Standard Units of Measure used in Physics:
Base Quantity
time
distance/length
velocity/speed
mass
Force
SI Base Units of Measure
Base Unit
seconds
meters
meters per second
kilograms
Newtons
Abbreviation
s
m
m/s
kg
N
SI Prefixes used in notation for simplifying written expressions:
Prefix
femto
pico
nano
micro
milli
centi
deci
kilo
mega
giga
tera
Abbreviation
f
p
n
μ
m
c
d
k
M
G
T
SI Prefixes
Decimal
Scientific Notation
0.000000000000001
× 10 -15
0.000000000001
× 10 -12
0.000000001
× 10 -9
0.000001
× 10 -6
0.001
× 10 -3
0.01
× 10 - 2
0.1
× 10 -1
1000
× 10 3
1000000
× 10 6
1000000000
× 10 9
1000000000000
× 10 12
Example
femtometer (fs)
picometer (pm)
nanometer (nm)
micrometer ( μ m)
millimeter (mm)
centimeter (cm)
decimeter (dm)
kilometer (km)
megameter (Mm)
gigameter (Gm)
terameter (Tm)
Conversions into standard units of measure:
The magnitude of numbers can be written in many forms and many times utilize SI prefixes to shorten the
written form. However, in order to work with these numbers more easily, they must be in similar units of
measure. If they are not, then we must convert them into the standard base units necessary for that specific
situation.
Examples:
1.1)
75cm 
75(.01)m  0.75m In this case the centi in centimeter is replaced by its decimal form
(0.01)and multiplied to the original value (75).
1.2)
5GL 
1.3)
8km / hr 
5(1000000000) L  5000000000 L
8(1000)m
 2.22m / s
3600 s
In this case the kilo is replaced by its decimal form
(1000) while the hours on the denominator are replaced
with its value in seconds (3600).
Scientific Notation:
Scientific notation is a way of writing very large or small numbers to shorten their appearance and make them
easier to compare with other numbers by utilizing a decimal rounded to no more than 2 decimal places
multiplied by a ten with an exponent that represents the number of places the decimal would have to move to
regenerate the original number. When writing the decimal, we must move the decimal in the original number
(left or right) until there is only one digit (1,2,…7,8,9) on the left side of the decimal. The number of times the
decimal was moved represents the number of tens that would need to be multiplied or divided into the original
number in order to make the decimal place change positions. The total number of tens multiplied or divided
into the original number then becomes the exponent for the notation form. If the decimal was moved to the
right—the exponent is positive. If the decimal was moved to the left—the exponent is negative.
Examples:
1.4)
Here the decimal was moved 8 times to the
right in order to get one digit (3) to the left
side of the decimal. Thus there were 8 tens
that would need to be multiplied to the
decimal in order to get the original number,
thus the exponent will be 8.
350,000,000
 3.5 10 10 10 10 10 10 10 10
 3.5 108
1.5)
This time the decimal was moved 9 times to
the right in order to get on digit (5) to the
left side of the decimal. Thus there were 9
tens that must be divided into the decimal in
order to get the original number.
0.00000000592
 5.92  10  10  10  10  10  10  10  10  10
 5.92  10 9
 3.95  1010
1.6)
39,452,087,000
1.7)
13379.2  10 9  1.34  10 4  10 9
1.8)
(3.52  10 6 )( 2.4  10 7 )  (3.52  2.4)(10 610 7 )
 1.34  10 5
 8.45  1013
Basics of Motion:
Distance- Total measureable length of how far an object has moved or traveled
Displacement- Shortest measureable distance between beginning and ending locations
Speed- The rate at which an object changes positions or covers a distance
Velocity- The speed at which an object is moving with an indicated direction
Acceleration- The rate at which an object changes its velocity (speed up, slow down, change directions)
v
d
t
a
v v  v0

t
t
Examples:
1.9)
1.10)
A car travels 75 m in 13 s. Find
its average velocity.
A ball rolling at an average
speed of 5 m/s rolls for 10 s.
How far did the ball travel?
d  75m
v
t  13s
v?
v  5m / s
t  10s
d ?
d 75m

