LEARNING LINEAR EQUATIONS
THROUGH KINEMATICS
THE GOAL OF THIS PRESENTATION IS TO INVESTIGATE THE
VARIETY OF MEANS BY WHICH THE MOTION OF OBJECTS CAN
BE DESCRIBED. THE VARIETY OF REPRESENTATIONS THAT WE
WILL INVESTIGATE INCLUDES VERBAL, NUMERICAL AND
GRAPHICAL REPRESENTATIONS. IN ALGEBRA CLASS, STUDENTS
SHOULD BE ABLE TO UTILIZE THEIR UNDERSTANDING OF
KINEMATICS TO LEARN LINEAR EQUATIONS.
By Prof. Ahmed Salama
PANTHER ACADEMY
• I HAVE SEARCHED FOR THE IMPACT OF PHYSICS TEACHERS ON
LEARNING PHYSICS THROUGH ALGEBRA.
• MOST PHYSICS TEACHERS HAVE INFORMED THEIR STUDENTS THAT IN
ORDER TO LEARN PHYSICS THEY NEED TO KNOW AT LEAST, LINEAR
EQUATIONS AND QUADRATIC EQUATIONS.
• AFTER THREE YEARS OF TEACHING PHYSICS WITH TWENTY SIX YEARS
OF TEACHING MATHEMATICS, I FOUND THAT THE OPPOSITE
STATEMENT COULD BE MORE SUCCESSFUL. STUDENTS NEED TO LEARN
PHYSICS AT AN EARLIER AGE BEFORE GOING THROUGH SERIOUS
ALGEBRA.
• BY DOING PHYSICS EXPERIMENTS AND INVESTIGATING KINEMATIC
GRAPHS; STUDENTS WILL BE ABLE TO PERFORM BETTER IN ALGEBRA
EXAMS.
• MOST EDUCATORS BELIEVE THAT THE LANGUAGE OF PHYSICS IS
MATHEMATICS. IN ORDER TO STUDY PHYSICS SERIOUSLY, ONE
NEEDS TO LEARN MATHEMATICS THAT TOOK GENERATIONS OF
BRILLIANT PEOPLE IN DIFFERENT CENTURIES TO WORK OUT.
ALGEBRA, FOR EXAMPLE, WAS CUTTING-EDGE MATHEMATICS
WHEN IT WAS BEING INVENTED BY OLD EGYPTIANS AND GREEKS
2500 YEARS AGO AND WAS DEVELOPED IN MESOPOTAMIA IN
THE 9TH CENTURY. BUT TODAY IN THE MODERN TIME WE NEED
TO TEACH STUDENTS PHYSICS BEFORE ALGEBRA 2, PRECALCULUS AND CALCULUS.
HOW DID WE CONCLUDE THAT WE
NEED TO TEACH PHYSICS
BEFORE/WITH ALGEBRA ?
I WILL SHARE WITH YOU MY CLASS
EXPERIENCES INCLUDING LECTURES, TEST
RESULTS, PICTURES AND VIDEOS.
• WHEN I STARTED TO TEACH PHYSICS I BEGAN WITH KINEMATICS 9TH
GRADE STUDENTS. MOST OF THEM HAD NOT LEARNED ALGEBRA 1 YET
• I GAVE MY STUDENTS A BASIC MATHEMATICS TEST. THE TEST CONSISTED
OF 10 MULTIPLE CHOICES AND TWO FREE RESPONSES WHICH CONSISTED
OF BASIC OPERATIONS, FRACTIONS, PERCENTAGES, AND RATIOS.
• STUDENT SCORES WERE AS FOLLOWS:
* MORE THAN 65% OF THE STUDENTS HAVE SCORED LESS THAN 40%
* LESS THAN 20% OF THE STUDENTS HAVE SCORED MORE THAN 75%.
NOTE: THE STUDENTS AT THAT TIME JUST BEGUN WITH ALGEBRA 1.
THE FIRST THREE CHAPTERS OF LEARNING PHYSICS
• KINEMATICS: DISTANCE, DISPLACEMENT, SPEED, VELOCITY AND
ACCELERATION.
• DYNAMICS: NEWTON’S LAWS OF MOTION, FORCE, MASS, WEIGH,
TENSION AND NORMAL FORCES, STATIC AND KINETIC
FRICTIONS.
• UNIFORM CIRCULAR MOTION: PERIOD, FREQUENCY,
ROTATIONAL VELOCITY AND APPLICATIONS.
WHAT HAVE STUDENTS LEARNED IN PHYSICS TO
ENHANCE THEIR ALGEBRA SCORES ?
EVERY STUDENT HAS DONE THREE DIFFERENT STAGES:
* MANIPULATING PHYSICS PROBLEMS: WE INCLUDED APPLYING
FORMULAS, CREATING EQUATIONS, GRAPHING FREE BODY DIAGRAMS AND
ANALYZING DATA.
* EXPERIMENTAL PHYSICS: BY DOING EXPERIMENTS IN EACH LESSON
STUDENTS WILL GAIN THE SENSE OF MATHEMATICAL RELATIONSHIP.
* DIGITAL LEARNING: MOST STUDENTS WILL WELCOME DEALING WITH THE
COMPUTERS EVEN CHROMEBOOK. THE STUDENT WILL LEARN HOW TO COLLECT,
ARRANGE, GRAPH AND ANALYZE DATA. THIS STAGE HELPED ME A LOT TO
DELIVER PHYSICS CONCEPTS AS WELL AS MATHEMATICAL HINTS.
