PHYS 1443 – Section 001
Lecture #2
Tuesday, June 7, 2011
Dr. Jaehoon Yu
•
•
•
Dimensional Analysis
Fundamentals
One Dimensional Motion: Average
Velocity; Acceleration; Motion under
constant acceleration; Free Fall
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
1
Announcements
• Homework registration and submissions
– 18/27 registered but only 1 submitted the answer!
– I will then have to approve your enrollment request
• So please go ahead and take an action as soon as possible
• The roster closes tomorrow, Wednesday!
• Quiz tomorrow at the beginning of the class!
– Problems will be on Appendices A and B!
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
2
Special Problems for Extra Credit
• Derive the quadratic equation for Bx2-Cx+A=0
 5 points
• Derive the kinematic equation v  vxi  2axx  x 
from first principles and the known kinematic
equations  10 points
• You must show your work in detail to obtain full
credit
• Due at the start of the class, Thursday, June 9
2
xf
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
2
f
i
3
Dimension and Dimensional Analysis
• An extremely useful concept in solving physical problems
• Good to write physical laws in mathematical expressions
• No matter what units are used the base quantities are the
same
– Length (distance) is length whether meter or inch is used to
express the size: Usually denoted as [L]
– The same is true for Mass ([M])and Time ([T])
– One can say “Dimension of Length, Mass or Time”
– Dimensions are treated as algebraic quantities: Can perform two
algebraic operations; multiplication or division
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
4
Dimension and Dimensional Analysis
• One can use dimensions only to check the
validity of one’s expression: Dimensional
analysis
– Eg: Speed [v] = [L]/[T]=[L][T-1]
•Distance (L) traveled by a car running at the speed
V in time T
– L = V*T = [L/T]*[T]=[L]
• More general expression of dimensional analysis
is using exponents: eg. [v]=[LnTm] =[L][T-1]
where n = 1 and m = -1
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
5
Examples
• Show that the expression [v] = [at] is dimensionally correct
• Speed: [v] =[L]/[T]
• Acceleration: [a] =[L]/[T]2
• Thus, [at] = (L/T2)xT=LT(-2+1) =LT-1 =[L]/[T]= [v]
•Suppose the acceleration a of a circularly moving particle with
speed v and radius r is proportional to rn and vm. What are n
andm?
a
r
v
a  kr n v m
Dimensionless
constant
Length
Tuesday, June 7, 2011
Speed
m
n L
1 2
LT
  L    LnmT  m
T 
m  2  m  2
n  m  n  2  1  n  1
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
2
v
a  kr 1v 2 
r
6
Some Fundamentals
• Kinematics: Description of Motion without understanding the cause
of the motion
• Dynamics: Description of motion accompanied with understanding
the cause of the motion
• Vector and Scalar quantities:
– Scalar: Physical quantities that require magnitude but no direction
• Speed, length, mass, height, volume, area, energy, heat, etc
– Vector: Physical quantities that require both magnitude and direction
• Velocity, Acceleration, Force, Momentum
• It does not make sense to say “I ran with velocity of 10miles/hour.”
• Objects can be treated as point-like if their sizes are smaller than
the scale in the problem
– Earth can be treated as a point like object (or a particle) in celestial problems
• Simplification of the problem (The first step in setting up to solve a problem…)
– Any other examples?
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
7
Some More Fundamentals
• Motions:Can be described as long as the position is
known at any given time (or position is expressed as a
function of time)
– Translation: Linear motion along a line
– Rotation: Circular or elliptical motion
– Vibration: Oscillation
• Dimensions
– 0 dimension: A point
– 1 dimension: Linear drag of a point, resulting in a line 
Motion in one-dimension is a motion on a line
– 2 dimension: Linear drag of a line resulting in a surface
– 3 dimension: Perpendicular Linear drag of a surface, resulting
in a stereo object
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
8
Displacement, Velocity and Speed
One dimensional displacement is defined as:
x  xf  xi
A vector quantity
Displacement is the difference between initial and final potions of the
motion and is a vector quantity. How is this different than distance?
Unit?
m
xf  xi x Displacement


