PHYS 1441 – Section 004
Lecture #6
Monday, Feb. 9, 2004
Dr. Jaehoon Yu
•
Chapter three: Motion in two dimension
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Monday, Feb. 9, 2004
Vector and Scalar
Properties of vectors
Vector operations
Components and unit vectors
2D Kinematic Equations
Projectile Motion
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
1
Announcements
• There will be a quiz this Wednesday, Feb. 11
– Will cover
• Sections A5 – A9
• Chapter 2
• Everyone is registered to homework system: Good
job!!!
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Monday, Feb. 9, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
2
Kinematic Equations of Motion on a
Straight Line Under Constant Acceleration
vxf t   vxi  axt
Velocity as a function of time
1
1
xf  xi  v x t  vxf  vxi t Displacement as a function
of velocities 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!!
Monday, Feb. 9, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
3
Vector and Scalar
Vector quantities have both magnitude (size)
and direction Force, gravitational pull, momentum
Normally denoted in BOLD letters, F, or a letter with arrow on top F
Their sizes or magnitudes are denoted with normal letters, F, or
absolute values: F or F
Scalar quantities have magnitude only
Can be completely specified with a value
and its unit Normally denoted in normal letters, E
Energy, heat,
mass, weight
Both have units!!!
Monday, Feb. 9, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
4
Properties of Vectors
• Two vectors are the same if their sizes and the directions
are the same, no matter where they are on a coordinate
system.
Which ones are the
same vectors?
y
D
A=B=E=D
F
A
Why aren’t the others?
B
x
E
Monday, Feb. 9, 2004
C
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
C: The same magnitude
but opposite direction:
C=-A:A negative vector
F: The same direction
but different magnitude
5
Vector Operations
•
Addition:
– Triangular Method: One can add vectors by connecting the head of one vector to
the tail of the other (head-to-tail)
– Parallelogram method: Connect the tails of the two vectors and extend
– Addition is commutative: Changing order of operation does not affect the results
A+B=B+A, A+B+C+D+E=E+C+A+B+D
A+B
B
A
•
A
=
B
A+B
OR
A+B
B
A
Subtraction:
– The same as adding a negative vector:A - B = A + (-B)
A
A-B
•
-B
Since subtraction is the equivalent to adding
a negative vector, subtraction is also
commutative!!!
Multiplication by a scalar is
increasing the magnitude A, B=2A
Monday, Feb. 9, 2004
B 2A
A
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
B=2A
6
Example of Vector Addition
A car travels 20.0km due north followed by 35.0km in a direction 60.0o west
of north. Find the magnitude and direction of resultant displacement.
Bsinq N
B 60o Bcosq
r
q
20
A



A2  B 2 cos 2 q  sin 2 q  2 AB cosq

A2  B 2  2 AB cosq

20.02  35.02  2  20.0  35.0 cos 60
 2325  48.2(km)
E
B sin 60
q  tan
1
 tan 1
35.0 sin 60
20.0  35.0 cos 60
30.3
 38.9 to W wrt N
37.5
 tan 1
Monday, Feb. 9, 2004
 A  B cosq 2  B sin q 2
r
A  B cos 60
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
Find other
ways to
solve this
problem…
7
Components and Unit Vectors
Coordinate systems are useful in expressing vectors in their components
y
Ay
(-,+)
Ax  A cos q
(+,+)
(Ax,Ay)
A
(+,-)
Ax
x
2

}
Components
} Magnitude
ur
ur
A
cos
q

  A sin q 

A  Ax  Ay
u
r
A 
q
(-,-)
Ay  A sin q
2
2
u
r 2
A
cos 2 q  sin 2 q

u
r
 A

• Unit vectors are dimensionless vectors whose magnitude are exactly 1
• Unit vectors are usually expressed in i, j, k or i, j , k
• Vectors can be expressed using components and unit vectors
ur
r ur
r
r
r
ur
So the above vector A
A  Ax i  Ay j  A cos q i  A sin q j
can be written as
Monday, Feb. 9, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
2
8
Examples of Vector Operations
Find the resultant vector which is the sum of A=(2.0i+2.0j) and B =(2.0i-4.0j)

 
r
r
r
r
ur ur ur
C  A  B  2.0i  2.0 j  2.0i  4.0 j
r
  2.0  2.0  i
r
  2.0  4.0  j
ur
C   4.02   2.02
 16  4.0  20  4.5(m)

r
r
 4.0i 2.0 j  m 
q
2.0
tan
 tan
 27o
Cx
4.0
1
Cy
1
Find the resultant displacement of three consecutive displacements:
d1=(15i+30j +12k)cm, d2=(23i+14j -5.0k)cm, and d3=(-13i+15j)cm
ur ur ur ur
r
r
r
r
r
r
r
r
D  d 1  d 2  d 3  15i  30 j  12k  23i  14 j  5.0k  13i  15 j
r
r
r
r
r
r
 15  23  13 i   30  14  15  j  12  5.0  k  25i 59 j 7.0k (cm)

Magnitude
Monday, Feb. 9, 2004
ur
D



 25  59   7.0  65(cm)
2
2
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
2
9
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:
Monday, Feb. 9, 2004
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 1441-004, Spring 2004
Dr. Jaehoon Yu
10
2-dim Motion Under Constant Acceleration
• Position vectors in x-y plane:
• Velocity vectors in x-y plane:
Velocity vectors
in terms of
acceleration
vector
r
r
r
r i  xi i  yi j
r
r
r
vi vxi i  v yi j
r
r
r
r f  xf i  yf j
r
r
r
v f  vxf i  v yf j
v xf  vxi  axt v yf  v yi  a y t
r
r
r r r
v f   vxi  axt  i  vyi  a yt j  vi  at


• How are the position vectors written in acceleration vectors?
1
1
x f  xi  vxi t  ax t 2
y f  yi  v yi t  a y t 2
2
2
r 
r
r 
r
r
1
1
2 
2 
r f  x f i  y f j   xi  vxi t  ax t  i   yi  v yi t  a y t  j
2
2





 



r
r
r
r
r 2 ur r
1r 2
1 r
 xi i  yi j  vxi i  vyi j t  ax i  a y j t  ri  v i t  at
2
2
Monday, Feb. 9, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
11
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 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 
r
r
r
r
r
Velocity vector
v(t )  vx  t  i v y  t  j   20  4.0t  i 15 j (m / s)
Compute the velocity and speed of the particle at t=5.0 s.
r
r
r
r
r
r
r
v t 5  vx ,t 5 i v y ,t 5 j   20  4.0  5.0  i 15 j  40i 15 j m / s

r
speed  v 
Monday, Feb. 9, 2004
 vx 
2
  vy 
2

 40
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
2

  15  43m / s
2
12
Example for 2-D Kinematic Eq. Cnt’d
Angle of the
Velocity vector
 vy
q  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. 9, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
13