Transcript Motion

Relative Motion
Frames of Reference
Object or point from which motion is
determined
 Most common is the
earth
 Motion is a change
in position relative to
a frame of reference

What is motion?

If you are standing in one place, and your friend
walks by you, are you moving relative to your
friend?
 Is your friend moving relative to you?
 Is either of you moving relative to the earth?
Answer:


You are moving relative to your friend, and your
friend is moving relative to you!
You are not moving relative to the earth, but
your friend is. You are both moving relative to
the sun!
What is motion?

If you and your friend are walking down the hall
together at the same speed, in the same direction,
are you moving relative to your friend?
 Is your friend moving relative
to you?
 Are either of you moving
relative to the earth?
Answer:

You are NOT moving relative to your
friend, and your friend is NOT moving
relative to you. You both are moving
relative to the earth.
Uniform motion
Uniform motion – occurs when an object is moving at
a constant speed/ velocity in a straight line.
Constant speed/velocity- means that the object is covering
the same distance per unit of time.
Scalar – any quantity that is represented by a
magnitude and a unit.
Vector – any quantity that is represented by a
Magnitude, a unit and a direction.
Distance and Displacement



Distance (d) is a
scalar measure of the
actual path between
two locations .
It has a magnitude
and a unit.
Ex: 50 m, 2.5 hrs.


Displacement (d) is a
vector measure of the
change in position
measured in a
straight line from a
starting reference
point.
Ex: 5 m [W]
Sign Convention

In physics we will
use a standard set
of signs and
directions.


Up, right, east and
north are positive
directions. ( + )
Down, left, west,
and south are
negative directions.
(-)
Distance – total trip
d
total
d
d
total
= d1 + d2 + d3 + d4
= 2m + 4m + 2m + 4m
total
= 20 m
Displacement – change in position
d
d
total
= +d1 + +d2 + -d3 + -d4
total
= +2m + +4m + -2m + -4m
d
total
=0m
Speed
Speed = Change in distance ÷ Time
Δ d_
V T
Example: A car travels 300km in 6 hours.
What is the speed of the car?

Answer:
 Speed
= distance ÷ time
 Speed = 300km ÷ 6 hours
 Speed = 50km/hr
More practice


1. How far can a plane travel if it flies
800km/hr for 9 hours?
2. How long does it take a ship to go
500 km if it travels at a speed of
50km/hr?
Answer
1.
Δd
V
T
Δd
800 9
800km ▪ 9hrs = 7200km
hr
Answer
2.
Δd
V
T
500
50 T
500km ÷ 50km = 10 hrs
hr
Instantaneous Speed


Instantaneous speed is speed at any
instant in time.
A speedometer measures speed in ‘real
time’ (the instantaneous speed).
Average Speed


Average speed is the average of all
instantaneous speeds; found simply by a
total distance/total time ratio
The average speed of a trip:
t ot aldist ance
averagespeed 
elapsed t ime
Vavg = d1 + d2 + d3 + d4........
t1 + t2 + t3 + t4 ..........
Velocity



Speed in a given direction is velocity
( vector).
What is the velocity of a boat that travels
from St. John’s, west to Longpond
(16 Km ) in 2.5 h ?
Answer
 Velocity
=
 Velocity =
 Velocity =
 Velocity =
displacement ÷ time
16 Km ÷ 2.5 h
6.4 km/h
6.4 km/h west
Change your answer to m/s!
=
6.4 km/h ÷ 3.6 = 1.8 m/s
Km/hr to m/s conversion trick
Km/hr
m/s
Divide by 3.6
m/s
multiply by 3.6
K/hr
Difference Between
Speed and Velocity
Scalar
Quantities
( Number and
unit)
Volume liters
Distance
Voltage
Speed (KM/h)
Vector Quantities
( Number, unit and
direction)
10 Km West
50 Km/hr south
100 newtons right
Velocity (Speed and Direction)
Speed is a
Velocity is a Vector
Scalar Quantity
Quantity
Distance-time graphs

On your paper, graph the following:

D (m)
0
5
10
15
T (sec)
0
7
14
21
Was your graph a straight line?
A
distance-time graph which is a
straight line indicates constant
speed.
 In constant speed, the object
does not speed up or slow down.
The acceleration is zero.
The Steeper the slope the faster the object is moving.
y2
y1
x1
x2
On a distance time graph for uniform motion the slope equals the average speed.
Vavg = Δd
Δt
What is the Vavg for this graph?
8 - 4 = 4 = 2m/s
4–2
2
Displacement Time Graphs

d
Like distance time graphs only displacement
can be either positive or negative, therefore
we need two quadrants.
t
Moving left away
from origin
d
t
d
Moving right toward
origin from left
t
Stopped right
of origin
d
t
Stopped left
of origin
Graphing !
d
B
A
t
C
A … Starts at home (origin) and goes right (+)
slowly
B … Stopped (position remains constant as time
progresses)
C … Turns around and goes in the (-) direction
quickly, passing by home
Explain what is happening for each leg of the trip.
Explain what is happening for each leg of the trip.
What is the velocity for each leg of the trip?
Hint: slope= rise/ run = Δ d
t
= d2 – d1 = Avg. velocity
t 2 – t1
Graph the following on a distancetime graph:

d (m)
0
5
20
45
80
125
t (s)
0
1
2
3
4
5
0 1 2 3 4 5
Was your graph a curve?
A
graph that curves on a
distance-time graph shows that
the object is accelerating ( nonuniform motion).
 Acceleration.
Distance-time graphs

