PHYS 1443 – Section 001
Lecture #3
Thursday, May 31, 2007
Dr. Jaehoon Yu
•
One Dimensional Motion:
–
–
•
Motion under constant acceleration
Free Fall
Motion in Two Dimensions
–
–
–
Vector Properties and Operations
Motion under constant acceleration
Projectile Motion
Today’s homework is HW #2, due 7pm, Monday, June 4!!
Thursday, May 31, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
1
Acceleration
Change of velocity in time (what kind of quantity is this?)
•Average acceleration:
vxf  vxi vx

ax 
t
tf  ti
analogs to
xf  xi x

vx 
tf  ti t
•Instantaneous acceleration:
vx dvx d  dx  d x
   2

dt
t
dt  dt  dt
2
ax 
lim
Δt 0
analogs to
vx  lim
Δt 0
x dx

t dt
• In calculus terms: The slope (derivative) of the
velocity vector with respect to time or the change
of slopes of position as a function of time
Thursday, May 31, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
2
Acceleration vs Time Plot
Thursday, May 31, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
3
Meanings of Acceleration
• When an object is moving in a constant velocity
(v=v0), there is no acceleration (a=0)
– Is there any acceleration when an object is not
moving?
• When an object speeds up as time goes on,
(v=v(t) ), acceleration has the same sign as v.
• When an object slows down as time goes on,
(v=v(t) ), acceleration has the opposite sign as v.
• Is there acceleration if an object moves in a
constant speed but changes direction? YES!!
Thursday, May 31, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
4
One Dimensional Motion
• Let’s start with the simplest case: acceleration is constant (a=a0)
• Using definitions of average acceleration and velocity, we can
draw equation of motion (description of motion, position wrt time)
vxf  vxi
ax 
tf  ti
ax 
If tf=t and ti=0
vxf  vxi
t
Solve for vxf
vxf  vxi axt
vxi  vxf 2vxi  axt
1
For constant acceleration,

 vxi  axt
vx 
simple numeric average
2
2
2
vx  xf  xi  If tf=t and ti=0
tf  ti
Resulting Equation of
Motion becomes
Thursday, May 31, 2007
xf  xi
vx 
t
Solve for xf
xf  xi  vxt
xf  xi  vxt  xi  vxit 
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
1 2
axt
2
5
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
Displacement as a function of
time, velocity, and acceleration
vxf  vxi  2axxf  xi 
2
2
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!!
Thursday, May 31, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
6
How do we solve a problem using a kinematic
formula for constant acceleration?
• Identify what information is given in the problem.
–
–
–
–
Initial and final velocity?
Acceleration?
Distance?
Time?
• Convert the units of all quantities to SI units to be consistent.
• Identify what the problem wants.
• Identify which kinematic formula is appropriate and easiest to
solve for what the problem wants.
– Frequently multiple formulae can give you the answer for the quantity you
are looking for.  Do not just use any formula but use the one that can
be easiest to solve.
• Solve the equation for the quantity wanted.
Thursday, May 31, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
7
Example 2.11
Suppose you want to design an air-bag system that can protect the driver in a headon collision at a speed 100km/hr (~60miles/hr). Estimate how fast the air-bag must
inflate to effectively protect the driver. Assume the car crumples upon impact over a
distance of about 1m. How does the use of a seat belt help the driver?
How long does it take for the car to come to a full stop?
As long as it takes for it to crumple.
100000m
vxi  100km / h 
 28m / s
The initial speed of the car is
3600 s
v xf  0m / s and xf  xi  1m
We also know that
Using the kinematic formula
The acceleration is
ax 
vxf
2
vxf 2  vxi 2 0   28m / s  390m / s 2


2  xf  xi 
2 1m
Thus the time for air-bag to deploy is
Thursday, May 31, 2007
2
v
xi
 2ax  xf  xi 