 5.77 m / s
t 13s
d
d
 5m / s 
t
10s
d  (5m / s)(10s)  50m
v
1.11)
A race car speeds up from 5
m/s to 32 m/s in 8.2 s. What is
the average acceleration
experienced by the car?
v0  5m / s
v  32m / s
t  8.2 s
a?
1.12)
A runner moving at 15 m/s
attempts to stop after running
passed the finish line by
slowing with an average
acceleration of  2.75m / s 2 .
How long does it take him to
come to a stop on the track?
v 0  15m / s
a
v 32(m / s)  5(m / s)

t
8.2 s
a
27m / s
 3.29m / s 2
8.2 s
a
v
t
v  0m/ s
a  2.75m / s 2
t ?
 2.75m / s 2 
t
0m / s  15m / s
t
 15m / s
 5.45s
 2.75m / s 2
Graphing Motion: Dot Diagrams
If we were to set up a camera and take pictures of a jogger running through the park each second of their jog
for 5 seconds and then superimpose those 5 pictures onto on photograph, we could see the motion of the
jogger in relation to their position as time passes. To simplify the picture we could replace the jogger with a
dot. We would still see the same information.
Examples:
1.13)
Constant Velocity
1.14)
Positive Acceleration
(speeding up)
1.15)
Negative Acceleration
(slowing down)
1.16)
All three forms represented
Distance vs Time Graphs:
Instead of the traditional x-y coordinate plane used in mathematics, physics uses a coordinate plane
appropriate for the example using specific variables. In this case the vertical axis is the distance or position of
the object in relation to the origin and the horizontal axis is the measure of time. As time passes, the distance
increases or decreases, based on the slope of the line.
Mathematically slope is described as “rise over run” (vertical
change divided by horizontal change) and in the graph to the
right the rise represents distance and the run represents time.
This means that the “rise over run” is distance over time which
is also known as velocity. This means that the slope of a
distance vs time graph represents the velocity of the object. In
this specific example the slope=5. With respect to physics the
distance traveled was 30 m and the time it took was 6.0 s. Using
both concepts, the velocity was 5 m/s.
Four Types of Slope:
Positive Slope
or
+ velocity
(moving forward)
Negative Slope
or
- velocity
(moving backwards)
Zero Slope
or
0 velocity
(not moving)
Undefined
Slope
or
not possible
Examples:
1.17)
The graph to the left represents an object’s motion for 12 seconds of
time.
a) Describe the object’s motion for each segment.
b) How fast was it moving during the first 4 seconds?
c) How fast was it moving during the last 6 seconds
d) How do you know the object was not moving from 4-6 s?
e) What was the total distance traveled?
f) What was the displacement of the object?
a) The object moves forward 40 meters in 4 seconds before coming to a stop for the next 2 seconds.
Finally it moves backward 20 meters during the last 6 seconds of the event.
b) slope 
c) v 
rise dist ance

 velocity
run
time
 v
d 40m

 10m / s
t
4s
d 20m

 3.33m / s
t
6s
d) There is no or zero slope for these two seconds.
e) 40 m + 0 m + 20 m = 60 m
f) 40 m + 0 m + (-20) m = 20 m
1.18) The graph to the right represents the motion of two
neighbor’s commute home from work on the freeway.
a) Who is traveling the fastest?
b) Who has the farthest distance to travel to get home?
c) What happens at the 45 second mark?
d) What is the approximate velocity of each driver?
a) Neighbor B (steeper slope)
b) Neighbor B (starts 50 meters behind A)
c) Neighbor B passes Neighbor A (190 meters)
d) Neighbor A v 
Neighbor B v =
Velocity vs Time Graphs:
d 190m

