1. Phy 2048C
Spring 2008
Lecture #2

2. Chapter 1 - Introduction
International System of Units
Conversion of units
Dimensional Analysis

3. International System of Units
POWER
PREFIX
ABBREVIATION
1015
peta P
1012
tera T
109
giga G
106
mega M
103
kilo k
102
hecto h
101
deka da
10-1
deci D
10-2
centi c
10-3
milli m
10-6
micro μ
10-9
nano n
10-12
pico p
10-15
femto f
QUANTITY
UNIT NAME
UNIT SYMBOL
Length
meter m
Time
second s
Mass
kilogram kg
Speed
m/s
Acceleration
m/s2
Force
Newton
N=kgm/s2
Pressure
Pascal
Pa = N/m2
Energy
Joule
J = Nm=kgm2/s2
Power
Watt
W = J/s
Temperature
Kelvin K

4. II. Conversion of units III. Dimensional Analysis
Chain-link conversion method: The original data are multiplied successively
by conversion factors written as unity. The units are manipulated like algebraic
quantities until only the desired units remain.
Example: feet/h  m/s
III. Dimensional Analysis
Dimension of a quantity: indicates the type of quantity it is; length [L],
mass [M], time [T]
Dimensional consistency: an equation is dimensionally consistent if each term in it has the same dimensions.
Example: x=x0+v0t+at2/2

5. Problem solving tactics
Explain the problem with your own words.
Make a good picture describing the problem.
Write down the given data with their units. Convert all data into S.I. system.
Identify the unknowns.
Find the connections between the unknowns and the data.
Write the physical equations that can be applied to the problem.
Solve those equations.
Check if the values obtained are reasonable  order of magnitude and units.

6. Chapter 2 - Motion along a straight line
MECHANICS  Kinematics
Chapter 2 - Motion along a straight line
Position and displacement
Velocity
Acceleration
Motion in one dimension with constant acceleration
Free fall

7. Position and displacement
Displacement: Change from position x1 to x2  Δx = x2-x (2.1)
- Vector quantity: Magnitude (absolute value) and direction (sign).
- Coordinate ≠ Distance  x ≠ Δx x x
Coordinate system x2
x1=x2 x1 t t
Δx >0
Δx = 0
Only the initial and final coordinates influence the displacement  many different motions between x1and x2 give the same displacement.

8. Displacement ≠ Distance
Distance: total length of travel.
- Scalar quantity
Displacement ≠ Distance
Example: round trip house-work-house  distance traveled = 10 km
displacement = 0
II. Velocity
Average velocity: Ratio of the displacement Δx that occurs during a particular time interval Δt to that interval.
Motion along x-axis
- Vector quantity  indicates not just how fast an object is moving but also in which direction it is moving.
- The slope of a straight line connecting 2 points on an x-versus-t plot is equal to the average velocity during that time interval.

9. Average speed: Total distance covered in a time interval.
Savg ≠ magnitude Vavg
Savg always >0
Scalar quantity
Example: A person drives 4 mi at 30mi/h and 4 mi and 50 mi/h  Is the
average speed >,<,= 40 mi/h ?
<40 mi/h
t1= 4 mi/(30 mi/h)=0.13 h ; t2= 4 mi/(50 mi/h)=0.08 h  ttot= h
 Savg= 8 mi/0.213h = 37.5mi/h
Instantaneous velocity: How fast a particle is moving at a given instant.
- Vector quantity

10. Instantaneous velocity:
Position
Slope of the particle’s position-time curve at a given instant of time. V is tangent to x(t) when Δt0
Time
When the velocity is constant, the average velocity over any time interval is
equal to the instantaneous velocity at any time.
Instantaneous speed: Magnitude of the instantaneous velocity.
Example: car speedometer.
- Scalar quantity
Average velocity (or average acceleration) always refers to an specific
time interval.
Instantaneous velocity (acceleration) refers to an specific instant of time.

