Transcript Lecture 1
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
P2
P1
A
P2
P1
B=-A
Vector Addition
Tip to tail method or
Parallelogram method
Vector addition is
commutative
F1 F2 F2 F1
(a)
(b)
Vector Components
Vector Components
A Ax Ay
Ax A cos
Ay A sin
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
D D1 D2 D3
D1 2.6km
D2 4.0km
D3 3.1km
D 9.7 km
Example: Young and Freedman Problem
1.31/1.38
D Dx D y D1 D2 D3
Dx D1x D2 x D3 x
D y D1 y D2 y D3 y
D1x 0km
D1 y 2.6km
D2 x 4.0km
D2 y 0km
D3 x 3.1 cos(45) 2.192km
D3 y 3.1sin(45) 2.192km
Dx 0 4 2.192 6.192km
Dy 2.6 0 2.192 4.792km
D D x2 Dy2 7.829 7.8km
Example: Young and Freedman Problem
1.31/1.38
DON’T FORGET
DIRECTION
Dx 6.192km
Dy 4.792km
4.792
tan
37.7
6.192
1
D 7.8km _ 38 NofE
Unit Vectors
Unit vectors are unitless
vectors with a magnitude
of 1.
Primarily used to point a
direction.
Represented by iˆ
A Ax iˆ Ay ˆj Az kˆ
Note scalar times vector
Example
G 2iˆ 4 ˆj
G
Dot Product
Or scalar product
Using components
A B A B cos
iˆ iˆ ˆj ˆj kˆ kˆ 1
iˆ ˆj ˆj kˆ iˆ kˆ 0
A B ( Axiˆ Ay ˆj Az kˆ) ( Bxiˆ By ˆj Bz kˆ)
Ax Bx Ay By Az Bz
Example
What is the angle
between the vectors
G 2iˆ 4 ˆj
H 3iˆ ˆj
H
Compute up to 3 sig figs.
Solution
G H G H cos 20 10 cos
Gx H x G y H y (2)(3) (4)(1) 2
2
cos (
) 81.9
200
1
G
Cross Product
Or vector product
C A B
C A B sin
Direction is dictated by
the right hand rule
Anti-commutative
A B B A
Cross Product by Components
iˆ iˆ ˆj ˆj kˆ kˆ 0
iˆ ˆj ˆj iˆ kˆ
ˆj kˆ kˆ ˆj iˆ
kˆ iˆ iˆ kˆ ˆj
A B ( Axiˆ Ay ˆj Az kˆ) ( Bxiˆ By ˆj Bz kˆ)
( Ay Bz Az By )iˆ ( Az Bx Ax Bz ) ˆj ( Ax By Ay Bx )kˆ
Determinant form
iˆ
A B Ax
Bx
ˆj
ˆ
k
Ay
By
Az
Bz
iˆ
A B Ax
Bx
ˆj
ˆ
k
Ay
By
Az
Bz
iˆ
A B Ax
Bx
ˆj
ˆ
k
Ay
By
Az
Bz
iˆ
A B Ax
Bx
ˆj
ˆ
k
Ay
By
Az
Bz
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 xyplane at a 30o angle with
the x-axis. Find AxB.
Solution
Let C A B
C A B sin
(5)(2) / 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
P1
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
vave
x2 x1 x
t2 t1
t
P2
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
x dx
v lim
t 0 t
dt
top speed of 431.072 km/h
(Sport version. Picture only shows
regular version)
x-t Graph
Average Velocity
x
Instantaneous Velocity
x
x2
x1
o
t1
t2
o
t
Average velocity is the slope of the
line between two points
t1
t
Instantaneous velocity is the slope of
the tangent line at a specific point
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
x 20.0m (1.39m s2 )t 2
a) Find the average velocity
from t=0s to t=10s
b) Find the average velocity
from t=5s to t=10s
c) Find the instantaneous
velocity at t=10s
Solution
x0 20m
x10 20m (1.39m s2 )(10)2 159m
a) vave
159 20 139
13.9 m s
10 0
10 2
x5 20m (1.39m s2 )(5) 54.75m
b)vave
159 54.75 104 .25
20.9 m s
10 5
5
m )t 2 ]
d
[
20
m
(
1
.
