mechanics. motion along a straight line ps 41 but first a review significant figures non-zero...

MECHANICS

Motion Along a Straight Line

Ps 41

But First a Review Significant Figures

Non-zero digits are always significant. Any zeros between two significant digits are

significant. A final zero or trailing zeros in the decimal portion

are significant. Ex. 0.002500 has 4 significant figures Ex. 2,500 has 2 significant figures Ex. 2.500 x 103 has 4 significant figures

Multiplication/Division – Determined by the LEAST number of significant figures

Addition/Subtraction – Determined by LEAST number of decimal of places in the decimal portion

Vectors Vectors are physical quantities with both

magnitude and direction and cannot be represented by just a single number Displacement vs. Distance Velocity vs. Speed

Represented by A The magnitude of A is represented by |A| or A

P1

P2A

P1

P2B=-A

Vector Addition Tip to tail method or

Parallelogram method

Vector addition is commutative

1221 FFFF (a)

(b)

Vector Components Vector Components

yx AAA cosAAx sinAAy

Vector Addition using Components

Example: Young and Freedman Problem 1.31/1.38 A postal employee

drives a delivery truck along the route shown. Determine the magnitude and direction of the resultant displacement.

Example: Young and Freedman Problem 1.31/1.38

kmD

kmD

kmD

kmD

DDDD

7.9

1.3

0.4

6.2

3

2

1

321

Example: Young and Freedman Problem 1.31/1.38

kmD

kmD

kmD

kmD

kmD

kmD

DDDD

DDDD

DDDDDD

y

x

y

x

y

x

yyyy

xxxx

yx

192.2)45sin(1.3

192.2)45cos(1.3

0

0.4

6.2

0

3

3

2

2

1

1

321

321

321

kmDDD

kmD

kmD

y

y

x

x8.7829.7

792.4192.206.2

192.6192.240

22

Example: Young and Freedman Problem 1.31/1.38 DON’T FORGET

DIRECTION

NofEkmD

kmD

kmD

y

x

38_8.7

7.37192.6

792.4tan

792.4

192.6

1

Unit Vectors Unit vectors are

unitless vectors with a magnitude of 1.

Primarily used to point a direction.

Represented by

Note scalar times vector

i

kAjAiAA zyxˆˆˆ

Example

jiG ˆ4ˆ2

G

Dot Product Or scalar product

Using components

cosBABA

1ˆˆˆˆˆˆ kkjjii0ˆˆˆˆˆˆ kikjji

)ˆˆˆ()ˆˆˆ( kBjBiBkAjAiABA zyxzyx

zzyyxx BABABA

Example What is the angle

between the vectors

Compute up to 3 sig figs.

Solution

jiG ˆ4ˆ2 jiH ˆˆ3

G

H

9.81)200

2(cos

2)1)(4()3)(2(

cos1020cos

1

yyxx HGHG

HGHG

Cross Product Or vector product

Direction is dictated by the right hand rule

Anti-commutative

BAC

sinBAC

ABBA

Cross Product by Components

0ˆˆˆˆˆˆ kkjjii

jkiik

ijkkj

kijji

ˆˆˆˆˆ

ˆˆˆˆˆ

ˆˆˆˆˆ

)ˆˆˆ()ˆˆˆ( kBjBiBkAjAiABA zyxzyx

kBABAjBABAiBABA xyyxzxxzyzzyˆ)(ˆ)(ˆ)(

Determinant form

zyx

zyx

BBB

AAA

kji

BA

ˆˆˆ

zyx

zyx

BBB

AAA

kji

BA

ˆˆˆ

zyx

zyx

BBB

AAA

kji

BA

ˆˆˆ

zyx

zyx

BBB

AAA

kji

BA

ˆˆˆ

Example Vector A has a

magnitude to 5 and lies in the direction of the x-axis. Vector B has a magnitude of 2 and lies along the xy-plane at a 30o angle with the x-axis. Find AxB.