Describe the motion of the object as shown in
the
graph.
From 0-8 sec,
constant speed:
(25 m/sec);
From 8-12 sec,
no motion
(stop);
From 12-16 sec,
acceleration;
From 16-20 sec,
constant speed
Speed-time graphs

Using the distance-time graph from the
last frame, draw a speed time graph. First
fill in the table below:
Average Speed (m/s)
25
____
0
____
37.5
____
Draw on board
Time (sec)
0 to 8
8 to 12
12 to 20
What does your graph look like?
 Constant
speed will be a
horizontal line on a speed time
graph.
 If the speed decreases, the line
will slant down.
 If the speed increases, the line
will slant up.
On a velocity - time graph the area between the
graphed line and the x-axis equals the displacement
Area = l x w = 6 s x 30 m/s = 180 m
This object is slowing down in a positive direction.
It is non-uniform motion. However we can still calculate
the displacement by finding the area of the triangle.
( ½ base x height )
Displacement = ½ base x height
= ½ 25.0 m/s x 25.0 s
= 312.5 m
Note how the units cancel.
Object is moving at a
constant speed for 5.0 s
then it speeds up for
the next 5.0 sec.
Displacement = area
=
= 10.0
=
=
of rectangle + area of a triangle
lxw
+ ½ base x height
s x 5.0m/s + ½ 10.0 m/s x 5.0 s
50 m
+
25 m
75m
Object moving right and speeding up. (+) Object moving right and slowing down. (+)
Object moving left and speeding up. (-)
Object moving left and slowing down (-)
The slope of the line on a velocity time graph equals the
average acceleration.
For uniform motion the graph is horizontal, therefore
the slope is zero and the acceleration is zero.
d
B
C
Graphing w/
Acceleration
t
A
D
A … Start from rest south of home; increase speed gradually
B … Pass home; gradually slow to a stop (still moving north)
C … Turn around; gradually speed back up again heading south
D … Continue heading south; gradually slow to a stop near the
starting point
Tangent
Lines
d
t
On a position vs. time graph:
The slope of a tangent line will give the velocity at
that point in time. ( instantaneous velocity )
SLOPE
VELOCITY
Positive
Negative
Zero
Positive
Negative
Zero
Graphing Tips
The same rules apply in making an acceleration graph from a
velocity graph. Just graph the slopes! Note: a positive constant
slope in blue means a positive constant green segment. The
steeper the blue slope, the farther the green segment is from
the time axis.
v
t
a
t
Area under a velocity graph
d
“forward area”
t
“backward area”
Area above the time axis = forward (positive) displacement.
Area below the time axis = backward (negative) displacement.
Net area (above - below) = net displacement.
Total area (above + below) = total distance traveled.
d
All 3 Graphs
t
v
t
a
t
What do the following speed-time
graphs depict?
Acceleration
 Change

in velocity
Can be change in speed or direction
 Acceleration
∆V

a
t
= ∆V/ ∆T
Acceleration problem

A roller coaster’s
velocity at the top of
a hill is 10m/s. Two
seconds later it
reaches the bottom of
the hill with a velocity
of 26m/s. What is
the acceleration of
the roller coaster?
Answer
 Acceleration
= ∆V/ ∆T
 a = 26m/s – 10m/s
2s
a = 16m/s
2s
a = 8m/s/s or 8m/s2
More acceleration problems



1. A car accelerates at a rate of
20Km/hr/s. How long does it take to
reach a speed of 80 Km/hr?
2. A car travels at 60 Kilometers per hour
around a curve. Is the car accelerating?
3. A car travels in a straight line at 60
Km/hr. Is the car accelerating?
Answers:
1.
∆V
a t
80Km/hr
20Km/hr/
t
4sec = t
2. yes! Because it’s changing direction!
3. no! It’s not changing speed or direction!
Deceleration
 Negative
acceleration
 Example: A car slows from
60Km/hr to 20 km/hr in 4
seconds. What is its acceleration?
Answer:
Acceleration = ∆V/ ∆T
 Acceleration = Vf – Vi
t
 a = 20km/hr – 60km/hr
4s
a = -40Km/hr
4s
a = -10Km/hr/s

Velocity & Acceleration Sign Chart
VELOCITY
A
C
C
E
L
E
R
A
T
I
O
N
+
-
+
-
Moving forward;
Moving backward;
Speeding up
Slowing down
Moving forward;
Moving backward;
Slowing down
Speeding up
Acceleration due to Gravity
Near the surface of the
Earth, all objects
accelerate at the same
rate (ignoring air
resistance).
This acceleration
vector is the
same on the way
up, at the top,
and on the way
down!
a = -g = -9.8 m/s2
9.8 m/s2
Interpretation: Velocity decreases by 9.8 m/s each second,
meaning velocity is becoming less positive or more
negative. Less positive means slowing down while going
up. More negative means speeding up while going down.
Motion Graphs – Position vs. Time
constant, rightward (+) velocity of +10 m/s
a rightward (+), changing velocity - that is, a car that is moving
rightward but speeding up or accelerating
Motion Graphs – Velocity vs. Time
constant, rightward (+) velocity of +10 m/s
a rightward (+), changing velocity - that is, a car that is moving
rightward but speeding up or accelerating
Review: Distance-time graph of
acceleration
Distance-time graph of deceleration
Review:Speed-time graph of
acceleration
Review: Speed-time graph of
deceleration
Review: Distance-time graph of
constant speed