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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