t Elapsed Time
tf  ti
Unit? m/s
A vector quantity
Displacement per unit time in the period throughout the motion
The average velocity is defined as: vx 
Total Distance Traveled
The average speed is defined as: v 
Total Elapsed Time
Unit?
Tuesday, June 7, 2011
m/s
A scalar quantity
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
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x2
x1
What is the displacement?
How much is the elapsed time?
Tuesday, June 7, 2011
x= x 2- x1
t= t - t 0
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
10
Displacement, Velocity and Speed
One dimensional displacement is defined as:
x  xf  xi
Displacement is the difference between initial and final potions of the
motion and is a vector quantity. How is this different than distance?
Unit?
m
xf  xi x Displacement


t Elapsed Time
tf  ti
Unit? m/s
Displacement per unit time in the period throughout the motion
The average velocity is defined as: vx 
Total Distance Traveled
The average speed is defined as: v 
Total Elapsed Time
Unit?
m/s
Can someone tell me what the difference between speed and velocity is?
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
11
Difference between Speed and Velocity
• Let’s take a simple one dimensional translation
that has many steps:
Let’s call this line as X-axis
Let’s have a
couple of motions
in a total time
interval of 20 sec.
+15m
+10m
-5m
-10m
+5m
-15m
Total Displacement: x  x  xi  xi  xi  0(m)
f
Average Velocity:
vx 
xf  xi x 0
 0(m / s)


tf  ti
t 20
Total Distance Traveled: D 10  15  5  15  10  5  60(m)
Average Speed: v  Total Distance Traveled  60  3(m / s)
Total Time Interval
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
20
12
Example 2.1
The position of a runner as a function of time is plotted as moving along the
x axis of a coordinate system. During a 3.00-s time interval, the runner’s
position changes from x1=50.0m to x2=30.5 m, as shown in the figure. What
was the runner’s average velocity? What was the average speed?
• Displacement:
x  x  xi  x2  x1  30.5  50.0  19.5(m)
• Average Velocity:
f
vx 
xf  xi x2  x1 x 19.5



 6.50(m / s )
t2  t1
t
3.00
tf  ti
• Average Speed:
Total Distance Traveled
v
Total Time Interval
50.0  30.5 19.5


 6.50(m / s)
3.00
3.00
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
13
Instantaneous Velocity and Speed
• Can average quantities tell you the detailed story of the
whole motion? NO!!
• Instantaneous velocity is defined as:
x
vx  lim
– What does this mean?
Δt 0
t
• Displacement in an infinitesimal time interval
• Average velocity over a very, very short amount of time
•Instantaneous speed is the size (magnitude) of the
velocity vector:
*Magnitude of Vectors
vx 
lim
Δt 0
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
x
t
are Expressed in
absolute values
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Instantaneous Velocity
Instantaneous Velocity
Average Velocity
Time
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
15
Position vs Time Plot
Position
It is helpful to understand motions to
draw them on position vs time plots.
x1
x=0
2
1
t=0
1.
2.
3.
Tuesday, June 7, 2011
t1
3
t2
t3
time
Running at a constant velocity (go from x=0 to x=x1 in t1,
Displacement is + x1 in t1 time interval)
Velocity is 0 (go from x1 to x1 no matter how much time changes)
Running at a constant velocity but in the reverse direction as 1. (go
from x1 to x=0 in t3-t2 time interval, Displacement is - x1 in t3-t2 time
interval)
Does this motion physically make sense?
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
16
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
17
Example 2.3
A jet engine moves along a track. Its position as a function of time is given
by the equation x=At2+B where A=2.10m/s2 and B=2.80m.
(a) Determine the displacement of the engine during the interval from t1=3.00s to t2=5.00s.
x1  xt1 3.00  2.10  (3.00)2  2.80  21.7m
Displacement is, therefore:
x2  xt 5.00  2.10  (5.00)2  2.80  55.3m
2
x x2  x1  55.3  21.7  33.6  m
(b) Determine the average velocity
x

vx 
during this time interval.
t
Tuesday, June 7, 2011
33.6
33.6

 16.8  m / s 
5.00  3.00 2.00
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
18
Example 2.3 cont’d
(c) Determine the instantaneous velocity at t=t2=5.00s.
 