11. III. Acceleration
Average acceleration: Ratio of a change in velocity Δv to the time interval
Δt in which the change occurs.
- Vector quantity
- The average acceleration in a “v-t” plot is the slope
of a straight line connecting points corresponding to
two different times. V t
Instantaneous acceleration: Rate of change of
velocity with time.
- Vector quantity
The instantaneous acceleration is the slope of the tangent line (v-t plot) at a particular time. t t

12. IV. Motion in one dimension with constant acceleration
If particle’s velocity and acceleration have same sign  particle’s speed increases. If the signs are opposite  speed decreases.
Positive acceleration does not necessarily imply speeding up, and negative acceleration slowing down.
Example: v1= -25m/s ; v2= 0m/s in 5s  particle slows down, aavg= 5m/s2
An object can have simultaneously v=0 and a≠0
Example: x(t)=At2  v(t)=2At  a(t)=2A ; At t=0s, v(0)=0 but a(0)=2A
IV. Motion in one dimension with constant acceleration
- Average acceleration and instantaneous acceleration are equal.

13. 1D motion with constant acceleration
tf – t0 = t

14. 1D motion with constant acceleration
In a similar manner we can rewrite equation for average velocity:
and than solve it for xf
Rearranging, and assuming

15. 1D motion with constant acceleration
Using
and than substituting into equation for final position yields
(1)
(2)
Equations (1) and (2) are the basic kinematics equations

16. 1D motion with constant acceleration
These two equations can be combined to yield additional equations.
We can eliminate t to obtain
Second, we can eliminate the acceleration a to produce an equation in which acceleration does not appear:

17. Equations of motion

18. Equations of motion

19. Kinematics with constant acceleration - Summary

20. Kinematics - Example 1
How long does it take for a train to come to rest if it decelerates at 2.0m/s2 from an initial velocity of 60 km/h?
Using we rearrange to solve for t:
Vf = 0.0 km/h, vi=60 km/h and a= -2.0 m/s2.
Substituting we have

21. - Equations for motion with constant acceleration:
t missing t t

22. PROBLEMS - Chapter 2
A red car and a green car move toward each other in adjacent lanes and parallel to
The x-axis. At time t=0, the red car is at x=0 and the green car at x=220 m. If the red car has
a constant velocity of 20km/h, the cars pass each other at x=44.5 m, and if it has a constant
velocity of 40 km/h, they pass each other at x=76.6m. What are (a) the initial velocity, and (b)
the acceleration of the green car? x
d=220 m
Xr1=44.5 m
Xr2=76.6m O
vr1=20km/h
vr2=40km/h
Xg=220m
ag = 2.1 m/s2
v0g = m/s

23. - The motion equations can also be obtained by indefinite integration:
V. Free fall
Motion direction along y-axis ( y >0 upwards)
Free fall acceleration: (near Earth’s surface)
a= -g = -9.8 m/s2 (in constant acceleration motion eqs.)
Due to gravity  downward on y, directed toward
Earth’s center

24. VI. Graphical integration in motion analysis
Approximations:
Locally, Earth’s surface essentially flat  free fall “a” has same direction
at slightly different points.
- All objects at the same place have same free fall “a” (neglecting air resistance).
VI. Graphical integration in motion analysis
From a(t) versus t graph  integration = area between
acceleration curve and time axis, from t0 to t1 v(t)
Similarly, from v(t) versus t graph  integration = area
under curve from t0 to t1x(t)

25. 1) Ascent  a0= 4m/s2 2) Ascent  a= -9.8 m/s2
B1: A rocket is launched vertically from the ground with an initial velocity of 80m/s. It ascends with a constant acceleration of 4 m/s2 to an altitude of 10 km. Its motors then fail, and the rocket continues upward as a free fall particle and then falls back down.
(a) What is the total time elapsed from takeoff until the rocket strikes the ground?
(b) What is the maximum altitude reached?
(c) What is the velocity just before hitting ground?
1) Ascent  a0= 4m/s2 x y
v0= 80m/s
t0=0
a1= -g
a0= 4m/s2
y2=ymax
y1= 10 km
v2=0, t2
+v1, t1 t1 t2
t3=t2 t4
a2= -g
2) Ascent  a= -9.8 m/s2
Total time ascent = t1+t2= s s= s
3) Descent  a= -9.8 m/s2 v3
ttotal=t1+2t2+t4= s + 2·29.96 s s= s
hmax= y2  y2-104 m = v1t2-4.9t22= (294 m/s)(29.96s)- (4.9m/s2)(29.96s)2 = 4410 m