SolutionLet

BAC

sinBAC

52/)2)(5(

Motion Along a Straight Line

Ps 41

Displacement Is a vector quantity,

usually denoted by x. Change in the position of

a point. (we can approximate objects to be a particle)

Remember, since it’s a vector, it’s important to note both magnitude and direction.

Define positive displacement to be a movement along the positive x-axis

Average Velocity At time t1 the car is

at point P1 and at time t2 the car is at point P2

We can define P1 and P2 to have coordinates x1 and x2 respectively

Δx=x2-x1

Average velocity

P1

P2

t

x

tt

xxvave

12

12

Velocity Velocity is the change

in displacement per unit time in a specific direction.

It is a vector quantity, usually denoted by v

Has SI unit of m/s Average velocity can

be useful but it does not paint the complete picture. The winner of a race has the

highest average velocity but is not necessarily the fastest.

Instantaneous Velocity Velocity at a specific

instant of time Define instant as an

extremely short amount of time such that it has no duration at all.

Instantaneous Velocity

top speed of 431.072 km/h

(Sport version. Picture only shows regular version)

dt

dx

t

xv

t

0lim

x-t Graph Average Velocity

Average velocity is the slope of the line between two points

Instantaneous Velocity

Instantaneous velocity is the slope of the tangent line at a specific point

x

t

x

t

x2

x1

o t1 t2o t1

Sample Problem A Bugatti Veyron is at rest

20.0m from an observer. At t=0 it begins zooming down the track in a straight line. The displacement from the observer varies according to the equation

a) Find the average velocity from t=0s to t=10sb) Find the average velocityfrom t=5s to t=10sc) Find the instantaneous velocity at t=10s

2)39.1(0.20 2 tmxs

m

Solution

a)

b)

c)

sm

avev 9.1310

139

010

20159

mmxs

m 159)10)(39.1(20 210 2

mx 200

mmxs

m 75.54)5)(39.1(20 25 2

sm

avev 9.205

25.104

510

75.54159

sm

sm tv 8.27))(78.2( 210

))(78.2(])39.1(20[

2

22

tdt

tmd

dt

dvs

msm

Acceleration Acceleration

describes the rate of change of velocity with time.

Average Acceleration

Vector quantity denoted by

Instantaneous acceleration

t

v

tt

vvaave

12

12

a

2

2

0lim

dt

xd

dt

dv

t

va

t

WARNING Just because

acceleration is positive (negative) does not mean that velocity is also positive (negative).

Just because acceleration is zero does not mean velocity is zero and vice versa.

Motion at Constant Acceleration Assume that

acceleration is constant.

Generally

ct

va

t

vva

0

tavv 0

00 ttiatvv 0

Feel Free to use vf, vi, v0 whatever notation you’re more comfortable with

BUT be consistent through out the entire problem

Motion at Constant Acceleration

t

x

tt

xxvave

0

0

22000 vatvvv

vave

atvvave 21

0 2

21

0 attvx 2

21

00 attvxx

Seat Work #1 Using

Derive

Hint: Eliminate time

atvv 0

221

00 attvxx

xavv 220

2

Giancoli Chapter 2 Problem 26 In coming to a stop a car leaves skid marks 92

m long on the highway. Assuming a deceleration of 7.00m/s2, estimate the speed of the car just before braking.

Chapter 2 Problem 26

Ignore negative

?

0

00.7

92

0

2

v

v

a

mx

sm

sm

sm

smv

v

v

xavv

3689.35

1288

)92)(00.7(20

2

0

20

20

20

2

Falling Objects Most common example of

constant acceleration is free fall.

Freely falling bodies are objects moving under the influence of gravity alone. (Ignore air resistance)

Attracts everything to it at a constant rate.

Note: because it attracts objects downwards acceleration due to gravity is

Galileo Galilei formulated the laws of motion for free fall

280.9s

mg

280.9s

mga

Freely Falling A freely falling body

is any body that is being influenced by gravity alone, regardless of initial motion.

