Download Unit 1 – Linear Motion

Physics Unit 1: Units of Measure & Basic Motion Review with Examples
Standard Units of Measure used in Physics:
SI Base Units of Measure
Base Quantity
Base Unit
time
Seconds
distance/length
meters
mass
kilograms
temperature
Kelvin
electric current
Amperes
Abbreviation
s
m
kg
K
A
SI Derived Units of Measure
Base Quantity
Base Unit
Force
Newtons
velocity/speed
meters/second
acceleration
meters/second2
Abbreviation
N
m/s
m/s2
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  5000000000L
8(1000)m
 2.22m / s
3600s
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)
0.00000000592
 5.92  10  10  10  10  10  10  10  10  10
 5.92  10 9
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.
 3.95 1010
1.6)
39,452,087,000
1.7)
13379.2 109  1.34 104 109  1.34 105
1.8)
(3.52  106 )(2.4  107 )  (3.52  2.4)(106107 )
 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
Examples:
d
t
a
v v  v0

t
t
1.9)
A car travels 75 m in 13 s. Find
its average velocity.
d  75m
t  13s
v?
v
1.10)
A ball rolling at an average
speed of 5 m/s rolls for 10 s.
How far did the ball travel?
v  5m / s
t  10s
d ?
v
1.11)
1.12)
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?
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  5m / s
v  32m / s
t  8. 2 s
a?
v0  15m / s
d
d
 5m / s 
t
10s
d  (5m / s )(10s )  50m
v 32(m / s)  5(m / s)

t
8.2 s
27m / s
a
 3.29m / s 2
8.2 s
a
a
v  0m/ s
a  2.75m / s 2
t ?
d 75m

 5.77m / s
t 13s
v
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 for Distance Graphs:
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 right 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
d 40m

 velocity  v  
 10m / s
run
time
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 =
d 190m

 4.22m / s
t
45s
d (190 + 50) m
=
= 5.33m / s
t
45 s
Velocity vs Time Graphs:
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 for Velocity Graphs:
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 changein 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

1
2
+ area of rectangle
bh  (4s)(20m / s)  lw  (4s)(20m / s) 
1
2

+ area of last triangle
1
2
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