PHYS 1443 – Section 003
Lecture #4
Wednesday, Sept. 1, 2004
Venkat Kaushik
1. One Dimensional Motion
Acceleration
Motion under constant acceleration
Free Fall
2. Motion in Two Dimensions
Vector Properties and Operations
Motion under constant acceleration
Projectile Motion
Wednesday, Sept. 1, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
1
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 
x dx

t
dt
Wednesday, Sept. 1, 2004
Δt0
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
lim
Δt0
2
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 dt  dt  dt
t
2
ax 
lim
Δt 0
analogs to
vx  lim
Δt0
x dx

t dt
• In calculus terms: A slope (derivative) of velocity
with respect to time or change of slopes of
position as a function of time
Wednesday, Sept. 1, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
3
Acceleration vs Time Plot
Wednesday, Sept. 1, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
4
Example 2.4
A car accelerates along a straight road from rest to 75km/h in 5.0s.
(km / h)
What is the magnitude of its average acceleration?
vxi  0 m / s
v xf 
75000m
 21 m / s
3600 s
Wednesday, Sept. 1, 2004
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 1443-003, Fall 2004
Dr. Jaehoon Yu
5
Example for Acceleration
vx (t )   40  5t 2  m / s
• Velocity, vx, is express in:
• Find average acceleration in time interval, t=0 to t=2.0s
vxi (ti  0)  40(m / s)
vxf (t f  2.0)   40  5  2.0 2   20( m / s)
ax 
vxf  vxi
t f  ti

v x
20  40

 10(m / s 2 )
t
2.0  0
•Find instantaneous acceleration at any time t and t=2.0s
Instantaneous
dvx d
ax  t  

40  5t 2  10t
Acceleration
dt dt
at any time

Wednesday, Sept. 1, 2004

PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
Instantaneous
Acceleration at
any time t=2.0s
ax(t  2.0)
 10  (2.0)
 20(m / s 2 )
6
Meanings of Acceleration
• When an object is moving in a constant velocity (v=v0), there is
no acceleration (a=0)
– Is there net acceleration when an object is not moving? No!
• 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. 1, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
7
Example 2.7
A particle is moving on a straight line so that its position as a function of
time is given by the equation x=(2.10m/s2)t2+2.8m.
(a) Compute the average acceleration during the time interval from t1=3.00s to t2=5.00s.
x dx d
v x  lim


2.10t 2  2.80  4.20t
t 0 t
dt dt


vxi  vx t  3.00s   4.20  3.00  12.6m / s 
vxf  vx t  5.00s   4.20  5.00  21.0m / s 
v x 21.0  12.6 8.40
ax 


 4.2 m / s 2
t
5.00  3.00 2.00


(b) Compute the particle’s instantaneous acceleration as a function of time.
ax  lim
Δt0
vx dvx d

 4.20t   4.20 m / s 2
t
dt dt
What does this mean?


The acceleration of this particle is independent of time.
This particle is moving under a constant acceleration.
Wednesday, Sept. 1, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
8
One Dimensional Motion
• Let’s start with the simplest case: acceleration is a constant (a=a0)
• Using 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. 1, 2004
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 1443-003, Fall 2004
Dr. Jaehoon Yu
9
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. 1, 2004
2

ax

vxf2  vxi2
2a x
v  v  2ax  x f  xi 
2
xf
2
xi
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu

t

10
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. 1, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
11
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?
• Identify what the problem wants.
• Identify which formula is appropriate and easiest to solve for
what the problem wants.
– Frequently multiple formula 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
Wednesday, Sept. 1, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
12
Example 2.11
Suppose you want to design an air-bag system that can protect the driver in a
head-on collision at a speed 100km/s (~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
Wednesday, Sept. 1, 2004
2
v
xi
 2ax  xf  xi 

2
t
vxf  vxi
a
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
0  28m / s


 0.07s
2
390m / s
13
Free Fall
• Falling motion is a motion under the influence of gravitational
pull (gravity) only; Which direction is a freely falling object
moving?
– A motion under constant acceleration
– All kinematic formula 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 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 direction are indicated as the variable “y”
• Thus the correct denotation of gravitational acceleration on
the surface of the earth is g=-9.80m/s2
Wednesday, Sept. 1, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
14
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)
Wednesday, Sept. 1, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
15
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 
Wednesday, Sept. 1, 2004
1
 ( 9.80)  (5.00) 2 27.5(m)
2
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
16
Coordinate Systems
• Makes it easy to express locations or positions
• Two commonly used systems, depending on convenience
– 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)
2
2
r
r

x
1

y
1


x1  r cosq
q
O (0,0)
Wednesday, Sept. 1, 2004
x1
+x
y1  r sin q
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
tan q 
y1
x1
17
Example
Cartesian Coordinate of a point in the xy plane are (x,y)= (-3.50,2.50)m. Find the polar coordinates of this point.
r
y

q
qs
(-3.50,-2.50)m
  3.50 
2
  2.50 
2

 18.5  4.30( m)
x
r
2
2
x

y


q  180  q s
tan q s 
2.50 5

3.50 7
o
1  5 
tan
qs 
   35.5
7
q  180  q s  180o  35.5o  216o
Wednesday, Sept. 1, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
18
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!!!
Wednesday, Sept. 1, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
19
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
Wednesday, Sept. 1, 2004
C
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
C: The same magnitude
but opposite direction:
C=-A:A negative vector
F: The same direction
but different magnitude
20
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
Wednesday, Sept. 1, 2004
B 2A
A
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
B=2A
21
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
 A  B cosq 2  B sin q 2
r
20
A

A2  B 2 cos 2 q  sin 2 q  2 AB cos q

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
Wednesday, Sept. 1, 2004


A  B cos 60
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
Find other
ways to
solve this
problem…
22
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
Wednesday, Sept. 1, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
23
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
Wednesday, Sept. 1, 2004
ur
D



 25  59   7.0  65(cm)
2
2
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
2
24
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:
Wednesday, Sept. 1, 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 1443-003, Fall 2004
Dr. Jaehoon Yu
25
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  ayt j  vi  at


• How are the position vectors written in acceleration vectors?
1
1
x f  xi  vxi t  axt 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
Wednesday, Sept. 1, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
26
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 
Wednesday, Sept. 1, 2004
 vx 
2
  vy 
2

 40
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
2

  15  43m / s
2
27
Example for 2-D Kinematic Eq. Cnt’d
Angle of the
Velocity vector
 vy
q  tan 
 vx
1

1  3 
o
1  15 

tan


21
tan


 


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 
Wednesday, Sept. 1, 2004
PHYS 1443-003, Fall 2004
Dr. Jaehoon Yu
28