Upload
cachez
View
33
Download
0
Tags:
Embed Size (px)
DESCRIPTION
Ch. 3 Motion in two or three dimensions. AP Physics. Parametric equations. - PowerPoint PPT Presentation
Citation preview
CH. 3 MOTION IN TWO OR
THREE DIMENSIONS
AP Physics
PARAMETRIC EQUATIONS Think of a curve being traced out over
time, sometimes doubling back on itself or crossing itself. Such a curve cannot be described by a function y=f(x). Instead, we will describe our position along the curve at time t by
x = x(t)y = y(t)
Then x and y are related to each other through their dependence on the parameter t.
FOR EXAMPLE: Suppose we
trace out a curve according to
x = t2−4ty = 3t
where t > 0. -6 -4 -2 0 2 4 60
2
4
6
8
10
12
14
16
X-Position (m)
Y-Posi-tion (m)
YOUR TRY: Sketch the
parametric curve for the following set of parametric equations from t = 0 to t = 5 s.
0 5 10 15 20 25 30 35
-2
0
2
4
6
8
10Y-Values
X-Position (m)
Y-P
osit
ion
(m
)
WARM-UP Given the following parametric
equations, graph the trajectory an object from t = 0 s to t = 3s:
33x t t
4 36 3 2 12y t t t
0 2 4 6 8 10 12 14 16 18 20
-50
0
50
100
150
200
250
300
350
400
450
X-Position (m)
Y-P
osit
ion
(m
)
CONSIDER PROJECTILE MOTION: Suppose a projectile is launched at an
initial speed v0, from a height h0, at an angle with the horizontal. It’s natural to consider the horizontal distance and the height of the projectile separately. Let
t represent time,x represent the horizontal distance
from the launching spot,y represent the height, andg the acceleration due to gravity,
in the appropriate units.
ANALYZING THE HORIZONTAL MOTION: If one was looking at the projectile from
above and had no depth perception, it would look as if the projectile was travelling in a
straight line at a constant speed equal to
v0x = v0 cos . Since the speed is constant, it should be
clear thatx = (v0 cos )t.
ANALYZING VERTICAL MOTION: If one looked at the projectile from
behind, in the plane of its motion, and had no depth perception, it would look as if theprojectile was first going straight up and then falling, with an initial upward speed of v0y = v0 sin but subject to gravity causing an acceleration g.
y =1/2gt2 + (v0 sin ) t + y0
PARAMETRIC PROJECTILE MOTION EQUATIONS:
x = (v0 cos )t
y =1/2gt2 + (v0 sin ) t + y0
SCHEDULE CHANGES-SEPT. 14TH-PERIOD 1 Day 3: Application of Parametric
EquationsAP Physics Practice Problem #1 Due (aka blue
notebook) Day 4: More on Position, Displacement,
Velocity, & Acceleration Vectors Day 5: Lab Report #1 Due; Lab #2;
Projectile Motion Day 6: Uniform Circular Motion & Relative
Motion Day 1: Practice Problem Solving Unit 1 Test: Thursday, Sept. 24th
SCHEDULE CHANGES-SEPT. 14TH-PERIOD 10 Day 3: Application of Parametric Equations
AP Physics Practice Problem #1 Due (aka blue notebook)
Day 4: More on Position, Displacement, Velocity, & Acceleration Vectors; Lab Report #1 Due; Lab #2
Day 5: Projectile Motion Day 6: Uniform Circular Motion & Relative
Motion Day 1: Practice Problem Solving; Forces
Baseline Inventory Unit 1 Test: Wednesday, Sept. 23rd
APPLICATION OF PARAMETRIC EQUATIONS: Your task: To determine the magnitude
and direction of the initial velocity of a soccer ball as it leaves your foot.
UNIT VECTORS Unit vectors are
defined for convenience to be a vector of magnitude 1 in a particular direction. Traditionally the unit vectors for the x, y, and z axes are called i, j and k and are just unit vectors that point in the positive direction of the x, y, and z axes. The unit vectors are often written with small hats ^.
POSITION VECTOR
ˆˆ ˆx y zA A i A j A k
ˆA A u
2 2 2x y zA A A A
POSITION The position of an object is described
by telling three things about it:
1) The location of a reference point, called the origin. This is usually defined in a drawing of the experimental situation which shows the location of the origin.2) The distance of the object from the origin.3) The direction in which you must move from the origin in order to go to the object.
