PHYS 1443 – Section 002
Lecture #6
Monday, Sept. 17, 2007
Dr. Jaehoon Yu
•
Motion in Two Dimensions
– Motion under constant acceleration
– Projectile Motion
– Maximum ranges and heights
Today’s homework is homework #4, due 7pm, Monday, Sept. 24!!
Monday, Sept. 17, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
1
Announcements
• E-mail distribution list: 66 of you subscribed to the list
so far
• There was a brief problem with the homework
system last night
– 3 of you noticed, yeah!!
– It is fixed now. So please submit your homework!!
• First term exam next Wednesday, Sept. 26
– Will be in class from 1pm
– Will cover Ch 1 to what we cover next Monday (~Ch4)
– Mixture of multiple choice and numeric essay problems
Monday, Sept. 17, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
2
Kinematic Quantities in 1D and 2D
Quantities
Displacement
Average Velocity
1 Dimension
2 Dimension
r
r 
x  x f  xi
vx 
r
r
r
r r r f  r i
v

t
t f  ti
x f  xi
x

t
t f  ti
Inst. Velocity
x dx
v x  lim

t  0 t
dt
Average Acc.
v x v xf  v xi
ax 

t
t f  ti
Inst. Acc.
Monday, Sept.What
17, 2007is
r r
r f  ri
r
r
r
r d r
v  lim

t 0 t
dt
r
r
r
r
v v f  vi
a

t
t f  ti
r
r
r
2
vx dvx d x r
v dv d r
a x  lim

 2 a  lim

 2
t 0 t
t 0 t
dt dt
dt dt
2
the difference
between
1D2007
and 2D quantities?
PHYS
1443-002, Fall
Dr. Jaehoon Yu
3
2-dim Motion Under Constant Acceleration
• Position vectors in x-y plane:
r
r
r
rf  x f i  y f j
r
r
r
ri  xi i  yi j
• Velocity vectors in x-y plane:
r
r
r
vi  vxi i  v yi j
r
r
r
v f  vxf i  v yf j
Velocity vectors in terms of the acceleration vector
X-comp
v xf  vxi  axt
Y-comp

v yf  v yi  a y t
 

r
r
r
r
r
r
r
v f   vxi  axt  i   v yi  a yt  j  vxi i  v yi j  ax i  a y j t 
r r
 vi  at
Monday, Sept. 17, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
4
2-dim Motion Under Constant Acceleration
• How are the 2D position vectors written in
acceleration vectors?
Position vector
components
Putting them
together in a
vector form
Regrouping
the above
Monday, Sept. 17, 2007
1 2
x f  xi  vxi t  ax t
2
r
r
r
rf  x f i  y f j 
1 2
y f  yi  v yi t  a y t
2
r 
r
1
1

2
2
  xi  vxit  axt  i   yi  v yit  a yt  j
2
2







r 2
r
r
r
r
1 r
 xi i  yi j  vxi i  v yi j t   ax i  a y j  t
2
r r
1r 2
 ri vit  at
2D problems can be
2
interpreted as two 1D
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
problems in x and y
5
Example for 2-D Kinematic Equations
A particle starts at origin when t=0 with an initial velocity v=(20i-15j)m/s.
The particle moves in the xy plane with ax=4.0m/s2. Determine the
components of the velocity vector at any time t.
vxf  vxiaxt  20 4.0t  m / s  v yf  v yi  a y t  15 0t  15  m / s 
Velocity vector
r
r
r
r
r
v  t   vx  t  i v y  t  j   20  4.0t  i 15 j ( m / s )
Compute the velocity and the speed of the particle at t=5.0 s.
r
r
r
r
r
r
r
vt 5  vx,t 5i v y ,t 5 j   20  4.0  5.0  i 15 j  40i  15 j m / s

r
speed  v 
Monday, Sept. 17, 2007
 vx 
2
  vy 
2

 40
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
2

  15  43m / s
2
6
Example for 2-D Kinematic Eq. Cnt’d
Angle of the
Velocity vector
 vy
  tan 
 vx
1

1  15 
1  3 
o

tan

tan


21

 


40
 8 



Determine the x and y components of the particle at t=5.0 s.
1 2
1
x f  vxi t  ax 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
r
r
r
r
r f  x f i  y f j  150i 75 j  m 
Monday, Sept. 17, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
7
Projectile Motion
• A 2-dim motion of an object under
the gravitational acceleration with
the following assumptions
– Free fall acceleration, g, is constant
over the range
of the motion
r
r
• g  9.8 j  m s 2 
– 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 )
Monday, Sept. 17, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
8
Show that a projectile motion is a parabola!!!
x-component
vxi  vi cos 
v yi  vi sin i
y-component
a axi  a y j   gj
ax=0
x f  vxi t  vi cos i t
t
xf
vi cos i
In a projectile motion,
the only acceleration is
gravitational one whose
direction is always
toward the center of the
earth (downward).
1 2
1
2

v
sin

t

gt
y f  v yi t    g  t
i
i
2
2
Plug t into
the above
xf

xf

 1 
y f  vi sin i 
  2 g  v cos  
i 
 i
 vi cos i 

g
y f  x f tan  i  
2
2
2
v
cos
i
 i
Monday, Sept. 17, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
 2
x f


2
What kind of parabola is this?
9
Projectile Motion
Monday, Sept. 17, 2007
The
only acceleration in this
PHYS 1443-002, Fall 2007
Dr. Jaehoon
motion.
It isYua constant!!
10
Example for Projectile Motion
A ball is thrown with an initial velocity v=(20i+40j)m/s. Estimate the time of
flight and the distance the ball is from the original position when landed.
Which component determines the flight time and the distance?
1
y f  40t    g  t 2  0m
2
Flight time is determined
by the y component,
t 80  gt   0
because the ball stops
moving when it is on the
ground after the flight.
So the possible solutions are…
Distance is determined by the x
component in 2-dim, because
the ball is at y=0 position when it
completed it’s flight.
Monday, Sept. 17, 2007
80
 t  0 or t 
 8 sec
g
Why isn’t 0
 t  8sec
the solution?
x f  vxi t  20  8  160  m
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
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