d
Ct n  nCt n 1 and
Calculus formula for derivative
dt
The derivative of the engine’s
equation of motion is
d
C   0
dt
x dx d
   At 2  B   2 At
vx  lim
t  0 t
dt dt
The instantaneous v t  5.00s 
2 A 5.00  2.10 10.0  21.0  m / s 

x
velocity at t=5.00s is
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
19
Displacement, Velocity and Speed
Displacement
x  xf  xi
Average velocity
xf  xi x
vx 

tf  ti t
Average speed
Total Distance Traveled
v
Total Time Spent
Instantaneous velocity
vx  lim
x dx

t dt
Instantaneous speed
vx 
lim
x dx

t
dt
Tuesday, June 7, 2011
Δt 0
Δt 0
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
20
Acceleration
Change of velocity in time (what kind of quantity is this?) Vector
•Definition of the average acceleration:
vxf  vxi vx

ax 
t
tf  ti
analogous to
xf  xi x

vx 
tf  ti t
•Definition of the instantaneous acceleration:
ax 
lim
Δt 0
vx dvx d  dx  d 2 x
  

dt
t
dt  dt  dt 2
analogous 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
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
21
Acceleration vs Time Plot
What does this plot tell you?
Yes, you are right!!
The acceleration of this motion
is a constant!!
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
22
Example 2.4
A car accelerates along a straight road from rest to 75km/h in 5.0s.
What is the magnitude of its average acceleration?
vxi  0 m / s
v xf 
75000m
 21 m / s
3600 s
Tuesday, June 7, 2011
v x
21  0 21



 4.2(m / s 2 )
ax 
t
5.0
5.0
t f  ti
vxf  vxi
4.2  3600 

 5.4  10 4 (km / h 2 )
1000
2
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
23
Few Confusing Things on 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 is moving faster as time goes on, (v=v(t) ),
acceleration is positive (a>0).
– Incorrect, since the object might be moving in negative direction initially
• When an object is moving slower as time goes on, (v=v(t) ),
acceleration is negative (a<0)
– Incorrect, since the object might be moving in negative direction initially
• In all cases, velocity is positive, unless the direction of the
movement changes.
– Incorrect, since the object might be moving in negative direction initially
• Is there acceleration if an object moves in a constant speed but
changes direction?
The answer is YES!!
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
24
Displacement, Velocity, Speed & Acceleration
x  xf  xi
Displacement
xf  xi x
vx 

tf  ti t
Total Distance Traveled
v
Total Time Spent
Average velocity
Average speed
Instantaneous velocity
Instantaneous speed
Average acceleration
Instantaneous acceleration
Tuesday, June 7, 2011
vx  lim
x dx

t dt
lim
x dx

t
dt
Δt 0
vx 
Δt 0
vxf  vxi vx

ax 
t
tf  ti
ax 
lim
Δt 0
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
vx dvx d  dx  d 2 x
  

dt dt  dt  dt 2
t
25
One Dimensional Motion
• Let’s focus on the simplest case: acceleration is a constant (a=a0)
• Using the definitions of average acceleration and velocity, we can
derive equations of motion (description of motion, velocity and
position as a function of time)
vxf  vxi
vxf  vxi
a

(If tf=t and ti=0) x
a x  ax 
t
tf  ti
For constant acceleration, average
velocity is a simple numeric average
vx 
xf  xi
tf  ti
(If tf=t and ti=0)
Resulting Equation of
Motion becomes
Tuesday, June 7, 2011
vxf  vxi  axt
vxi  vxf 2vxi  axt
1

 vxi  axt
vx 
2
2
2
xf  xi
vx 
t
xf 
xf  xi  v t
x
1 2
xi  v xt  xi  vxit  axt
2
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
26
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
2
2
1 2
xf  xi  vxit  axt
2

vxi t
Displacement as a function
of velocities and time
Displacement as a function of
time, velocity, and acceleration
vxf 2  vxi 2  2ax xf  xi  Velocity as a function of
Displacement and acceleration
You may use different forms of Kinematic equations, depending on
the information given to you in specific physical problems!!
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
27
How do we solve a problem using a kinematic
formula under constant acceleration?
• Identify what information is given in the problem.
–
–
–
–
Initial and final velocity?
Acceleration?
Distance, initial position or final position?
Time?
• Convert the units of all quantities to SI units to be consistent.
• Identify what the problem wants
• Identify which kinematic formula is most 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 equations for the quantity or quantities wanted.
Tuesday, June 7, 2011
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
28
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
2
Thus the time for air-bag to deploy is
Tuesday, June 7, 2011
2
v
xi
 2ax  xf  xi 

t
vxf  vxi
a
PHYS 1443-001, Spring 2011 Dr.
Jaehoon Yu
0  28m / s


 0.07s
2
390m / s
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