PHYS 1441 – Section 002
Lecture #4
Wednesday, Sept. 15, 2010
Dr. Jaehoon Yu
•
One Dimensional Motion
•
•
•
•
•
•
Wednesday, Sept. 15,
2010
Instantaneous Velocity and Speed
Acceleration
Motion under constant acceleration
One dimensional Kinematic Equations
How do we solve kinematic problems?
Falling motions
PHYS 1441-002, Fall 2010
Dr. Jaehoon Yu
1
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Wednesday, Sept. 15,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
2
Reminder: Special Problems for Extra Credit
• Derive the quadratic equation for yx2-zx+v=0
 5 points
• Derive the kinematic equation v  v0  2a  x  x0 
from first principles and the known kinematic
equations  10 points
• You must show your OWN work in detail to obtain
the full credit
2
2
• Must be in much more detail than in pg. 19 of this lecture note!!!
• Due Monday, Sept. 27
Wednesday, Sept. 15,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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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
The average velocity is defined as: vx 


t Elapsed Time
tf  ti
Unit? m/s
Displacement per unit time in the period throughout the motion
The average speed is defined as: v 
Unit?
Wednesday, Sept. 15,
2010
m/s
Total Distance Traveled
Total Elapsed Time
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
5
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
Wednesday, Sept. 15,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
6
Example Distance Run by a Jogger
How far does a jogger run in 1.5 hours (5400 s) if his
average speed is 2.22 m/s?
Average speed 
Distance
Elapsed time
Distance   Average speed  Elapsed time  
  2.22 m s  5400 s   12000 m
Wednesday, Sept. 15, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
7
Example: The World’s Fastest Jet-Engine Car
Andy Green in the car ThrustSSC set a world record of 341.1 m/s in 1997.
To establish such a record, the driver makes two runs through the course,
one in each direction to nullify wind effects. From the data, determine the
average velocity for each run.
r
1609 m
r x

 339.5m s
v
t
4.740 s
r
What is the speed? v  v  339.5m s
r
1609 m
r x

 342.7 m s
v
t
4.695 s
r
What is the speed? v  v  342.7 m s
Wednesday, Sept. 15,
2010
PHYS
342.7
m Fall
s 2010 Dr. Jaehoon
1441-002,
Yu
8
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 short amount of time
•Instantaneous speed is the size (magnitude) of the
velocity vector:
*Magnitude of Vectors
vx 
lim
Δt 0
Wednesday, Sept. 15,
2010
x
t
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
are Expressed in
absolute values
9
Position vs Time Plot
Position
It is useful to understand motions to
draw them on position vs time plots.
x1
x=0
2
1
t=0
1.
2.
3.
Wednesday, Sept. 15,
2010
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)
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Wednesday, Sept. 15,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Velocity vs Time Plot
Wednesday, Sept. 15,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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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
Instantaneous speed
vx 
Wednesday, Sept. 15,
2010
Δt 0
x
t
lim
Δt 0
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
x
t
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Acceleration
Change of velocity in time (what kind of quantity is this?)
Vector!
•Average acceleration:
vxf  vxi v
x  xi x
ax 


analogs to vx 
t
t
t  ti
tf  ti
f
x
f
Dimension? [LT-2]
Unit? m/s2
•Instantaneous acceleration: Average acceleration
over a very short amount of time.
ax  lim
Δt 0
Wednesday, Sept. 15,
2010
vx
t
analogs to
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
vx  lim
Δt 0
x
t
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Acceleration vs Time Plot
What does this plot tell you?
Yes, you are right!!
The acceleration of this motion
is a constant!!
Wednesday, Sept. 15,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
15
Example 2.3
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
Wednesday, Sept. 15,
2010
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 1441-002, Fall 2010 Dr. Jaehoon
Yu
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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!!
Wednesday, Sept. 15,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
17
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
Wednesday, Sept. 15,
2010
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 1441-002, Fall 2010 Dr. Jaehoon
Yu
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One Dimensional Motion cont’d
Average velocity
Since
 vxi  vxf
x f  xi  v xt  xi  
 2
vxi  vxf
vx 
2
vxf  vxi
ax 
t
Solving for t
vxf  xxi
t
ax
Substituting t in the
 vxf  vxi   vxf  vxi 
x f  xi  
  xi 

above equation,

Resulting in
Wednesday, Sept. 15,
2010
2

ax

vxf2  vxi2
2a x
v  v  2ax  x f  xi 
2
xf

t

2
xi
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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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!!
Wednesday, Sept. 15,
2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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