PHYS 1443 – Section 001
Lecture #5
Monday, February 7, 2011
Dr. Jaehoon Yu
•
Motion in Two Dimensions
•
•
•
•
Two dimensional kinematic quantities
Motion under constant acceleration
Projectile Motion
Maximum ranges and heights
Today’s homework is homework #4, due 10pm, Tuesday, Feb. 15!!
Announcements
• The first term exam is to be on this Wednesday,
Feb. 9
– Time: 1 – 2:20pm
– Will cover CH1 – what we finish today (CH3.8?),
Feb. 7 + Appendices A and B
– Mixture of multiple choices and free response
problems
– 135 practice problems prepared and distributed
today
• No Department colloquium this week!
• Bring your special project front after the class if
you haven’t already
Monday, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
2
Special Project for Extra Credit
• Show that the trajectory of a projectile motion
is a parabola!!
– 20 points
– Due: next Wednesday, Feb. 16
– You MUST show full details of your OWN
computations to obtain any credit
• Beyond what was covered in page 27 of this
lecture note!!
Monday, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
3
Displacement, Velocity, and Acceleration in 2-dim
• Displacement:
• Average Velocity:
• Instantaneous
Velocity:
• Average
Acceleration
• Instantaneous
Acceleration:
Monday, Feb. 7, 2011
r r
r
r  r f  r i
r
r
r
r
r f  ri
r

v
t f  ti
t
r
r
r
r
dr
v  lim

t  0 t
dt
r
r
r
r
v f  vi
v

a
t f  ti
t
How is each of
these quantities
defined in 1-D?
r
r
r
r
2
r
v dv d  d r   d r
a  lim


2


dt
dt
t 0 t
dt
dt


PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
4
2D Displacement
r
ro  initial position
r
r  final position
Displacement: rr  rr  rr
o
Monday, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
5
2D Average Velocity
Average velocity is the
displacement divided by
the elapsed time.
r r
r
r r  ro r
v

t  to
t
Monday, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
6
The instantaneous velocity indicates how fast the car
moves and the direction of motion at each instant of time.
r
r
r
v  lim
t0 t
Monday, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
7
2D Average Acceleration
r
v
r
v
r
vo
Monday, Feb. 7, 2011
r r
r
r v  vo
v
a

t  to
t
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
8
Kinematic Quantities in 1D and 2D
Quantities
Displacement
Average Velocity
Inst. Velocity
1 Dimension
x  x f  xi
x x f  xi
vx 

t
t f  ti
x dx
vx  lim

t 0 t
dt
vxf  vxi
vx
ax 

t
t f  ti
2 Dimension
r
r
r
f  ri
r
r
r
r
r 
r
r
r f  ri
v

t
t f  ti
r
r
r
r
dr
v  lim

t 0 t
dt
r
r
r
r
v
v f  vi
a

Average Acc.
t
t t
r f ri
r
vx dvx
v dv
a

lim

Inst. Acc.
a  lim

x
t0 t
t0 t
dt
dt
What is the difference
between 1D and 2D quantities?
Monday, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
9
Jaehoon Yu
A Motion in 2 Dimension
This is a motion that could be viewed as two motions
combined into one. (superposition…)
Monday, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
10
Motion in horizontal direction (x)
v

vx  vxo  ax t
x
v  v  2ax x
x  vxot  ax t
2
x
2
xo
Monday, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
1
2
xo
 vx t
1
2
2
11
Motion in vertical direction (y)
v y  v yo  a y t
y
1
2
v
yo

 vy t
v  v  2a y y
2
y
2
yo
y  v yot  a y t
1
2
Monday, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
2
12
A Motion in 2 Dimension
Imagine you add the two 1 dimensional motions on the left.
It would make up a one 2 dimensional motion on the right.
Monday, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
13
Kinematic Equations in 2-Dim
x-component
y-component
vx  vxo  ax t
v y  v yo  a y t
x
1
2
v


v
t
xo
x
2
y
2
xo
x  vxot  axt
1
2
Monday, Feb. 7, 2011
v


v
t
yo
y
v  v  2a y y
v  v  2ax x
2
x
y
1
2
2
2
yo
y  v yot  a y t
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
1
2
14
2
Ex. A Moving Spacecraft
In the x direction, the spacecraft in zero-gravity zone has an initial velocity
component of +22 m/s and an acceleration of +24 m/s2. In the y direction, the
analogous quantities are +14 m/s and an acceleration of +12 m/s2. Find (a) x
and vx, (b) y and vy, and (c) the final velocity of the spacecraft at time 7.0 s.
Monday, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
15
How do we solve this problem?
1. Visualize the problem  Draw a picture!
2. Decide which directions are to be called positive (+) and
negative (-).
3. Write down the values that are given for any of the five
kinematic variables associated with each direction.
4. Verify that the information contains values for at least
three of the kinematic variables. Do this for x and y
separately. Select the appropriate equation.
5. When the motion is divided into segments in time,
remember that the final velocity of one time segment is
the initial velocity for the next.
6. Keep in mind that there may be two possible answers to
a kinematics problem.
Monday, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
16
Ex. continued
In the x direction, the spacecraft in a zero gravity zone has an initial velocity
component of +22 m/s and an acceleration of +24 m/s2. In the y direction, the
analogous quantities are +14 m/s and an acceleration of +12 m/s2. Find (a) x
and vx, (b) y and vy, and (c) the final velocity of the spacecraft at time 7.0 s.
x
ax
vx
vox
t
?
+24.0 m/s2
?
+22.0 m/s
7.0 s
y
ay
vy
voy
t
?
+12.0 m/s2
?
+14.0 m/s
7.0 s
Monday, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
17
First, the motion in x-direciton…
x
ax
vx
vox
t
?
+24.0 m/s2
?
+22 m/s
7.0 s
x  vox t  ax t