POSITION VECTOR The distance and direction information
is combined into the position vector, drawn here with an arrow above a letter. (The arrow reminds us that there is a direction associated with this algebraic variable.) Vectors are represented pictorially as arrows. The direction of the arrow represent s the direction of the vector. For position, the length of the arrow represents the distance from the origin.
DISPLACEMENT VECTOR When an object moves from one place to another, it is
said to be "displaced." When it moves, the position of the object changes. The displacement vector is defined to be the vector which connects the old position to the new position, as sketched in the figure. Displacement differs from position in that no origin is needed to specify a displacement.
In the figure, the subscript "f" represents the final position, and the subscript "i" represents the initial position. The drawing represents the displacement "Delta R" associated with changing from R(initial) to R(final).
Since the displacement represents the difference between two positions, it is written algebraically as a subtraction.
AVERAGE VELOCITY VECTOR
f iavg
f i
r rrv
t t t
INSTANTANEOUS VELOCITY VECTOR
0limt
r drv
t dt
ˆˆ ˆdx dy dzv i j k
dt dt dt
AVERAGE ACCELERATION VECTOR
avg
va
t
INSTANTANEOUS ACCELERATION VECTOR
0lim
ˆˆ ˆ
t
yx z
v dva
t dtdvdv dv
a i j kdt dt dt
WB-PRACTICE PROBLEM Using the parametric
equations, calculate: (express each answer in unit vector notation)
The displacement vector from t = 1 s to t = 3s,
Average velocity from t = 1 s to t = 3 s,
Instantaneous velocity at t = 1 s and t = 3 s,
Average acceleration from t = 1 s to t = 3 s, and
Instantaneous acceleration at t = 1s and t = 3 s.
33x t t
4 36 3 2 12y t t t
PARALLEL & PERPENDICULAR COMPONENTS OF ACCELERATION
The tangential component of acceleration or the component parallel to the velocity vector changes the magnitude of the velocity, but not its direction.
The perpendicular component of the acceleration or the radial component changes the direction of the velocity vector.
PROJECTILE MOTION http://www.glenbrook.k12.il.us/gbssci/Ph
ys/mmedia/vectors/mzi.html
COMPARE & CONTRAST: A projectile
launched horizontally with a dropped object and an object moving with a constant speed.
PROJECTILE MOTION
Horizontal MotionConstant motionvo = vox
ax = 0
Vertical MotionUniform
accelerated motion
ay = - g = - 9.8 m/s2
PRACTICE PROBLEM #1: If the horizontal
speed is 12 m/s and the height of the table is 0.95 m,How long does it
take to hit the floor?
How far from the edge of the table does it land on the floor?
PRACTICE PROBLEM #2: A projectile is
launched a with speed of 25 m/s at a 70o angle above the horizontal from the top of a building that 50 m tall. How long does it take to land on the ground?
PRACTICE PROBLEM #3: The soccer
player kicks the ball at 30 m/s at an angle of 40o angle below the horizontal. If the ball lands 55 m from the top of the hill, how high is the hill?
PRACTICE PROBLEM #4: The quarterback
releases the football with a speed of 15 m/s at an angle of 35o above the horizontal when the receiver is 10 m away from him. If the receiver catches the ball at the same height as it was released, how much further did he need to run?
UNIFORM CIRCULAR MOTION
The two triangles are similar; therefore,
For one complete revolution,
2va
r
2 rv
MORE ON CIRCULAR MOTION:
When the motion is non-uniform circular motion, the speed and the direction change.
2
tan
rad
va
rd v
adt
PRACTICE PROBLEM If girl on the
swing makes one complete revolution in 10.0 seconds,What is her
constant speed?What is the
centripetal acceleration?
RELATIVE VELOCITY: Velocity seen by
a particular observer is said to be relative to that observer.
RELATIVE VELOCITY-2D:
RIVER BOAT PROBLEM #1: Predict the path
that the boat will travel.
http://www.physics.mun.ca/~jjerrett/relative/relative.html
RIVER BOAT PROBLEM #2: Predict its
motion. http://www.physi
cs.mun.ca/~jjerrett/relative/relative.html
WARM-UP PROBLEM The truck still moves at 40 km/hr west,
but the car turns on to a road going 40 degrees south of east, and travels at 30 km/hr. What is the velocity of the car relative to the truck now?
The relative velocity equation for this situation looks like this:
vCT = vCG + vGT
KINEMATICS GRAPHIC ORGANIZER
1-D
Uni.Accn. Non-Uni Accn.
Vectors Applications
2-or 3-D