1. KINEMATICS
The Study of Motion
Copyright Sautter 2003

2. Measuring Motion
The study kinematics requires the measurement of three properties of motion.
(1) displacement – the straight line distance between two points (a vector quantity)
(2) velocity – the change in displacement with respect to time (a vector quantity)
(3) acceleration – the change in velocity with respect to time (a vector quantity)
The term distance like displacement, refers to the change in position between two points, but not in a straight line. Distance is a scalar quantity. Speed refers to change in position with respect to time but unlike velocity, does not require straight line motion. Speed is a scalar quantity.

3. Speed ( a scalar ) & Velocity ( a vector )
Velocity = Displacement from A to B/ time
Distance traveled
from A to B
x B
Lake Tranquility
Displacement
from A to B
A x
Speed = Distance from A to B/ time

4. VELOCITY & ACCELERATION
OBJECTS IN MOTION MAY MOVE AT CONSTANT VELOCITY (COVERING EQUAL DISPLACEMENTS IN EQUAL TIMES) OR BE ACCELERATED (COVER INCREASING OR DECREASING DISPLACEMENTS IN EQUAL TIMES).
VELOCITY MEASUREMENTS MAY BE OF TWO TYPES, AVERAGE VELOCITY (VELOCITY OVER A LARGE INTERVAL TIME) OR INSTANTANEOUS VELOCITY (VELOCITY OVER A VERY SHORT INTERVAL OF TIME).
ACCELERATION MAY BE UNIFORM OR NON UNIFORM. UNIFORM OR CONSTANT ACCELERATION REQUIRES THAT THE VELOCITY INCREASE OR DECREASE AT A CONSTANT RATE WHILE NON UNIFORM ACCELERATION DISPLAYS NO REGULAR PATTERN OF CHANGE.

5. Uniformly Accelerated Motion
Constant Velocity
1 sec
2 sec
3sec
4sec
5 sec
EQUAL DISPLACEMENTS IN EQUAL TIMES
Uniformly Accelerated Motion
1 sec
2 sec
3sec
4sec
CLICK
HERE
REGULARLY INCREASING DISPLACEMENTS IN EQUAL TIMES

6. Displacement vs Time for a Uniformily Accelerated Body S POSITIVE
ACCELERATION
Equal time
intervals result
in increasingly
larger displacements S S t t t
time

7. Average Velocity for a Uniformily
Accelerated Body D I S P L A C E M N T
Average velocity
between t1 and t2
Is the slope of the
Secant line =
S/ t s2 S
Secant
line s1 t t1 t2
time

8. Instantaneous Velocity for a Uniformily Accelerated Body
Draw a tangent line at the point
Finding velocity
at point t1, s1
(instantaneous velocity) D I S P L A C E M N T S
Find the slope of
the tangent line s1 t
Instantaneous velocity
equals the slope of
the tangent line t1
time

9. DISPLACEMENT, VELOCITY & CONSTANT ACCELERATION
The velocity of an object at an instant can be found by determining the slope of a tangent line drawn at a point to a graph of displacement versus time for the object.
If several instantaneous velocities are found and plotted against time the graph of velocity versus time is a straight line if the object is experiencing constant acceleration.
The slope of the straight line velocity versus time graph is constant and since acceleration can be determined by the slope of a velocity – time graph, the acceleration is constant.
The graph acceleration versus time for a constant acceleration system is a horizontal line. (A slope of zero since constant acceleration means that acceleration is not changing with time!)

10. Finding Velocity & Acceleration from Displacement vs time
Slope of a tangent drawn to a point on
a displacement vs time graph gives
the instantaneous velocity at that point A C E L R T I O N
Time
PLOT OF INSTANTANEOUS
VELOCITIES VS TIME V E L O C I T Y
Time v
Slope of a tangent drawn to a point on
a velocity vs time graph gives the
instantaneous acceleration at that point t

11. MEASURING VELOCITY & ACCELERATION
VELOCITY IS MEASURED AS DISPLACEMENT PER TIME. UNIT FOR THE MEASUREMENT OF VELOCITY DEPEND ON THE SYSTEM USED. IN THE MKS SYSTEM (METERS, KILOGRAMS, SECONDS) IT IS DESCRIBED IN METERS PER SECOND.
IN THE CGS SYSTEM (CENTIMETERS, GRAMS, SECONDS - ALSO METRIC) IT IS MEASURED IN CENTIMETERS PER SECOND.
IN THE ENGLISH SYSTEM IT IS MEASURED AS FEET PER SECOND.
ACCELERATION IN THE MKS SYSTEM IS EXPRESSED AS METERS PER SECOND PER SECOND OR METERS PER SECOND SQUARED.
IN CGS UNITS IT IS CENTIMETERS PER SECOND PER SECOND OR CENTIMETERS PER SECOND SQUARED. IN THE ENGLISH SYSTEM FEET PER SECOND PER SECOND OR FEET PER SECOND SQUARED ARE USED.

12. GRAVITY & CONSTANT ACCELERATION
Gravity is the most common constant acceleration system on earth. As object fall under the influence of gravity (free fall) they continually increase in velocity until a terminal velocity is reached.
Terminal velocity refers to the limiting velocity caused by air resistance. In an airless environment the acceleration provided by gravity would allow a falling object to increase in velocity without limit until the object landed.
In most problems in basic physics air resistance is ignored. In actuality, terminal velocity is related to air density, surface area, the velocity of the object and the aerodynamics of the object (the drag coefficient).

