Do now – on a new piece of paper. The diagrams below represent two types motions. One is constant motion, the other, accelerated motion. Which one is constant motion and which one is accelerated motion? explain your answer. Objectives. 1. Distinguish between a scalar and a vector.
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1. Distinguish between a scalar and a vector.
2. Add and subtract vectors using the graphical method.
3. Multiply and divide vectors by scalars.
Homework – vector 1
Pointing the Way
Vectors
Vectors on paper are simply arrows
Direction represented by the way the ARROW POINTS
Magnitude represented by the ARROW LENGTH
Examples of Vectors
Displacement
Velocity
Acceleration
Vector Diagrams
head
tail
Vectors can be moved parallel to themselves in a diagram
Reference Vector
Uses due EAST as the 0 degree reference,
all other angles are measured from that point
20 meters at 190°
34 meters at 48°
Compass Point
20 meters at 10° south of west
34 meters at 42° east of north
90°
N
180°
W
E
0°
S
270°
What is the reference vector angle for a vector that points 50 degrees east of south?
What is the reference vector angle for a vector that points 20 degrees north of east?
50°
20°
270° + 50° = 320°
20°
ADD/SUBTRACTwith a vector
To produce a NEW VECTOR
MULTIPLY/DIVIDEby a vector or a scalar
To produce a NEW VECTOR or SCALAR
The resultant is the vector sum of two or more vectors. It is the result of adding two or more vectors together
+
= ?
A
B
A
B
A
B
C
A
B
C
+
=
+
=
(Resultant)
A
B
A
B
30o
60o
school
The resultant is not from the head to tail, it is from beginning to end.
Adam’s house
resultant
Frank’s house
Magnitude: measure with ruler, determine using scale
Direction: measure with protractor with East
A
B
C
Parallelogram method
R
R
Parallelogram method
Head and tail method
match the graphs with the following motion
position
position
position
position
velocity
velocity
velocity
velocity
Time
Time
Time
Time
Time
Steps for adding vectors using head and tail method
A + B = B + A

+
(
)
=
B
A
B
A

=
+

=
?
Resultant A + B
A
Difference A  B
B
Add vectors mathematically
The procedure is restricted to the addition of two vectors that make right angles to each other.
opp.
opp.
tanθ =
adj.
adj.
(
)
θ = tan1
Hyp.
opp.
θ
adj.
R2 = (5.0km)2 + (10km)2
R = 11 km
Or at 26 degrees south of west
B
A
B
A
R
R
A
Equilibrant
Equilibrant
B
Head to tail
Parallelogram
Vectors can be broken into COMPONENTS
XY system of components
AX = A cos θ
AY = A sin θ
Example
vi = 5.0 m/s at 30°
vix = 5.0 m/s (cos 30°) = 4.33 m/s
viy = 5.0 m/s (sin 30°) = 2.5 m/s
Any vector can be broken into unlimited sets of components
EXAMPLE
Calculate the x and y components of the following vectors.
a. A = 7 meters at 14°
b. B = 15 meters per second at 115°
c. C = 17.5 meters per second2 at 276°
Vectors can be added together by adding their COMPONENTS
Results are used to find
RESULTANT MAGNITUDE
RESULTANT DIRECTION
Adding Vectors Algebraically
Add vectors D and F by following the steps below.
a. Calculate the components of vectors D and F.
D = 35 meters at 25° F = 55 meters at 190°
b. Calculate the sum of the xcomponents of vectors D and F.
c. Calculate the sum of the ycomponents of vectors D and F.
d. Sketch the resultant x and y vectors on the axes below.
e. Calculate the length of the resultant generated by the resultant components
f. Calculate the direction of the resultant generated by the resultant components
A bus heads 6.00 km east, then 3.5 km north, then 1.50 km at 45o south of west. What is the total displacement?
A: 6.0 km, 0° CCWB: 3.5 km, 90° CCWC: 1.5 km, 225° CCW
+
+
A
B
C
Cx = Ccos225o = 1.06 km
Cy = Csin225o =  1.06 km
10/10 do now
A
B
+
= ?
The wrong diagrams
Draw the right diagram for A + B
Checking Your References
Relative Motion
and
2D Kinematics
objectives
Homework – castle learning
Homework quiz – Friday
Lab – Thu. / Fri. – Dress warm
How would Homer know that he is hurtling
through interstellar space if his speed were
constant?
Without a window, he wouldn’t!
All of the Laws of Motion apply within his
FRAME of REFERENCE
Do you feel like you are motionless
right now?
The only way to define motion is by changing position…
The question is changing position relative to WHAT?!?
MORE MOTION!!!
YOU'RE NOT!
