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