 4.22m / s
t
45s
d (190 + 50) m
=
= 5.33m / s
t
45 s
In this case the vertical axis is the velocity of the object and the horizontal axis is the measure of time. As time
passes, the velocity increases or decreases, based on the slope of the line. Since slope is described as “rise over
run” then the rise represents velocity and the run represents time. This means that the “rise over run” is
velocity over time which is also known as acceleration. Therefore, the slope of a velocity vs time graph
represents the acceleration of the object. In this specific example the slope=1.67. With respect to physics the
velocity was 20 m/s and the time it took was 12.0 s. Using both concepts, the acceleration was 1.67 m/s.
Furthermore, since the distance an object travels is defined by the velocity at which it travels, we can find the
distance an object has moved by finding the area under the curve of the graph. In the above graph, the shape
generated by the line is a triangle with a height of 20 and a base of 12. Therefore the area under the curve is
half of 20 m/s multiplied by 12 s which is 12 bh  12 (12s)( 20m / s)  120m
Four Types of Slope:
Positive Slope
or
+ acceleration
(speeding up)
Negative Slope
or
- acceleration
(slowing down)
Zero Slope
or
0 acceleration
(constant velocity)
Undefined
Slope
or
not possible
Examples:
1.19)
The graph to the left represents an object’s motion for 16 s of time.
a) Describe the object’s motion for each segment.
b) Find the average acceleration during the first 4 seconds?
c) Find the average acceleration during the last 8 seconds
d) How far did the car move during the first 4 seconds?
e) What was the total distance traveled?
a) For the first 4 seconds the car speeds up from 0m/s to 20 m/s. It then goes at a constant velocity
for another 4 seconds before slowing to a stop over the last 8 seconds.
b) slope 
rise change in velocity
v 20m / s  0m / s

 acceleration  a 

 5m / s 2
run
time
t
4s
v 0m / s  20m / s

 2.5m / s 2
t
8s
d) area of a triangle= 12 bh  12 (4s)( 20m / s)  40m
c)
a
e) area of first triangle
+ area of rectangle
12 bh  12 (4s)(20m / s)  lw  (4s)(20m / s)
+ area of last triangle

12 bh  12 (8s)(20m / s) 
40m  80m  80m  200m
1.20) The graph to the right represents 12 seconds of travel.
a) Describe what Kyle & Jill do during the 12 seconds.
b) What is the average acceleration for Kyle & Jill?
c) What happens at the 6 second mark?
d) What is the distance traveled by each person?
a) Kyle is moving at a constant velocity of 20 m/s while Jill
started from rest and speeds up to 40 m/s.
v 20m / s  20m / s

 0m / s 2
t
12 s
v 40m / s  0m / s

 3.33m / s 2
Jill= a 
t
12 s
b) Kyle= a 
c) Both Jill and Kyle have the same velocity at this moment.
d) Kyle (rectangle)= lw  (12s )( 20m / s )  240m & Jill (triangle)=
1
2
bh  12 (12s)( 40m / s)  240m
Vectors:
The direction and magnitude of an event in physics can be represented by arrows (vectors). The arrow points
the direction of the event and the length of the arrow shows it magnitude. Vectors can also be put in
combination with each other. This is called vector addition. When combining (adding) vectors together
graphically, we draw them tip-to-tail. This means that from the tip of the first vector we begin drawing the tail
of the next.
For example: a boy walks to his friends house by going 1 block north,
2 blocks east, another block north, west 1 block, and 1 more block
north. In the picture to the right we see in (a) the path taken and his
resulting displacement vector drawn in as a dashed line-d. Using
these same set of component vectors but in a different order (b) we
can still arrive at the friend’s house which will still constitute the
exact same resultant displacement vector.
Examples:
1.21) Using the component vectors to the below find the following:
a) A + B
b) B + A
c) A + C + D
d) A – B
a)
b)
1.22) An airplane traveling at 500 m/s north runs into a
headwind of 50 m/s. What is its velocity with
respect to the ground?
c)
d)
Since a headwind is
where it blows opposite
the direction of motion,
you subtract the vectors.
500m/s-50m/s=450m/s
1.23) The same airplane traveling at 500 m/s experiences a
tailwind of 50 m/s. What is its velocity with respect
to the ground?
Since a tailwind is where
it blows in the same
direction of motion, you
add the vectors.
500m/s+50m/s=550m/s
R 2  A2  B 2
1.24) A hiker leaves his home and heads 5 km north before
turning due east for another 7 km. What is the
magnitude of his displacement?
R 2  5 2  7 2  74
R 2  74
R  8.6km
R 2  A2  B 2
1.25) A boat with a maximum speed of 15 m/s in still water
attempts to cross a river flowing at 3 m/s. What is
the speed of the boat with respect to the shore?
R 2  3 2  15 2  234
R 2  234
R  815.3m / s
Trigonometry for Right Triangles:
The Pythagorean Theorem ( A 2  B 2  C 2 ) allows us to find the length of one side of a right triangle if we
know the other two sides. However, if we are missing two of the sides, it does not allow us to find the missing
sides, nor does it help in finding angles inside the triangle. To do these two things we must utilize the basic trig
functions: Sine (sin), Cosine (cos), and Tangent (tan) which are found on your
calculator. For the triangle to the right, we would like to focus on the angle identified
by  . In consequence each side of the triangle has names in regards to their location
with respect to this angle. If the angle we are searching for changes location, the
names of each of the sides (opposite or adjacent) also changes to represent their
location with respect to the angle. However, the longest (angled) side of the triangle is
always the hypotenuse.
The three trig functions of these angles are congruent to the ratio of two of the sides of the triangle as
identified by the acronym SOH CAH TOA (Sine of the angle = the ratio of the Opposite side divided by the
Hypotenuse side…Cosine of the angle = the ratio of the Adjacent side divided by the Hypotenuse side…Tangent
of the angle = the ratio of the opposite side divided by the Adjacent side) more easily written
sin  
opp
hyp
cos  
adj
hyp
tan  
opp
adj
Examples:
1.25) Find the missing side (x).
In relation to the 35 angle, x is the adjacent side and 12 is the
hypotenuse. This means we need to use the Cosine function to
find the missing side:
cos 35 
1.26) Find the missing side (x).
x
12
 12(cos 35)  x