You are moving at about 1000 miles per hour relative to the
center of the Earth!
The Earth is hurtling around the Sun at over
66,000 miles per hour!
A train is moving east at 25 meters per second. A man on the train gets up and walks toward the front at 2 meters per second.
How fast is he going?
Depends on what we want to relate his speed to!!!
+2 m/s (relative to a fixed point on the train)
+27 m/s (relative to a fixed point on the Earth)
vperson = +2 m/s
vtrain = +25 m/s
What happens to the aircraft’s forward speed when the wind changes direction?
Wind is now slowing the plane
somewhat AND pushing it
SOUTH.
Wind is now working against
the aircraft thrust, slowing it
down, but causing no drift.
Wind is now NOT having any
effect on forward movement,
but pushes plane SOUTH.
Wind is still giving the plane
extra speed, but is also
pushing it SOUTH.
Wind in same direction
as plane – adds to overall
velocity!
No wind – plane
moves with velocity
that comes from engines
vthrust
vwind
100 km/hr)2 + (25 km/hr)2 = R2
10 000 km2/hr2 + 625 km2/hr2 = R2
10 625 km2/hr2 = R2
SQRT(10 625 km2/hr2) = R
103.1 km/hr = R
tan(θ) = (opposite/adjacent)
tan (θ) = (25/100)
θ = tan1(25/100)
θ = 14.0o – CCW 256o
Suppose that the river was moving with a velocity of 3 m/s, North and the motorboat was moving with a velocity of 4 m/s, East. What would be the resultant velocity of the motorboat (i.e., the velocity relative to an observer on the shore)?
(4.0 m/s)2 + (3.0 m/s)2 = R2
16 m2/s2 + 9 m2/s2 = R2
25 m2/s2 = R2
SQRT (25 m2/s2) = R
5.0 m/s = R
tan(θ) = (opposite/adjacent)
tan(θ) = (3/4)
θ = tan1 (3/4)
θ = 36.9o
5 m/s
time = distance /(ave. speed)
time = (80 m)/(4 m/s) = 20 s
80 m
5 m/s
distance = ave. speed * time
d
distance = = (3 m/s) * (20 s)
distance = 60 m
Critical variable in multi dimensional problems is TIME.
We must consider each dimension SEPARATELY, using TIME as the only crossover VARIABLE.
Now, assume that as the swimmer moves ACROSS the river, a current pushes him DOWNSTREAM at 0.1 meter per second.
200 m
vs = 0.5 m/s
How long will it take the swimmer
to get across?
t =0
vc= 0.1 m/s
The time to cross is unaffected! The
swimmer still arrives on the other bank
in 400 seconds. What IS different?
t = 400 s
The arrival POINT will be
shifted DOWNSTREAM!
A swimmer moving at 0.5 meters per second swims across a 200 meter wide river.
h = ?
35o
2 m
4 people groups
35o
h = ?
2 m
1.68 m
7 m/s
d = ?
4 m/s
Homework – castle learning
Fire Away!!!
Ground Launched Projectiles
An object that is launched into the air with some INITIAL VELOCITY
Can be launched at ANY ANGLE
In FREEFALL after launch (no outside forces except force of gravity)
The path of the projectile is a PARABOLA
What is the horizontal part of
the soccer ball’s initial velocity?
What is the vertical part of
the soccer ball’s initial velocity?
Pythagorean Theorem
vi2 = vix2 + viy2
vix = vi cos θ
viy = vi sin θ
12 m/s
6 m/s
30°
10.4 m/s
Initial vertical velocity = vi sin θ
Acceleration = 9.81 m/s2
Vertical speed will be 0 at the maximum height
Time to top = HALFtotal time in the air
Find time to top using final velocity equation
Vertical Distance – Max height
Use time to top and solve vertical distance equation
Initial vertical velocity = vi cos θ
Acceleration = 0 (if we assume no air resistance)
Horizontal Distance – Range
Use total time and solve horizontal distance equation
t1/2
vfy = 0 (at top)
x y
a = 0 a = 9.81 m/s2
vix = vi cos θ viy = vi sin θ
vfy = 0 (at top)
ttot = 2t1/2
ttot
Effect of air resistance
A projectile is fired with initial horizontal velocity at 10.00 m/s, and vertical velocity at 19.62 m/s. Determine the horizontal and vertical velocity at 1 – 5 seconds after the projectile is fired.
h = ?
35o
2 m
4 people groups
7 m/s
d = ?
