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```Two-Dimensional Motion and
Vectors
Chapter 3
SCALARS
Column 1
Money
Mass
Speed
Temperature
Volume
Column 2
VECTORS
Displacement
Velocity
Acceleration
Force
Position
Two-Dimensional Motion and Vectors
• Vectors
– are quantities that have magnitude and direction.
– Displacement, velocity, acceleration, force, and momentum
are examples of vectors.
– are shown with boldface symbols.
– use arrows to show direction and size..
• A scalar is a physical quantity that has magnitude but no
direction.
How do we draw vectors?
Arrows, scaled drawings, and directions.
http://www.physicsclassroom.com/class/vectors/U3L1a.cfm
How do we designate vector
direction?
1. The direction of a vector is often expressed
as an angle of rotation of the vector about its
“tail” from east, west, north, or south. 30o N or
W or 60o W of N
2. The direction of a vector is
often expressed as a
counterclockwise angle of
rotation of the vector about its
“tail” from due East.
http://www.physicsclassroom.com/class/vectors/U3L1a.cfm
Two-Dimensional Motion and Vectors
A resultant vector represents the sum of two or more
vectors. It is the result of adding vectors.
Two-Dimensional Motion and Vectors
How do we determine
the resultant
vector when the
individual vectors
are at angles to
each other?
1) Graphically head-to-tail using a scaled diagram, or
2)
Mathematically, using Pythagorean
theorem and a little trigonometry.
Two-Dimensional Motion and Vectors
• Vectors can be added graphically.
Draw the first vector to scale at the
correct angle using a ruler and a
protractor. Place the tail of the
second vector on the tip of the first
vector and then draw it to scale at
the appropriate angle. Continue
doing this until all vectors have
been drawn. The resultant vector
is the final vector (arrow) drawn
from the tail of the first vector
(origin) to the tip of the last vector.
Measure its length and angle.
Chapter 3
Properties of Vectors
• Vectors can be added in any order.
A + B = C and B + A = C
• To subtract a vector, add its opposite.
A + (-B) = D
• Multiplying or dividing vectors by scalars
results in vectors.
• Video 1
• Video 2
Two-Dimensional Motion and Vectors
• To determine the resultant vector.
vectors can be moved parallel to themselves.
vectors can be added in any order.
Two-Dimensional Motion and Vectors
3.2 Vector Operations
1. Determine the resultant magnitude and the
resultant direction.
2. Resolve any vector into components.
and then determine the magnitude of the
resultant using the Pythagorean theorem and
direction of the resultant using SOH CAH TOA.
3.2 Vector Operations
–
To add vectors that are at right angles use the
Pythagorean theorem to determine the resultant
magnitude and inverse tangent to determine the
resultant direction.
Two-Dimensional Motion and Vectors
3.2 Determining Resultant Magnitude and Direction
Eric leaves the base camp and hikes 11 km, north and then
hikes 11 km east. Determine Eric's resulting
displacement.
Two-Dimensional Motion and Vectors
R = 11 km
R = 50 km
Two-Dimensional Motion and Vectors
3.2 Determining Resultant Direction
Chief Soh Cah Toa
sin q
=
Soh
opp
hyp
cos q
=
hyp
Cah
tan q
=
opp
Toa
Two-Dimensional Motion and Vectors
3.2 Determining Resultant Magnitude and Direction
12 m/s E
cos q
8 m/s N
R = 14.4 m/s
=
sin q
=
hyp
q = sin-1
=
=
8 m/s
14.4 m/s
8 m/s
= 56o
14.4 m/s
tan q
12 m/s
hyp
q = cos-1
q
opp
=
opp
=
12 m/s
8 m/s
14.4 m/s
q = tan-1 12 m/s
12 m/s
14.4 m/s
= 56o
8m/s
= 56o
Two-Dimensional Motion and Vectors
Determine the direction of the resultant.
R = 11 km
R = 50 km
q = 27o W of N
R = 53o S of W
Two-Dimensional Motion and Vectors
Practice A p. 89
Two-Dimensional Motion and Vectors
All vectors can be resolved into a horizontal (x) component and a
vertical (y) component.
The components of a vector are the projections of the vector along
the axes of a coordinate system.