1
2
2

 24 m s 7.0 s  740 m
 22 m s 7.0 s 
2
1
2
v x  vox  ax t


 22m s  24m s
Monday, Feb. 7, 2011
2
2
7.0 s 190m s
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
18
Now, the motion in y-direction…
y
ay
vy
voy
t
?
+12.0 m/s2
?
+14 m/s
7.0 s
y  voy t  a y t
1
2


2
 12 m s 7.0 s  390 m
 14 m s 7.0 s 
2
1
2
v y  voy  a y t


 14m s  12m s
Monday, Feb. 7, 2011
2
2
7.0 s 98m s
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
19
The final velocity…
v
v y  98m s

vx  190 m s

  98m s
98 190 27
v  190 m s
  tan
1
Yes, you are right! Using
components and unit vectors!!
Monday, Feb. 7, 2011
2
2
o
 210 m s
A vector can be fully described when
the magnitude and the direction are
given. Any other way to describe it?


r
r
r
r
r
v  vx i  v y j  190 i  98 j m s
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
20
If we visualize the motion…
Monday, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
21
2-dim Motion Under Constant Acceleration
• Position vectors in x-y plane:
r
r
r
ri  xi i  yi j
r
r
r
rf  x f i  y f 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, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
22
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, Feb. 7, 2011
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 1443-001, Spring 2011 Dr.
Jaehoon Yu
problems in x and y
23
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
v  vx,t 5 i v y ,t 5 j  20  4.0  5.0 i 15 j  40i  15 j m / s

t 5
r
speed  v 
Monday, Feb. 7, 2011
v   v 
2
2
x


y

 40
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
2

  15  43m / s
2
24
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, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
25
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, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
26
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, Feb. 7, 2011
 2
x f


PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
2
What kind of parabola is this?
27
Projectile Motion
Monday, Feb. 7, 2011
The
only acceleration in this
PHYS 1443-001, Spring 2011 Dr.
Jaehoon
motion.
It isYu a constant!!
28
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, Feb. 7, 2011
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 1443-001, Spring 2011 Dr.
Jaehoon Yu
29
Horizontal Range and Max Height
• Based on what we have learned, one can analyze a
projectile motion in more detail
– Maximum height an object can reach
– Maximum range
What happens at the maximum height?
At the maximum height the object’s vertical
motion stops to turn around!!
v yf  v yi  a y t
vi
h
 vi sin   gt A  0

Solve for tA
Monday, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
vi sin 
t A 
g
30
Horizontal Range and Max Height
Since no acceleration is in x direction, it still flies even if vy=0.
 vi sin i 
R  vxi t  vxi  2t A   2vi cos i 

g


 vi 2 sin 2 i 


Range
R

y f  h  v yi t 
Height
Monday, Feb. 7, 2011
g


1
 vi sin i  1  vi sin  i 
2
  g  t  vi sin i 

  g
g
g
2

 2 

 vi 2 sin 2 i 
yf  h  

2g


PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
31
2
Maximum Range and Height
• What are the conditions that give maximum height and
range of a projectile motion?
 vi 2 sin 2  i
h  
2g

 vi 2 sin 2 i 

R  

g


Monday, Feb. 7, 2011




This formula tells us that
the maximum height can
be achieved when θi=90o!!!
This formula tells us that
the maximum range can
be achieved when 2θi=90o,
i.e., θi=45o!!!
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
32
Example for a Projectile Motion
• A stone was thrown upward from the top of a cliff at an angle of 37o
to horizontal with initial speed of 65.0m/s. If the height of the cliff is
125.0m, how long is it before the stone hits the ground?
vxi  vi cos   65.0  cos 37  51.9m / s
v yi  vi sin i  65.0  sin 37  39.1m / s
1 2
y f  125.0  v yi t  gt
2
Becomes
gt 2  78.2t  250  9.80t 2  78.2t  250  0
t
78.2 
 78.22  4  9.80  (250)
2  9.80
t  2.43s or t  10.4s
t Monday,
10.Feb.
4s7, 2011
Since negative time does not exist.
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
33
Example cont’d
• What is the speed of the stone just before it hits the ground?
v xf  v xi  vi cos   65.0  cos 37  51.9m / s
v yf  v yi  gt  vi sin i  gt  39.1  9.80 10.4  62.8m / s
v  vxf  v yf  51.9   62.8  81.5m / s
2
2
2
2
• What are the maximum height and the maximum range of the stone?
Do these yourselves at home for fun!!!
Monday, Feb. 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
34