PHYS 1443 – Section 003
Lecture #5
Wednesday, Sept. 10, 2003
Dr. Jaehoon Yu
•Motion in Two Dimensions
•Vector Components
•2D Motion under constant acceleration
•Projectile Motion
Today’s homework is homework #3, due noon, next Wednesday!!
Wednesday, Sept. 10, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
1
Announcements
• Your lab-sessions began Monday, Sept. 8. Be sure to attend
the lab classes.
• Quiz #2 next Wednesday, Sept. 17
• Homework: 100% of you have signed up
– Very good!!!
– If you are new to the class, please come see me after the lecture
• e-mail distribution list:26 of you have subscribed so far.
– There will be negative extra credit from this week
• -1 point if not done by 5pm, Friday, Sept. 12
• -3 points if not done by 5pm, Friday, Sept. 19
• -5 points if not done by 5pm, Friday, Sept. 26
– A test message will be sent next Wednesday for verification
purpose
• Please use office hours or send me e-mail for appointments
– 2:30 – 3:30pm, Mondays and Wednesdays
Wednesday, Sept. 10, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
2
Kinetic Equations of Motion on a Straight
Line Under Constant Acceleration
Velocity as a function of time
v t   vxi  axt
xf
1
1
xf  xi  v x t  vxf  vxi t Displacement as a function
of velocity and time
2
2
1 2
xf  xi  vxit  axt
2
vxf  vxi  2axxf  xi 
2
2
Displacement as a function of
time, velocity, and acceleration
Velocity as a function of
Displacement and acceleration
You may use different forms of Kinetic equations, depending on the
information given to you for specific physical problems!!
Wednesday, Sept. 10, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
3
Vector Components
• Coordinate systems are useful in expressing vectors in their
components (Sizes along coordinate axes…)
– Expressing vectors in their components makes vector algebra easier
– Well but then there is something missing….
Something that tells you the direction…
y
Ay
(-,+)
Ax  A cos q
(+,+)
(Ax,Ay)
A
Wednesday, Sept. 10, 2003
(+,-)
Ax
x
2

PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
Components
2
2
2
A
}
} Magnitude
 A cosq    A sin q 
A  Ax  Ay
A 
q
(-,-)
Ay  A sin q
cos
2
2
q  sin 2 q   A
4
Unit Vectors
• Unit vectors are the ones that tells us the
directions of the components
• Dimensionless
• Magnitudes are exactly 1
• Unit vectors are usually expressed in i, j, k or
i, j , k
So the vector A can
be re-written as
Wednesday, Sept. 10, 2003
A  Ax i  Ay j  A cosq i  A sin q j
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
5
Examples of Vector Additions
Find the resultant vector which is the sum of A=(2.0i+2.0j) and B =(2.0i-4.0j)


C  A  B  2.0i  2.0 j  2.0i  4.0 j
OR
C 
4.02   2.02
q  tan 1
Cy
Cx
 tan 1


 16  4.0  20  4.5(m)
 2.0
 27 
4.0
Magnitude
Direction
Find the resultant displacement of three consecutive displacements:
d1=(15i+30j +12k)cm, d2=(23i+14j -5.0k)cm, and d1=(-13i+15j)cm


 2.0  2.0i  2.0  4.0 j  4.0i  2.0 j m


D  d 1  d 2  d 3  15i  30 j  12k  23i  14 j  5.0k  13i  15 j

 15  23 13i  30  14  15 j  12  5.0k  25i  59 j  7.0k (cm)
D 
252  592  7.02
Wednesday, Sept. 10, 2003
 65(cm)
Magnitude
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
6
Displacement, Velocity, and Acceleration in 2-dim
• Displacement:
r  r f  r i
• Average Velocity:
r
r f  ri
v

t
t f  ti
• Instantaneous
Velocity:
• Average
Acceleration
• Instantaneous
Acceleration:
Wednesday, Sept. 10, 2003
v  lim
t 0
r
dr

t
dt
How is each of
these quantities
defined in 1-D?
v
v f  vi
a

t
t f  ti
v d v d  d r  d 2 r
a  lim


 2


t 0 t
dt dt  dt  dt
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
7
Kinetic Quantities in 1d and 2d
Quantities
1 Dimension
2 Dimension
Displacement
x  x f  xi
r  r f  r i
Average Velocity
x x f  xi
vx 

