LINEAR MOTION
FAST
Speed
•depends on distance and time
Average speed
•uses total distance and total time
dtot
v av =
t tot
•Use this when an object travels at different speeds
SCALARS VS. VECTORS
•Scalars: measure the amount (magnitude)
•ex: distance traveled, temperature, speed
limits
•Vectors: measure both amount and direction
•vector: scalar with a direction
•ex: weight, velocity of a car
•In which direction is the 10kg weight directed?
Vector or Scalar?
Categorize each quantity as being either a vector or a
scalar.
Category
a.
b.
c.
d.
e.
f.
Scalar
___________________
Vector
___________________
Vector
___________________
___________________
Scalar
___________________
Scalar
___________________
Scalar
Quantity
5m
30 m/s East
10 mi. North
20 degrees Celsius
256 bytes
4000 Calories
DISTANCE VS. DISPLACEMENT
•distance: total miles traveled (scalar)
•displacement: change in position (vector)
•distance from start to end
2a. What is the displacement and distance of
runners when they finish a one-mile race on
an oval track?
1 mi
Displacement: 0 mi
Distance:
DISTANCE VS. DISPLACEMENT
•distance: total miles traveled (scalar)
•displacement: change in position (vector)
•distance from start to end
2b. What is your displacement and distance if
you walk 3m north and then 5m south?
3m
3m + 5m = 8m
Displacement: 3m - 5m = 2m south
Distance:
5m
SPEED VS. VELOCITY
•Speed ( v ) = distance / time
•Velocityv( ) = displacement / time
arrows mean they are vectors
d
v av =
t
d = distance
d = displacement
FRAME OF REFERENCE
•Speed is Relative
•F.o.R. is something to compare speed to
•How fast are you moving now?
•Earth is rotating at 1,000 mph
FRAME OF REFERENCE
•Earth is orbiting sun at 66,000 mph
•Everything in universe is moving
FRAME OF REFERENCE
•So, if you drive 55 mph, you are going
55 mph relative to the earth
•But the earth is rotating at 1000 mph!
•So, Relative to outer space, you are
moving 1055 mph!
55 mph
1000 mph
1055 mph
55 mph
1000 mph
945 mph
FRAME OF REFERENCE
•Is this car moving?
•Speed Limit of the Universe: light speed!
(3.0 x 108 m/s)
Time to Practice
Go to pg. 248
GRAPHING RULES
Distance
(m)
1. Use a ruler
(straightedge)!
2. Label your axes!
•
(units in parentheses)
•
time is always the x-axis
Time (s)
Variable
Unit
GRAPHING RULES
Distance vs.
Time
Distance
(m)
3.Title the graph!
(Y vs. X)
Time (s)
GRAPHING RULES
4.SCALE.
Stretch out your
axes!
GRAPHING RULES
5. Use a Pencil!!
6. Do not just connect the dots!
Line of best fit
curve: smooth
line: ruler
The line might not touch dots
GRAPHING RULES
Drawing tangent lines
Distance
(m)
Distance vs.
Time
drawn at a point
“balance” ruler on curve
 perpendicular with normal
Ahh. Just right!
make it long enough
to find the slope
Time (s)
GRAPHING: START WORK!
PG 254
rise
Dy
slope =
=
=
run
Dx
y2 - y1
x2 - x1
velocity = slope =
y2 - y1
x2 - x1
Distance (m)
Distance vs.
Time
(x2, y2)
(0.15, y1)
Time (s)
0.15 s
MOVIES
And now for a short movie
EUREKA: INERTIA
EUREKA: MASS
EUREKA: SPEED
ACCELERATION
•acceleration  the rate of change of velocity
v f - vi
aav =
t
Velocity
Speed
Direction
arrows mean …
• v f = final velocity
• v i = initial velocity
•refers to speeding up and slowing down or…
EXAMPLE
A car moving at 20 m/s comes to a stop in four
seconds. What was the car’s acceleration?
Given:
v f = 0 m /s
v i = 20 m /s
t = 4.0 s
Unknown:
aav
EXAMPLE
solve for acceleration
v f - vi
aav =
t
0 - 20 m /s
m /s
m
aav =
= -5
= -5 2
4.0 s
s
s
said “negative five meters per second per second”
negative acceleration means… slowing down
ACCELERATION
•You “feel” speed when you accelerate
•This includes speeding up, slowing down and
• sharp turns at constant speed!
•All three are accelerations
EUREKA: ACCELERATION I
DISTANCE VS. TIME GRAPHS
• AKA velocity!
