AP Physics
 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.
16
 Suppose
we trace
out a curve
according to
x
= t2−4t
 y = 3t
14
12
YPosition
(m)
10
8
6
4
2
where t > 0.
0
-5
0
5
X-Position (m)
10
 Sketch
Y-Values
10
Y-Position (m)
the
parametric curve
for the following
set of parametric
equations from t =
0 to t = 5 s.
8
6
4
2
0
0
-2
20
X-Position (m)
40
 Given
the following parametric equations,
graph the trajectory an object from t = 0 s to
t = 3s:
x  3t  t 3
y  6t 4  3t 3  2t  12
450
400
350
300
Y-Position (m)
250
200
150
100
50
0
0
-50
5
10
X-Position (m)
15
20
 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, and
g the acceleration due to gravity, in the
appropriate units.
 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 that
x = (v0 cos )t.
 If
one looked at the projectile from behind,
in the plane of its motion, and had no depth
perception, it would look as if the
projectile 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
x
= (v0 cos )t
y
=1/2gt2 + (v0 sin ) t + y0
 Day

3: Application of Parametric Equations
AP 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

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

 Your
task: To determine the magnitude and
direction of the initial velocity of a soccer
ball as it leaves your foot.

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 ^.
A  Axiˆ  Ay ˆj  Az kˆ
A  A uˆ
A  Ax 2  Ay 2  Az 2
 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.
 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.

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.
vavg
r rf  ri
 
t t f  ti
r
dr
v  lim

t  0  t
dt
v
dx ˆ dy ˆ dz ˆ
i
j k
dt
dt
dt
aavg
v

t
v dv
a  lim

t 0 t
dt
dvx ˆ dv y ˆ dvz ˆ
a
i
j
k
dt
dt
dt

x  3t  t 3


y  6t  3t  2t  12
4
3



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.


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.
 http://www.glenbrook.k12.il.us/gbssci/Phys
/mmedia/vectors/mzi.html
A
projectile
launched
horizontally with a
dropped object
and an object
moving with a
constant speed.
 Horizontal



Motion
Constant motion
vo = vox
ax = 0
 Vertical


Motion
Uniform accelerated
motion
ay = - g = - 9.8 m/s2
 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?
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?
 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?

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?
 The
two triangles
are similar;
therefore,
v2
a
r
 For
one complete
revolution,
v 
2 r

 When
the motion
is non-uniform
circular motion,
the speed and the
direction change.
arad
atan
v2

r
d v

dt
 If
girl on the swing
makes one
complete
revolution in 10.0
seconds,


What is her
constant speed?
What is the
centripetal
acceleration?
 Velocity
seen by a
particular observer
is said to be
relative to that
observer.
 Predict
the path
that the boat will
travel.
 http://www.physic
s.mun.ca/~jjerrett
/relative/relative.
html
 Predict
its motion.
 http://www.physic
s.mun.ca/~jjerrett
/relative/relative.
html
 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
1-D
Uni.Accn.
Non-Uni
Accn.
2-or 3-D
Vectors
Applications