PHYS 1441 – Section 001
Lecture #4
Monday, June 2, 2008
Dr. Jaehoon Yu
•
•
One Dimensional Motion:
– Free Fall
Two dimensional Motion
– Motion under constant acceleration
– Vector recap
– Projectile Motion
– Maximum ranges and heights
Today’s homework is homework #3, due 9pm, Friday, June 6!!
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
1
Announcements
• E-mail Distribution list
– 40 out of 55 registered as of this morning!!
– 3 points extra credit if done by next Monday, June 2
• First term exam
– 8 – 10am, this Wednesday, June 4, in SH103
– Covers CH1 – what we finish next Tomorrow +
appendices
• No after-class office hour today
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
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 work in detail to obtain the
full credit
• Due tomorrow, Tuesday, June 3
2
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
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3
Kinematic Equations of Motion on a
Straight Line Under Constant Acceleration
v  t   v0  at
Velocity as a function of time
1
1
x  x0  vt   v  v0  t
2
2
1 2
x  x0  v0t  at
2
v2  v02  2a  x  x0 
Displacement as a function
of velocities and time
Displacement as a function of
time, velocity, and acceleration
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!!
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
4
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); when the
vertical directions are indicated with the variable “y”
• Thus the correct denotation of the gravitational acceleration
2
2
g


32.2ft
s
on the surface of the earth is g= - 9.80m/s
Monday, June 2, 2008
Note the negative sign!!
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
5
1D Motion Under Gravitational Acceleration
v  t   v0  gt
Velocity as a function of time
1
1
y  y0  vt   v  v0  t
2
2
1 2
y  y0  v0t  gt
2
v  v  2g  y  y0 
2
2
0
Displacement as a function
of velocities and time
Displacement as a function of
time, velocity, and acceleration
Velocity as a function of
Displacement and acceleration
All you need to do is to replace a with g=-9.8m/s2.
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
6
Free Falling Bodies
Down to the center of the Earth
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
r
g 
9.80 m s
7
2
Example 10
A stone is dropped from the top of a tall building. After 3.00s of
free fall, what is the displacement y of the stone?
y
a
?
-9.8 m/s2
Monday, June 2, 2008
v
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
vo
t
0 m/s
3.00 s
8
y
a
v
?
-9.80 m/s2
vo
t
0 m/s
3.00 s
Which formula would you like to use?
y  vot  at  vot  gt
1
2
2
  0m s  3.00 s  
1
2
1
2
2
 9.80m s  3.00 s 
2
 44.1 m
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
9
2
Example 12 How high does it go?
The referee tosses the coin up with
an initial speed of 5.00m/s. In the
absence if air resistance, how high
does the coin go above its point of
release?
y
?
Monday, June 2, 2008
a
v
vo
-9.8 m/s2
0 m/s
+ 5.00 m/s
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
t
10
y
a
v
vo
?
-9.80 m/s2
0 m/s
+5.00 m/s
v  v  2ay
2
2
o
v v
y
2a
2
Solve for y
t
2
o
 0 m s    5.00 m s   1.28 m
v v
y

2
2  9.80 m s 
2a
2
Monday, June 2, 2008
2
o
2
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
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Conceptual Ex. 14 Acceleration vs Velocity
• There are three parts to the motion of the coin.
– On the way up, the coin has a vector velocity that is
directed upward and has decreasing magnitude.
– At the top of its path, the coin momentarily has zero
velocity.
– On the way down, the coin has downward-pointing velocity
with an increasing magnitude.
• In the absence of air resistance, does the acceleration
of the coin, like the velocity, change from one part to
another?
Nope!
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
12
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)
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
13
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 
Monday, June 2, 2008
1
 ( 9.80)  (5.00) 2 27.5(m)
2
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
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Conceptual Example 15 Taking Advantage of Symmetry
Does the pellet in part b strike the ground beneath the cliff with
a smaller, greater, or the same speed as the pellet in part a?
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
The same!!
15
2 Dimensional Kinematic Quantities
r
ro  initial position
r
r  final position
r r r
r  r  ro  displacement
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
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2D Average Velocity
Average velocity is the
displacement divided by
the elapsed time.
r r
r
r r  ro r
v  t t 
t
o
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
17
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, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
18
2D Instantaneous Velocity
r
r
r
v  lim
t 0 t
Slope of the trajectory
as a function of time!!
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
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2D Average Acceleration
r
v
r
v
r
vo
Monday, June 2, 2008
r r
r
v
r v  vo