39
s2
c)dv
(2.78 m s 2 )(t )
dt
dt
v10 (2.78m s 2 )(t ) 27.8 m s
Acceleration
Acceleration describes the
rate of change of velocity
with time.
Average Acceleration
aave
v2 v1 v
t2 t1 t
Vector
quantity denoted
by a
Instantaneous acceleration
v dv d x
a lim
2
t 0 t
dt dt
2
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.
v
a
c
t
v v0
a
t
v v0 at
Generally
ti t0 0
v v0 at
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
x x0 x
vave
t t0
t
v v0 v0 at v0
vave
2
2
vave v0 12 at
2
1
x v0t 2 at
x x0 v0t 2 at
1
2
Seat Work #1
Using
v v0 at
x x0 v0t 12 at2
Derive
v v 2ax
2
2
0
Hint: Eliminate time
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
x 92m
a 7.00 m s 2
v 0ms
v0 ?
v 2 v02 2ax
0 v02 2(7.00)(92)
v02 1288
Ignore negative
v0 35.89 m s 36 m s
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.
g 9.80 m s 2
Note: because it attracts
objects downwards
acceleration due to gravity is
a g 9.80m s 2
Galileo Galilei formulated
the laws of motion for free
fall
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
x 11m
a 9.80 m s 2
v0 18.0 m s
v?
t ?
v 2 v02 2ax
v 2 (18) 2 2(9.80)(11)
v 108.4
v 10.4 m s
2
We can’t ignore negative
Giancoli 2-42
x x0 v0t
1
2
at
2
11.0 18.0t 1 2 (9.80)t 2
(4.90)t 18.0t 11 0
2
18 (18) 2 4(4.9)(11)
t
2(4.9)
t 0.775s
t 2.90s
b b 4ac
2a
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!
v v0 at
1.
2.
x x0 v0t 2 at
3.
v2 v02 2ax
4.
1
vave
v v0
2
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.
Start
95 kph
1 hour break
End
55 kph
Chapter 2 Problem 14
Average speed = change in distance / change in time
speed ave
For first leg
d
t
250km
250km
t1
2.6316hr
95kph
95
kph
t
1
For return
250km
250km
t 2
4.5455hr
55kph
55kph
t 2
Total time
ttotal t1 1 t2 8.1771hr
Average speed d
500
speed ave
61.15kph 61kph
t 8.1771
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
x 110 m
vave
22 m s
t
5s
v 22 m s
aave
5.5 m s 2
t
4s
1g
m
aave 5.5 s 2 (
) 0.56g ' s
9.80 m s 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
vcurrent
x 8900 m 1 min
5.49 m sec v0
t 27 min 60 sec
The runner will then accelerate for t seconds covering
some distance d,
v v0 at 5.49 0.20(t )
d v0t 12 at2 5.49(t ) 12 (0.20)t 2
and will then cover the remaining distance in (180-t). (a is
now zero)
x 1100 d
v
t
180 t
Chapter 2 Problem 31
v(180 t ) 1100 d
(v0 at)(180 t ) 1100 d
180v0 180at v0t at 2 1100 d
1100 180v0 180at v0t at d
2
1100 180v0 180at v0t at v0t at
2
1100 180v0 180at at
1
2
1
2
2
2
1100 180(5.49) 180(0.20)t (0.5)(0.20)t 2
0.10t 36t 111.8 0
2
Chapter 2 Problem 31
Almost there Quadratic Equation
b b 4ac
2a
2
Results in t=357s or t=3.11s
t<180 seconds so t=3.11s
whew
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
x0 0
v0 0 m s
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!!!
Chapter 2 Problem 44
x x0 v0t 1 2 at 2
x
1
2
(
9
.
80
)
t
2
x 4.90t 2
xw 4.90t w2
xw 2.2 4.90(t w 0.28) 2
xw 2.2 4.90(t w2 0.56t w 0.0784)
xw 2.2 4.90t w2 2.744t w 0.38416
2.2 2.744t w 0.38416
t w 0.662s
xw 2.1m