4 m/s
d = ½ (vi + vf)t
vf = vi + at
d = vit + ½at2
vf2 = vi2 + 2ad
Vertical
Horizontal
viy = visinθ
vfy = 0 (at top)
vfy =  viy (at same height)
y = ½ (viy + vfy)t
vfy = viy + ayt
y = viyt + ½ayt2
vfy2 = viy2 + 2ayy
vix = vicosθ
x = vix∙t
vi2 = vix2 + viy2
SOH CAH TOA
sinθ = viy / vi
viy = visinθ
cosθ = vix / vi
vix = vicosθ
vi
viy
θ
vix
Special case: horizontally launched projectile:
θ = 0o:viy = visinθ = 0; vix = vicosθ = vi
Due to the symmetrical nature of a projectile's trajectory: viy =  vfy
Solve for t – use vertical information:
vfy = viy + ayt
t = 3.61 s
Solve for x – use horizontal information:
x = vix•t + ½ •ax•t2
x = 63.8 m
Solve for peak height – use vertical information:
vfy2 = viy2 + 2ayypeak
At the top, vfy = 0
ypeak = 15.9 m
t = 1.1 s
x = 12.2 m
ypeak = 1.6 m
Over the Edge
Horizontal Projectiles
A red ball rolls off the edge of a table
What does its path look like as it falls?
Parabolic path
Thus, the projectile travels with a constant horizontal velocity and a downward vertical acceleration.
As the red ball rolls off the edge, a green
ball is dropped from rest from the same
height at the same time
Which one will hit the ground first?
They will hit
at the SAME
TIME!!!
The green ball falls from rest
and has no initial
velocity IN EITHER
DIRECTION!
viy and vix = 0
The red ball has an initial
HORIZONTAL velocity (vix)
But does not have any initial
VERTICAL velocity (viy = 0)
vix
Both balls begin with no VERTICAL VELOCITY
Both fall the same DISTANCE VERTICALLY
Find time of flight by solving in the appropriate dimension
We can find an object’s displacement in
EITHER DIMENSION using TIME
A bullet is fired horizontally from a gun that is 1.7 meters above the ground with a velocity of 55 meters per second.
At the same time that the bullet is fired, the shooter drops an identical bullet from the same height.
Which bullet hits the ground first?
Both hit the ground at the same time
Horizontal motion is constant: velocity is constant.
Vertical: same as drop the ball from rest: velocity is increasing by 9.81 m/s every second
Horizontal motion is constant: x = vix∙t
Vertical: same as free fall with initial zero velocity: y = ½ a∙t2 (a =  9.81 m/s2)
Vertical
Horizontal
θ = 0
d = ½ (vi + vf)t
vf = vi + at
d = vit + ½at2
vf2 = vi2 + 2ad
viy = = 0
y = ½ (viy + vfy)t
vfy = viy + ayt
y = viyt + ½ayt2
vfy2 = viy2 + 2ayy
vix = vicosθ = vi
x = vix∙t
x and y has the same t
An airplane making a supply drop to troops behind enemy lines is flying with a speed of 300 meters per second at an altitude of 300 meters.
How far from the drop zone should the aircraft drop the supplies?
Need time from vertical
dy = viyt + ½ ayt2
300 m = 0 + ½ (9.81 m/s2)t2
t = 7.82 s
Use time in horizontal
dx = vixt + ½ axt2
dx = (300 m/s)(7.82 s) + 0
dx = 2346 m
A stuntman jumps off the edge of a 45 meter tall building to an air mattress that has been placed on the street below at 15 meters from the edge of the building.
What minimum initial velocity does he need in order to make it onto the air mattress?
Need time from vertical
dy = viyt + ½ ayt2
45 m = 0 + ½ (9.81 m/s2)t2
t = 3.03 s
Use time to find v
v = d / t
v = 15 m / 3.03 s
v = 4.95 m /s
A CSI detective investigating an accident scene finds a car that has flown off the edge of a cliff. The car is 79 meters from the edge of the 25 meter high cliff.
What was the car’s initial horizontal velocity as it went off the edge?
Example #4
Need time from vertical
dy = viyt + ½ ayt2
25 m = 0 + ½ (9.81 m/s2)t2
t = 2.26 s
Use time in horizontal
dx = vixt + ½ axt2
79 m = vix (2.26 s) + 0
vix = 34.96 m/s
The path of a stunt car driven horizontally off a cliff is represented in the diagram below. After leaving the cliff, the car falls freely to point A in 0.50 second and to point B in 1.00 second.
PURPOSE
MATERIALS
PRELIMINARY QUESTIONS
3.A pair of computerinterfaced Photogates can be used to accurately measure the time interval for an object to break the beam of one Photogate and then another. If you wanted to know the velocity of the object, what additional information would you need?
OBJECTIVES
Materials
10/10 do now
+
= ?
The wrong diagrams
What is the right diagram ?
7 m/s
d = ?
4 m/s
10. N
10. N