2
t
vxf  vxi
a
0  28m / s


 0.07s
2
390m / s
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
8
Free Fall
• Free fall is a motion under the influence of gravitational pull
(gravity) only; Which direction is a freely falling object moving?
Down to the center of the Earth!
– A motion under constant acceleration
– All kinematic formulae we learned can be used to solve for falling
motions.
• Gravitational acceleration is inversely proportional to the
distance between the object and the center of the earth
• The magnitude of the gravitational acceleration is
|g|=9.80m/s2 on the surface of the Earth, most of the time.
• The direction of gravitational acceleration is ALWAYS toward
the center of the earth, which we normally call (-y); where up
and down directions are indicated with the variable “y”
• Thus the correct denotation of gravitational acceleration on
the surface of the earth is g= - 9.80m/s2
Thursday, May 31, 2007
Note the negative sign!!
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
9
Example for Using 1D Kinematic
Equations on a Falling object
Stone was thrown straight upward at t=0 with +20.0m/s
initial velocity on the roof of a 50.0m high building,
What is the acceleration in this motion?
g=-9.80m/s2
(a) Find the time the stone reaches at the maximum height.
What is so special about the maximum height?
V=0
v f vyi  ayt  20.0  9.80t  0.00m / s
Solve for t
t
20.0
 2.04s
9.80
(b) Find the maximum height.
1 2
1
yf  yi  vyit  ayt  50.0  20  2.04   (9.80)  (2.04) 2
2
2
 50.0  20.4  70.4(m)
Thursday, May 31, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
10
Example of a Falling Object cnt’d
(c) Find the time the stone reaches back to its original height.
t  2.04  2  4.08s
(d) Find the velocity of the stone when it reaches its original height.
vyf  vyi  ayt  20.0  (9.80)  4.08  20.0(m / s)
(e) Find the velocity and position of the stone at t=5.00s.
Velocity
Position
vyf  vyi  ayt  20.0  (9.80)  5.00  29.0(m / s )
1
yf  yi  vyit  ayt 2
2
 50.0  20.0  5.00 
Thursday, May 31, 2007
1
 ( 9.80)  (5.00) 2 27.5(m)
2
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
11
Coordinate Systems
• They make it easy to express locations or positions
• Two commonly used systems, depending on convenience, are
– Cartesian (Rectangular) Coordinate System
• Coordinates are expressed in (x,y)
– Polar Coordinate System
• Coordinates are expressed in (r,q)
• Vectors become a lot easier to express and compute
+y
How are Cartesian and
Polar coordinates related?
y1
(x1,y1)=(r,q)
r
x1  r cosq
r   x12  y12 
q
O (0,0)
Thursday, May 31, 2007
x1
+x
y1  r sinq
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
y1
tanq 
x1
12
Example
Cartesian Coordinate of a point in the xy plane are (x,y)= (-3.50,-2.50)m.
Find the equivalent polar coordinates of this point.
r
y

q
qs
(-3.50,-2.50)m
2
 y2 
  3.50 
2
  2.50 
2

 18.5  4.30( m)
x
r
x
q  180  q s
tan q s 
2.50 5

3.50 7
1
5
q s  tan    35.5
7
o
o
o
q  180  q s  180  35.5  216
Thursday, May 31, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
o
13
Vector and Scalar
Vector quantities have both magnitudes (sizes)
and directions Force, gravitational acceleration, 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 magnitudes only
Can be completely specified with a value
and its unit Normally denoted in normal letters, E
Energy, heat,
mass, weight
Both have units!!!
Thursday, May 31, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
14
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
Thursday, May 31, 2007
C
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
C: The same magnitude
but opposite direction:
C=-A:A negative vector
F: The same direction
but different magnitude
15
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
Thursday, May 31, 2007
B 2A
A
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
B=2A
16
Example for 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
Thursday, May 31, 2007
 A  B cosq 2  B sin q 2
r
A  B cos 60
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
Find other
ways to
solve this
problem…
17
Components and Unit Vectors
Coordinate systems are useful in expressing vectors in their components
y
Ay
(-,+)
(+,+)
(Ax,Ay)
A
(+,-)
Ay  A sin q
Ax
x
2

Components
} Magnitude
ur
ur
A
cos
q

  A sin q 

A  Ax  Ay
u
r
A 
q
(-,-)
}
Ax  A cos 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
Thursday, May 31, 2007
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
18
2
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
Thursday, May 31, 2007
ur
D



 25  59   7.0  65(cm)
2
2
2
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
19
Displacement, Velocity, and Acceleration in 2-dim
• Displacement:
r r f  ri
• Average Velocity:
r f  ri
r

v
t f  ti
t
• Instantaneous
Velocity:
• Average
Acceleration
• Instantaneous
Acceleration:
Thursday, May 31, 2007
v  lim
t  0
r
dr

t
dt
How is each of
these quantities
defined in 1-D?
v f  vi
v

a
t
t f  ti
d  dr  d2r
v d v

 2


a  lim
t  0  t
dt  dt  dt
dt
PHYS 1443-001, Summer 2007
Dr. Jaehoon Yu
20