t
t f  ti
v
Inst. Velocity
x dx
v x  lim

t  0 t
dt
Average Acc.
v x v xf  v xi
ax 

t
t f  ti
r d r
v  lim

t 0 t
dt
Inst. Acc.
What
is the
Wednesday, Sept.
10, 2003
a
r
r f  ri

t
t f  ti
v
v f  vi

t
t f  ti
vx dvx d 2 x
v d v d 2 r
ax  lim

 2 a  lim

 2
t 0 t
t 0 t
dt dt
dt dt
difference
between
1D2003
and 2D quantities?
PHYS
1443-003, Fall
Dr. Jaehoon Yu
8
2-dim Motion Under Constant Acceleration
• Position vectors in x-y plane:
r i  xi i  yi j
r f  xf i  yf j
• Velocity vectors in x-y plane:
v i  v xi i  v yi j
v f  v xf i  v yf j
Velocity vectors in terms of acceleration vector
Velocity vector
components
Putting them
together in a
vector form
v xf  v xi  a x t
v yf  v yi  a y t
v f  vxi  a x t i  v yi  a y t  j v f  vi  at
Wednesday, Sept. 10, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
9
2-dim Motion Under Constant Acceleration
• How are the position vectors written in acceleration
vectors?
Position vector
components
Putting them
together in a
vector form
Regrouping
above results in
1 2
x f  xi  vxi t  a x t
2
1 2
y f  yi  v yi t  a y t
2
1
1


2
2
r f   xi  v xi t  a x t i   yi  v yi t  a y t  j
2
2

 






1
r f  xi i  yi j  v xi i  v yi j t  a x i  a y j t 2
2
1 2
r f  ri  v i t  at
2
Wednesday, Sept. 10, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
10
Example in 2-D Kinetic EoM
A particle starts at origin when t=0 with an initial velocity v=(20i15j)m/s. The particle moves in the xy plane with ax=4.0m/s2.
Determine the components of velocity vector at any time, t.
vxf  vxi  a x t  20  4.0t m / s  v yf  v yi  a y t  15m / s 


v(t )  20  4.0t i 15 j m / s
Compute the velocity and speed of the particle at t=5.0 s.




v  20  4.0  5.0i 15 j m / s  40i 15 j m / s
speed  v 
vx 2  v y 2  402  152  43m / s
Wednesday, Sept. 10, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
Magnitude
11
Example in 2-D Kinetic EoM cont’d
 vy
Direction q  tan 
 vx
1

  tan 1   15   tan 1   3   21o
 40 
 8 

Determine the x and y components of the particle at t=5.0 s.
1
1
2
x f  v xi t  a x t  20  5   4  52  150(m)
2
2
y f  v yi t  15  5  75 (m)
Can you write down the position vector at t=5.0s?


r f  x f i  y f j  150i  75 j m
Wednesday, Sept. 10, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
12
Projectile Motion
• A 2-dim motion of an object under
the gravitational acceleration with
the assumptions
– Free fall acceleration, -g, is constant
over the range of the motion
– Air resistance and other effects are
negligible
• A motion under constant
acceleration!!!!  Superposition
of two motions
– Horizontal motion with constant
velocity ( no acceleration )
– Vertical motion under constant
acceleration ( g )
Wednesday, Sept. 10, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
13
Show that a projectile motion is a parabola!!!
x-component
vxi  vi cos q
v yi  vi sin q i
y-component
a  ax i  a y j   g j
ax=0
x f  v xi t  vi cos q i t
t
xf
vi cosq i
In a projectile motion,
the only acceleration is
gravitational one whose
direction is always
toward the center of the
earth (downward).
1
1 2
2
y f  v yi t   g  t  vi sin q i t  gt
2
2
Plug t into
the above
xf

y f  vi sin q i 
 vi cos q i
xf
 1 
  g 
 2  vi cos q i

g
y f  x f tan q i  
2
2
2
v
cos
qi
 i
Wednesday, Sept. 10, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
 2
x f





2
What kind of parabola is this?
14