x  9.83
In relation to the 15 angle, 4 is the opposite side and x is the
adjacent side. Use the Tangent function to find x:
tan 15 
4

x

( x) tan 15
4( x)

(tan 15)
x(tan 15)
4
x
 14.93
tan 15
In relation to the angle(   ), x 3 is the opposite side and 14 is
the hypotenuse. This means we need to use the Sine function
to find the missing angle:
1.27) Find the missing angle (  ).
sin  
3
14

3
sin 1 (sin  )  sin 1      12.4
 14 
1.28) A hiker leaves his home and heads
due west for 5 km before turning
and heading due south for another 3
km. What is the angle the hiker
should follow to return home?
tan  
3
5
3
tan 1 (tan  )  tan 1  
5
  30.96
Motion Equations:
When the two basic equations for determining motion do not provide enough detail for manipulation,
variations of their combinations can be derived as follows: (Note that any time a 0 is written as a subscript—
below the normal level or writing—it means initial or beginning…..example d 0 means initial distance)
Equation
Values Included
Missing Value
v, v0 , a, t
d, d0
d , d 0 , v0 , a, t
v  v0  at
d  d 0  v0 t  at
1
2
2
1
2
d  d 0  (v  v0 )t
d , d 0 , v, v0 , t
v
a
v 2  v02  2a(d  d 0 )
d , d 0 , v, v0 , a
t
Examples:
1.29)
1.30)
A marble is placed at the top of an incline
and released to roll down to the bottom. If
it takes 3.6 s and has a final speed of 1.2
m/s, what acceleration does the marble
experience?
A police car moving at 35 m/s accelerates
at a rate of 1.6m / s 2 for 5.5 s. How far
does the police car travel during this time?
v 0  0m / s
v  v0  at
t  3.6s
1.2m / s  0m / s  a(3.6s )
v  1.2m / s
a?
a
1.2m / s
 .0.33m / s 2
3.6s
v0  35m / s
d  d 0  v0 t  12 at 2
a  1.6m / s 2
t  5.5s
d ?
d  0  (35)(5.5)  12 (1.6)(5.5) 2
d  192.5  24.2  216.7m
1.31)
1.32)
A family on vacation going 30 m/s in their
mini-van has 287 km left of their trip. If
they slowly speed up to 40 m/s over the
course of the remaining trip, how long will
it take them to reach their destination?
An airplane attempting to take off from
rest must reach a speed of 155 m/s. If the
plane experiences an average acceleration
of 15m / s 2 , what is the minimum length
the runway must be in order for the plane
to take off properly?
v0  30m / s
d  d 0  12 (v  v0 )t
v  40m / s
d  287 km
 287,000m
t ?
287000  0  12 (40  30)t
287000  35t
t  8200 s or 2.28hr
v 0  0m / s
v 2  v02  2a (d  d 0 )
v  155m / s
155 2  0 2  2(15)( d  0)
a  15m / s 2
d ?
d
24025  30d
24025
 800.83m
30