Distance (m)
m
•units are =
s
Time (s)
Distance (m)
rise distance
•Slope =
=
run
time
Constant speed
Time (s)
Increasing Speed
SPEED VS. TIME GRAPHS
rise speed
•Slope =
=
run time
m
/
s
m
•units are =
= 2
s
s
• AKA acceleration!
m
rise -20 m /s
= -5 2
=
slope =
s
4s
run
(same answer as example)
Speed vs. Time Graphs
•So, to summarize the
graphs:
•For distance vs. time:
•slope = speed
•For speed vs. time:
•slope = acceleration
•area between line and x-
axis = distance covered
EUREKA: ACCELERATION II
FREEFALL
•freefall  objects moving
under only force of gravity
• a due to gravity = g
•g = 9.8 m/s2
•Terminal velocity is the
fastest an object can fall
•terminal velocity  when air resistance
becomes equal to gravity
FREEFALL
•let’s look at the motion of three objects
•An object dropped from rest
•An object thrown downwards
•An object thrown upwards
All have the same acceleration!
•All of these motions are types of… freefall!
EUREKA: GRAVITY
LAB: Acceleration due to
Gravity pg. 330A-D
1. Make sure the motion detector
only “sees” the ball
Not your arms
Not a table or the wire basket
2. Start the motion detector after
you hear beeping for 30 s
3. Make sure your graph has a
smooth curve
LAB: Acceleration due to
Gravity pg. 330A-D
Safety
Don’t stand on things with
wheels
Cover the motion detector with
Important!
a wireoffbasket
• Turn
your motion detector
when you are done gathering
data!
• Use the data table on the
handheld computer to see the
data you need to copy
Don’t forget to….
Lab Questions
Displacement, Velocity & acceleration
graphs:
http://www.youtube.com/watch?v=_E
S1JJ7ErzI
Slow Motion Ball:
http://www.youtube.com/watch?v=1Py
jLXIYMzI&feature=related
PUTTING IT TOGETHER
•Let’s use what we know about graphs to
make two more formulas.
•Let’s look at the graph from ti to tf
ti
tf
PUTTING IT TOGETHER
vi
vf
•Each time matches up with a velocity
•Initial velocity is vi
•final velocity is vf
ti
tf
PUTTING IT TOGETHER
vi
vf
•To find distance:
• area between the line and the x-axis
•d = area of rectangle + area of triangle
ti
tf
PUTTING IT TOGETHER
•d = area of rectangle + area of triangle
•area of rectangle = v i ´ t
•area of triangle = 12 [(v f - v i ) ´ t ]
v f - vi
a=
Þ v f - v i = at
t
]
1
2
vi
1
Þ d = v i t + [( at ) t ]
2
vf
[
1
d = v i t + (v f - v i ) t
2
1 2
Þ d = v i t + at
2
[( v
f
- vi ) ´ t
vi ´ t
ti
tf
]
PUTTING IT TOGETHER
1 2
d = v i t + at
2
we now have a connection between a and d
PUTTING IT TOGETHER
•solve for t from first a equation
•substitute into second a equation
•a little fancy algebra and…
v f - vi
v f - vi
a=
Þt=
t
a
2
æ v f - vi ö 1 æ v f - vi ö
d = v iç
÷ + aç
÷
è a ø 2 è a ø
v = v + 2ad
2
f
2
i
nice if you do not have
t
PUTTING IT TOGETHER
•use equation 1 only if acceleration is zero
•use equations 2-4 only if constant acceleration
PUTTING IT TOGETHER
•notice there are no arrows
•However, the variable are ALL still vectors
PUTTING IT TOGETHER
æ mö
ç ÷
s è s2 ø
(s) (m)( s ) ( s ) ( )
m
m
m
•vectors mean that direction is important
•ex. positive represents up, negative for down
EXAMPLE
A spear is thrown down at 15 m/s from the top
of a bridge at a fish swimming along the surface
below. If the bridge is 55 m above the water,
how long does the fish have before it gets stuck?
start
v i = -15m/s
d = -55m
a = -9.8m/s2
end
EXAMPLE
A spear is thrown down at 15 m/s from the top
of a bridge at a fish swimming along the surface
below. If the bridge is 55 m above the water,
how long does the fish have before it gets stuck?
Given:
v i = -15 m /s
d = -55 m
a = -9.8 m /s
Unknown:
t
why negative?
2
EXAMPLE
Which equation has vi, d, a and t?
•Eqn 3 works, but…you would need quadratic (bleh!)
1 2
d = v i t + at
2
•Eqn 2 would work if we had vf. Eqn 4 can get us vf!
v f - vi
a=
t
v = v + 2ad
2
f
2
i
EXAMPLE
First, eqn. 4
v = v + 2ad
2
f
2
i
Þ v f = v + 2ad
2
i
v f = (-15 m /s) + 2(-9.8 m /s )(-55 m)
2
v f = ± 36.1 m /s
v f = -36.1 m /s
2
positive or negative?
EXAMPLE
•solve for t in eqn 2.
•substitute vf into eqn 2.
v f - vi
a=
t
v f - vi
t=
a
(-36 m /s) - (-15 m /s)
t=
= 2.2 s
2
(-9.8 m /s )
POPPER LABETTE!
Pg 336-337
* It is easier to avoid using quadratic
equations in your calculations…