### 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  at

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  2ax
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  2ax
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  2ax
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  2ax

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
```