Resolving a vector allows you to analyze the motion in each
direction.
Chapter 3
Section 2 Vector Operations
Resolving Vectors into Components
Consider an airplane flying at 95 km/h at an angle of
20o to the ground.
• The hypotenuse (vplane) is the resultant vector that
describes the airplane’s total velocity.
• The adjacent leg represents the x component (vx),
which describes the airplane’s horizontal speed.
•
The opposite leg represents
the y component (vy),
which describes the
airplane’s vertical speed.
Resources
Chapter 3
Section 2 Vector Operations
Resolving Vectors into Components, continued
• The sine and cosine functions can be used to
find the components of a vector.
• The sine and cosine functions are defined in terms
of the lengths of the sides of right triangles.
opposite leg
sine of angle q =
hypotenuse
cosine of angle q =
hypotenuse
Resources
Two-Dimensional Motion and Vectors
3.2 Resolving Vectors into Components
Suppose you kick a ball into the
air with an initial velocity of 14
m/s at a 40o angle. The ball has
an upward (vertical motion) and a
horizontal motion. What are
these velocity components?
vx
v =14 m/s
vy
40o
vx = (14 m/s)(cos 40o) = 11 m/s
v =14 m/s
vy
40o
vx
vy = (14 m/s)(sin 40o) = 9 m/s
Two-Dimensional Motion and Vectors
3.2 Resolving Vectors into Components
Practice B p. 92
Two-Dimensional Motion and Vectors
3.2 Adding Multiple Vectors that are not Perpendicular
1. Resolve each vector into its x- and y- components.
2. Add all x-components to get a net x-component for
the resultant vector.
3. Add all y-components to get a net y-component for
the resultant vector.
4. Draw and label the x- and y- components of the
resultant and sketch the resultant vector.
5.
Use the Pythagorean Theorem to determine the
magnitude of the resultant and tan-1q to determine
the direction.
Section 2 Vector Operations
Chapter 3
Adding Vectors That Are Not Perpendicular
• Suppose that a plane travels first 5 km at an angle
of 35°, then climbs at 10° for 22 km, as shown
below. How can you find the total displacement?
• Because the original displacement vectors do not
form a right triangle, you cannot directly apply the
tangent function or the Pythagorean theorem.
d2
d1
Resources
Chapter 3
Section 2 Vector Operations
Adding Vectors That Are Not Perpendicular,
• You can find the magnitude and the direction of
the resultant by resolving each of the plane’s
displacement vectors into its x and y components.
• Then the components along each axis can be
As shown in the figure, these sums will
be the two perpendicular components
of the resultant, d. The resultant’s
magnitude can then be found by using
the Pythagorean theorem, and its
direction can be found by using the
inverse tangent function.
Resources
Chapter 3
Section 2 Vector Operations
Adding Vectors That Are Not Perpendicular
Resources
Chapter 3
Section 2 Vector Operations
Sample Problem
A hiker walks 27.0 km from her base camp at 35°
south of east. The next day, she walks 41.0 km in
a direction 65° north of east and discovers a forest
ranger’s tower. Find the magnitude and direction
of her resultant displacement
Resources
Chapter 3
Section 2 Vector Operations
Sample Problem, continued
1 . Select a coordinate system. Then sketch and
label each vector.
Given:
d1 = 27.0 km
d2 = 41.0 km
q1 = –35°
q2 = 65°
Tip: q1 is negative, because clockwise
movement from the positive x-axis
is negative by convention.
Unknown:
d=?
q=?
Resources
Chapter 3
Section 2 Vector Operations
Sample Problem, continued
2 . Find the x and y components of all vectors.
Make a separate sketch of the
displacements for each day. Use the cosine
and sine functions to find the components.