a
t  to
t
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
20
Displacement, Velocity, and Acceleration in 2-dim
• Displacement:
• Average Velocity:
• Instantaneous
Velocity:
• Average
Acceleration
• Instantaneous
Acceleration:
Monday, June 2, 2008
r r
r
r  r f  r i
r
r
r
rf
r

v
tf
t
r
r
r
v  lim
t  0 t
r
r
r
vf
v

a
tf
t
r
 ri
 ti
How is each of
these quantities
defined in 1-D?
r
 vi
 ti
r
r
v
a  lim
t  0 t
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
21
Kinematic Quantities in 1D and 2D
Quantities
Displacement
Average Velocity
1 Dimension
x  x f  xi
vx 
x f  xi
x

t
t f  ti
Inst. Velocity
x
vx  lim
t  0  t
Average Acc.
v x v xf  v xi
ax 

t
t f  ti
Inst. Acc.
Monday, June 2,What
2008
v x
ax  lim
t  0  t
2 Dimension
r
r 
r
r
r
r r r f  r i
v

t
t f  ti
r
r
r
v  lim
t 0 t
r
r
r
r
v v f  vi
a

t
t f  ti
r
r
v
a  lim
t 0 t
PHYS 1441-001,
Summer
is the difference
between
1D 2008
and 2D quantities?
Dr. Jaehoon Yu
r r
r f  ri
22
Kinematic Equations
v  vo  at
x
1
2
 vo  v  t
v  v  2ax
2
2
o
x  vot  at
1
2
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
2
23
A Motion in 2 Dimension
This is a motion that could be viewed as two motions
combined into one.
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
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Motion in horizontal direction (x)
 vxo  vx  t
vx  vxo  axt
x
v  v  2a x x
x  vxot  ax t
2
x
2
xo
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
1
2
1
2
2
25
Motion in vertical direction (y)
v y  v yo  a y t
y
1
2
v
yo
 vy  t
v  v  2ay y
2
y
2
yo
y  vyot  ayt
1
2
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
2
26
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, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
27
Kinematic Equations in 2-Dim
x-component
y-component
vx  vxo  axt
v y  v yo  a y t
x
1
2
 vxo  vx  t
2
y
2
xo
x  vxot  ax t
1
2
Monday, June 2, 2008
v
yo
 vy  t
v  v  2ay y
v  v  2a x x
2
x
y
1
2
2
2
yo
y  vyot  ayt
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
1
2
2
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Ex.1 A Moving Spacecraft
In the x direction, the spacecraft 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, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
29
How do we solve this problem?
1. Visualize the problem  Make a drawing.
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, remember
that the final velocity of one segment is the initial velocity
for the next.
6. Keep in mind that there may be two possible answers to
a kinematics problem.
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
30
Ex.1 continued
In the x direction, the spacecraft 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, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
31
First, the motion in x-direciton…
x
ax
vx
vox
t
?
+24.0 m/s2
?
+22 m/s
7.0 s
2
1
v
t

a
t
x  ox 2 x
  22m s  7.0 s  
1
2
 24m s  7.0 s 
2
vx  vox  axt
  22 m s    24 m s
Monday, June 2, 2008
2
2
 740 m
  7.0 s   190 m s
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
32
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
 14m s  7.0 s  
1
2
12m s  7.0 s 
2
v y  voy  a y t
 14 m s   12 m s
Monday, June 2, 2008
2
2
 390 m
  7.0 s   98 m s
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
33
The final velocity…
v
v y  98 m s

vx  190 m s
190 m s    98 m s   210 m s
A vector can be fully described when
1
  tan  98 190   27 the magnitude and the direction are
v
2
2
given. Any other way to describe it?
Yes, you are right! Using
components and unit vectors!!
Monday, June 2, 2008
v  vx i  vy j  190i  98 j  m s
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
34
If we visualize the motion…
Monday, June 2, 2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
35