects under study and what the time interval is.

3. Draw a diagram and choose coordinate axes.

4. Write down the known (given) quantities, and then the unknown ones that you need to find.

5. What physics applies here? Plan an approach to a solution.

Ch 2 39

Problem Solving Tips6. Which equations relate the known and unknown quantities? Are they valid in this situation? Solve algebraically for the unknown quantities, and check that your result is sensible (correct dimensions).

7. Calculate the solution and round it to the appropriate number of significant figures.

8. Look at the result – is it reasonable? Does it agree with a rough estimate?

9. Check the units again.

ch 2 1 chapter 2 kinematics in one dimension giancoli, physics,6/e © 2004. electronically...

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