For day 1 :
x1 = d1 cosq1 = (27.0 km)(cos –35) = 22 km
y1 = d1 sin q1 = (27.0 km)(sin –35) = –15 km
For day 2 :
x2 = d2 cosq 2 = (41.0 km)(cos 65) = 17 km
y2 = d2 sin q 2 = (41.0 km)(sin 65) = 37 km
Resources
Chapter 3
Section 2 Vector Operations
Sample Problem, continued
3 . Find the x and y components of the total
displacement.
xtot = x1  x2 = 22 km + 17 km = 39 km
ytot = y1  y2 = –15 km + 37 km = 22 km
4 . Use the Pythagorean theorem to find the
magnitude of the resultant vector.
d 2 = (xtot )2  (ytot )2
d = (xtot )2  (ytot )2 = (39 km)2  (22 km)2
d = 45 km
Resources
Chapter 3
Section 2 Vector Operations
Sample Problem, continued
5 . Use a suitable trigonometric function to find
the angle.
 y 
–1  22 km 
q = tan   = tan 


x
39
km
 


q = 29 north of east
–1
Resources
Two-Dimensional Motion and Vectors
3.2 Adding Vectors that are not Perpendicular
A
B
A
By
B
Bx
Rx = Ax + (-Bx)
Ry
Rx
Ry = Ay + By
Ay
Ax
Two-Dimensional Motion and Vectors
3.2 Vector Operations Practice
p. 89
p. 92
p. 94
Two-Dimensional Motion and Vectors
3.3
Projectile Motion
– the curved path that an object follows
when thrown, launched, or otherwise
projected near the surface of the earth
– 2-dimensional motion
– has a vertical (y) component and a
horizontal (x) component
– follows parabolic paths
– vx constant (horizontal component is
assumed constant)
– ay =-g
– is freefall combined with an initial
horizontal velocity
– punted football, thrown ball, water drop
Two-Dimensional Motion and Vectors
3.3 Projectile Motion
– Projectiles can be launched horizontally
or at an angle.
– Projectiles launched horizontally do not
have any initial vertical (y-direction)
velocity. The initial velocity is only in the
horizontal direction (x-direction).
– Projectiles launched at an angle have an
initial vertical velocity (y-direction) and
an initial horizontal velocity (x-direction).
– So, when analyzing projectile motion we
have to look at the vertical and
horizontal motions. These motions are
independent of each other.
– We use our motion equations to analyze
projectiles.
Horizontally Launched Projectiles
Horizontally Launched Projectiles
physicsphenomena.com
Horizontally Launched Projectiles
physicsclassroom.com
Two-Dimensional Motion and Vectors
3.3
Motion Equations Reminder
x
=
v=
t
xf - x i
t
vf = vi + at
vf 2 = vi 2 + 2ax
x = vit + ½at2
Two-Dimensional Motion and Vectors
3.3
Vertical Motion of a Projectile
Launched Horizontally vy,i = 0
vy,f = vy,i + at
vy,f = ayt
vy,f 2 = vy,i 2 + 2ayy
vy,f 2 = 2ayy
y = vy,it + ½ayt2
y = ½ ayt2
Two-Dimensional Motion and Vectors
3.3 Horizontal Motion of a Projectile
Remember vx is constant so
x
vx =
t
vx = vx,i = constant
x = vxt
x
t = v
x
or
.
Two-Dimensional Motion and Vectors
Suppose that a baseball leaves the hand of the pitcher, traveling
horizontally with a velocity of 15 m/s, and suppose that it is 2.0 m
above the ground at that instant. Where will it hit the ground?
v = 15 m/s = vx
2.0 m
y = ½
t =
What do we want to know and what do we know?
v = 15 m/s = vx
x = ?
y = - 2.0 m
tx = ?
vy,i = 0
tx = ty= ?
ay = -9.81 m/s2
x
vx =
t
ayt2
2y
x = vxt
x = (15 m/s) (0.64 s)
x = 9.6 m
ay
(2)(- 2.0 m)
t =
- 9.81
m/s2
= 0.64 s
The ball will hit the ground
9.6 m in front of the pitcher.
Two-Dimensional Motion and Vectors
Solving projectile problems:
1. Use one of the independent motions (horizontal or
vertical) to find the time in flight, then use it to determine
the position of impact.
2. If projectile impacts a vertical surface (wall) before hitting
time in flight.
3. If projectile impacts a horizontal surface (ground) first
4. Be careful to use the correct initial velocity component.
Two-Dimensional Motion and Vectors
A ball is thrown with a velocity of 30 ft/s at an angle of 37o above the
horizontal. It leaves the pitcher’s hand 4.0 ft above the ground and
15 ft from a wall. (g = 32 ft/s2) (1) At what height above the ground
will it hit the wall? (2) Will it still be going up just before it hits, or will
it already be on its way down?