Objects thrown upward or downward or simply released are all freely falling

Example Giancoli 2-42 A stone is thrown vertically upward with a

speed of 18.0 m/s. (a) How fast is it moving when it reaches a height of 11.0m? (b) How long will it take to reach this height? (c) Why are there two answers for b?

Giancoli 2-42

We can’t ignore negative

?

?

0.18

80.9

11

0

2

t

v

v

a

mx

sm

sm

smv

v

v

xavv

4.10

4.108

)11)(80.9(2)18(

2

2

22

20

2

Giancoli 2-42

0110.18)90.4(

)80.9(0.180.112

221

221

00

tt

tt

attvxx

a

acbb

2

42

st

st

t

90.2

775.0

)9.4(2

)11)(9.4(4)18(18 2

Giancoli 2-42 Why were there 2 answers to b?

Summary These 4 equations

will allow you to solve any problem dealing with motion in one direction as long as acceleration is CONSTANT!

1.

2.

3.

4.

atvv 0

221

00 attvxx

20vv

vave

xavv 220

2

Problems from the Book (Giancoli 6th ed) 14- Calculate the average speed and average

velocity of a complete round-trip, in which the outgoing 250 km is covered at 95km/hr, followed by a 1 hour lunch break and the return 250km is covered at 55km/hr.

95 kph

55 kph

Start

End

1 hour break

Chapter 2 Problem 14 Average speed = change in distance / change in

time

For first leg

For return

Total time

Average speed

t

dspeedave

1

25095

t

kmkph

hr

kph

kmt 6316.2

95

2501

2

25055

t

kmkph

hr

kph

kmt 5455.4

55

2502

hrttttotal 1771.81 21

kphkpht

dspeedave 6115.61

1771.8

500

Chapter 2 Problem 14 What was the cars average velocity?

Chapter 2 Problem 19 A sports car moving at constant speed travels

110m in 5.0s. If it then brakes and comes to a stop in 4.0 s, what is its acceleration in m/s2? Express the answer in terms of g’s where g=9.80 m/s2.

Chapter 2 Problem 19 First find v

sm

ave s

m

t

xv 22

5

110

25.54

22s

msm

ave st

va

sgg

as

msm

ave '56.0)80.9

1(5.5

2

2

Chapter 2 Problem 31 A runner hopes to complete a 10,000m run in

less than 30.0 min. After exactly 27.0 min, there are still 1100m to go. The runner must then accelerate at 0.20m/s2 for how many seconds in order to achieve the desired time?

Chapter 2 Problem 31 Find average v at 27 min

The runner will then accelerate for t seconds covering some distance d,

and will then cover the remaining distance in (180-t). (a is now zero)

0sec49.5sec60

min1

min27

8900v

m

t

xv mcurrent

)(20.049.50 tatvv 2

212

21

0 )20.0()(49.5 ttattvd

t

d

t

xv

180

1100

Chapter 2 Problem 31

08.1113610.0

)20.0)(5.0()20.0(180)49.5(1801100

1801801100

1801801100

1801801100

1100180180

1100)180)((

1100)180(

2

2

221

0

221

02

00

200

200

0

tt

tt

atatv

attvattvatv

dattvatv

dattvatv

dtatv

dtv

Chapter 2 Problem 31 Almost there Quadratic Equation

Results in t=357s or t=3.11s

t<180 seconds so t=3.11s

whew

a

acbb

2

42

Chapter 2 Problem 44 A falling stone takes 0.28s to travel past a

window 2.2m tall. From what height above the top of the window did the stone fall?

Chapter 2 Problem 44 Let

tw be the time it takes for the stone to reach the top of the window.

xw be the height above the window stone was dropped

At t=tw+0.28s the stone is now xw-2.2m!!!

smv

x

0

0

0

0

Chapter 2 Problem 44

mx

st

t

ttx

ttx

tx

tx

tx

tx

attvxx

w

w

w

www

www

ww

ww

1.2

662.0

38416.0744.22.2

38416.0744.290.42.2

)0784.056.0(90.42.2

)28.0(90.42.2

90.4

90.4

)80.9(

2

2

2

2

2

221

221

00