13. Acceleration Due to Gravity & Free Fall
g = 9.8 meters / second 2
g = 980 centimeters / second 2
g = 32 feet / second 2
CLICK
HERE

14. Free Fall 78.4 m 44.1 m 19.6 m 19.6 m/s 2.0 sec 29.4 m/s 3.0 sec

15. CALCULATING AVERAGE VELOCITY
Average velocity for an object moving with uniform (constant) acceleration can be calculated in two ways.
(1) average velocity = the change in displacement (displacement traveled, s) divided by the change in time ( t). (s is the symbol used for displacement)
(2) average velocity = the sum of two velocities divided by two (an arithematic average). 
V = s t 
ave 2 1
v v
V =
ave

16. CALCULATING INSTANTANEOUS VELOCITY
Instantaneous velocity can be found by taking the slope of a tangent line at a point on a displacement vs. time graph (as previously discussed).
Instantaneous velocity can also be determined from an acceleration vs. time graph by determining the area under the curve.
For constant acceleration systems, the acceleration times the time (a x t) plus the original velocity (v0) also gives the instantaneous velocity.
V = V + a t o i

17. GIVES THE INSTANTANEOUS
Area Under an Acceleration vs. Time
Curve Gives the
Instantaneous Velocity
AREA UNDER THE CURVE
(acceleration x time)
GIVES THE INSTANTANEOUS
VELOCITY AT TIME t1 A C E L R T I O N t1
Time

18. CALCULATING DISPLACEMENT
Displacement of a body in constant acceleration can be found in two ways.
Displacement is given by the area under a velocity vs. time graph.
Displacement can also be found using the follow equation where si = instantaneous displacement, vo = the original velocity of the object, a = the constant acceleration and t = elapsed time.
s = v t a t o
1/2 2 i

19. GIVES THE INSTANTANEOUS
Area Under a Velocity vs. Time
Curve Gives the
Instantaneous Displacement
AREA UNDER THE CURVE
(velocity x time)
GIVES THE INSTANTANEOUS
DISPLACEMENT AT TIME t1 V E L O C I T Y
Time t1

20. CALCULATING VELOCITY & ACCELERATION FROM DISPLACEMENT VS. TIME
The instantaneous velocity of an object can be found from a displacement versus time graph by measuring the slopes of tangent lines drawn to points on the graph.
Since the derivate of an equation gives the formula for calculating slopes, the derivative of the displacement versus time equation will give the equation for velocity versus time.
Additionally, the slope of an velocity versus time curve is the acceleration. Therefore, the derivative of the velocity versus time equation gives the acceleration versus time relationship.

21. Velocity from Displacement vs. time
s = v t a t o
1/2 2 i
The first derivative of displacement versus time
gives the instantaneous velocity in terms of time.
1 -1 = 0
s = v t a t o
1/2 2 i
v = d
/ dt
- 1 = 1
1 x
2 x
v = v a t o i

22. Acceleration from velocity vs. time
v = v a t o i
The first derivative of velocity versus time gives
the instantaneous acceleration in terms of time. t
1 - 1 = 0
v = v a t o i
a = d / d t
0 x
1 x
a = a ( 1 ) acceleration is constant

23. CALCULATING VELOCITY & DISPLACEMENT FROM ACCELERATION
The instantaneous velocity of an object can be determined from the area under an acceleration versus time graph.
Since the integration of an acceleration versus time equation gives the area under the curve, it also gives the velocity.
The area under a velocity versus time graph gives the displacement. Therefore, the integral of the velocity versus time equation gives the displacement versus time equation.

24. Velocity from Acceleration vs. time
a = a x t
V = a x t dt  i
The constant
is the original
velocity (V0)
0 + 1
v = a x t i
+ C
0 + 1
v = a t + v i

25.  Displacement from Velocity vs. time v = a t + v i s = a t + v i
s = a t v i
( t ) dt
s = a t v t C
1 + 1
0 + 1 i
s = v t a t o
1/2 2 i
+ C
The constant C is the original displacement of the object
If displacement is not measured from zero

26. Displacement, Velocity & Accelerationvs
time
Acceleration
vs.
time
slope
Velocity vs
time
slope
Area
under
curve
Area
under
curve
derivative
derivative
Acceleration
vs.
time
Displacement vs
time
Velocity vs
time
integral
integral

27. ACCELERATED MOTION SUMMARY
VAVERAGE = s/ t = (V2 + V1) / 2
VINST. = VORIGINAL + at
SINST = V0 t + ½ at2
Instantaneous velocity at a point equals the slope of a tangent line drawn at that time point on a displacement vs. time graph
Instantaneous acceleration at a point equals the slope of a tangent line drawn at that time point on a velocity vs. time graph.
The derivative of a displacement vs. time equation gives the instantaneous velocity.
The derivative of a velocity vs. time equation gives the instantaneous acceleration.
The integral of an acceleration vs. time equation gives the instantaneous velocity.
The integral of an velocity vs. time equation gives the instantaneous displacement.

28. PUTTING EQUATIONS TOGETHER
Often problems involving uniformly accelerated motion do not contain a time value. When this occurs these problems can be solved by combining equations which are already known.
To simplify the algebra, V0 will assumed to be zero. Therefore, Vi = VO + at becomes Vi = at and Si= V0 t + ½ at2 becomes Si = ½ at2.
Solving Vi = at for t we get t =Vi/a. Substituting into Si = ½ at2 gives Si = ½ a(Vi/a)2 or by simplifying the equation Si = ½ (Vi 2/a)
If V0 is not equal to zero the equation becomes
Si = ½ (Vi 2 – Vo2) /a (time is not required to solve this equation!)

29. In the next program the equations and relationships developed here will be used to solve one dimensional, uniform acceleration problems.
Free fall problems will be included since they are the most common examples of constant acceleration .
Problem involving variable acceleration and use to derivatives and integrals for their solution will be covered.
Math concepts are required and if the program on Math for Physics has not yet been viewed, it may be a good idea to do so!

30. The End