(1) When the ball travels a horizontal distance of
15 ft, how high will it be above the ground?
37o
Know or can calculate: g, vx, vy,i, tx, y for tx
4 ft
15 ft
Horizontal Motion
vx = (30 ft/s)(cos 37o) = 24 ft/s
x
vx =
t
x
15 ft
t =
= 24 ft/s = 0.63 s
vx
Vertical Motion
Vy,i = (30 ft/s)(sin 37o) = 18 ft/s
y = vy,i t + ½ ay t2
y = 18 ft/s(0.63s) + ½ (-32 ft/s2)(0.63s)2
y = 5 ft above starting point
The ball hits the wall 9 ft above the
ground.
Two-Dimensional Motion and Vectors
(2) Will it still be going up just before it hits, or will it already be on its
way down?
Use the final velocity equation
and calculate vy,f
at 0.63 s.
37o
4 ft
vy,f = vy,i + at
vy,f = vy,i + at
vy,f = 18 ft/s + (-32 ft/s)(0.63s)
vy,f = -2.2 ft/s
The negative sign tells us that the
velocity at 0.63s is in the negative
direction which is DOWN!
Two-Dimensional Motion and Vectors
(2) Will it still be going up just before it hits, or will it already be on its
way down? Another way
How long will it take the ball
reach max. height? Look at the
vertical motion equations.
37o
4 ft
vy,i = - at
- vy,i
t =
a
18 ft/s
t =
= 0.56 s
-32 ft/s2
0
vy,f = vy,i + at
The ball will be on its way down
since it is in the air 0.63 s before
hitting the wall and it reaches its
maximum height at 0.56 s.
Two-Dimensional Motion and Vectors
Vertical Motion
vy,f = vy,i + at
(vy,f) 2 = vy,i 2 + 2ayy
y = vy,it + ½ayt2
vy,i = vi.sinq
a = -g = -9.81m/s2 = -32ft/s2
Two-Dimensional Motion and Vectors
Horizontal Motion
vx,i = vi.cosq
vx,i = vi.cosq = x/t
3.4 Relative Motion
• On occasion objects move within a medium
that is moving with respect to an observer.
– An airplane usually encounters a wind - air that is moving with
respect to an observer on the ground below.
– A motorboat in a river is moving amidst a river current - water
that is moving with respect to an observer on dry land.
• In such instances as this, the magnitude of the
velocity of the moving object (whether it be a
plane or a motorboat) with respect to the
observer on land will not be the same as the
3.4 Relative Motion
Frames of Reference
• If you are moving at 80 km/h north and a car
passes you going 90 km/h, to you the faster car
seems to be moving north at 10 km/h.
• Someone standing on the side of the road would
measure the velocity of the faster car as 90 km/h
toward the north.
• This simple example demonstrates that velocity
measurements depend on the frame of reference
of the observer.
3.4 Relative Motion
Frames of Reference (continued)
Consider a stunt dummy dropped from a plane.
(a) When viewed from the plane, the stunt dummy falls straight
down.
(b) When viewed from a stationary position on the ground, the
stunt dummy follows a parabolic projectile path.
3.4 Relative Motion Relative Velocity
• When solving relative velocity problems, write down the
information in the form of velocities with subscripts.
• Using our earlier car example, we have:
• vse = +80 km/h north (se = slower car with respect to Earth)
• vfe = +90 km/h north (fe = fast car with respect to Earth)
unknown = vfs (fs = fast car with respect to slower car)
• Write an equation for vfs in terms of the other velocities. The
subscripts start with f and end with s. The other subscripts start
with the letter that ended the preceding velocity:
• vfs = vfe + ves
• ves = –vse
• Thus, this problem can be solved as follows:
• vfs = vfe + ves = vfe – vse
• vfs = (+90 km/h n) – (+80 km/h n) = +10 km/h n
Chapter 3
Relative Velocity, continued
• A general form of the relative velocity equation is:
• vac = vab + vbc
http://www.physicsclassroom.com/mmedia/vectors/plane.cfm
Two-Dimensional Motion and Vectors
3.4 Relative Motion
p. 105 #2
```
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