SAMPLE PHYSICS PROBLEMS
CONTRIBUTED TO IMPROVING THE
PERFORMANCE OF STUDENTS IN
ALGEBRA
KINEMATICS AND LINEAR
EQUATIONS
• THE POSITION VERSUS TIME GRAPH, BELOW, DESCRIBES THE MOTION OF THREE DIFFERENT
CARS MOVING ALONG THE X-AXIS.
1) DESCRIBE, IN WORDS, THE VELOCITY OF EACH OF THE CARS. MAKE SURE YOU DISCUSS
EACH CAR’S SPEED AND DIRECTION.
2) CALCULATE THE VELOCITY OF EACH OF THE CARS.
3) DRAW, ON ONE SET OF AXES, THE VELOCITY VERSUS TIME GRAPH FOR EACH OF THE
THREE CARS.
THE GRAPH REPRESENTS THE POSITION VS TIME FOR CAR1,
2, 3 (RED, GREEN, BLACK) ASSUME THAT ACCELERATION IS
ZERO, ANSWER THE FOLLOWING QUESTIONS:
• 1) DETERMINE INITIAL POSITIONS OF CARS 1, 2, 3 (X0 ) (IN
ALGEBRA FIND Y INTERCEPTS (b)).
• 2) DETERMINE INITIAL VELOCITIES OF CARS 1,2,3 (V0) (ALGEBRA,
FIND SLOPES (m)).
• 3) WRITE THE KINEMATIC EQUATION OF EACH CAR
X=X0 +V0t+1/2at2
X=X0 +V0t (ALGEBRA, y= mx+ b )
a=0
• 4) FIND OUT WHEN AND WHERE DO CARS 2, 3 MEET (t. x) ?
( LATER IN ALGEBRA INTERSECTION POINT (x, y).
- Hint solve the equations of both cars. Look at the next graph it will show
the positions of all three cars.
• The Answers:
• 1) Car 1 stands 10 meters to the right of the station. (y-intercept)
Car2 starts moving at 10 meters to the right. (y-intercept)
Car3 starts moving at 5 meters to the left. (y-intercept)
• 2) Car 1 does not move because its position never change over the time.
( in Algebra, slope is zero for horizontal line.).
Car 2 moves with a velocity of-5m/s ( in algebra, slope (rise/run) is -5)
Car 3 moves with a velocity of5m/s ( in algebra, slope (rise/run) is 5)
• 3) CAR 1
CAR 2
CAR 3
X=X0 +V0t = 10 ( IN ALGEBRA y= mx+b = 10 )
x=x0 +v0t = 10 -5T ( IN ALGEBRA y= -5x+ 10)
x=x0 +v0t = -5 +5t ( IN ALGEBRA y= 5x-5)
• 4) CAR 2 AND CAR 3 WILL MEET AS SHOWN ON THE FIGURE AT
DISTANCE 2.5 METERS. AND TIME t= 1.5 SECOND.
(IN ALGEBRA WE SOLVE BOTH EQUATIONS AS FOLLOWs:
y= -5x+10 and y= 5x-5
-5x+ 10 = 5x-5
equity property
10x= 15 , x= 1.5
addition and division properties
y= -5(1.5) + 10= 2.5 substitution property ).
THE POSITION VS. TIME GRAPH OF A MOVING OBJECT IS SHOWN TO THE NEXT
SLIDE. USE THIS GRAPH TO ANSWER QUESTIONS.
1 .WHAT IS THE AVERAGE SPEED FROM 0 S TO 4 S?
A. 0.5 M/S
B. 1 M/S
C. 2 M/S
D. 3 M/S
E. 4 M/S
2. WHAT IS THE AVERAGE SPEED FROM 4 S TO 8 S?
A. 0.5 M/S
B. 1 M/S
C. 2 M/S
3. WHAT IS THE OBJECT’S POSITION AT 6 S?
D. 3 M/S
E. 4 M/S
A. 2 M
B. 1 M
C. 3 M
D. 7 M
E. 9 M
4. WHAT IS THE AVERAGE ACCELERATION FROM 4 S TO 8 S?
A. 0 M/S2
B. 1 M/S2
C. 2 M/S2
5. GRAPH THE VELOCITY VS TIME FOR THIS SITUATION.
D. 3 M/S2
E. 4 M/S2
FREE FALL AND LINEAR EQUATION
FOR ANY OBJECT THROWN STRAIGHT UP INTO THE
AIR. THE GRAPH REPRESENT VELOCITY VS TIME
• V= V0+at = 39.2+(-9.8)(8)= -39.2m/s ACCELERATION DUE TO GRAVITY IS 9.8m/s2 (recall linear equation y=mx+b)
• The object will stop momentary at the highest point v=0 ( In algebra, vertex point.
• Students will learn acceleration as the change in velocity over the change in time
v  v0 v 0  39.2
a


 9.8m / s 2
t  t0 t
40
y  y0 y
m

x  x0 x
• Students will learn velocity direction in both sides in the way up and down.
In Algebra students will know that once the curve go down under the x axis the sign
will be changed.
•
STUDENTS COULD FIND THE MAXIMUM RANGE (DISTANCE) BY
DOING PHYSICS AND GEOMETRY:
• SECOND KINEMATIC EQUATION, X = X0 + V0t +1/2 gt2
x = x0 + V0t +1/2 gt2
This the half range
x = 0+ 0+1/2(-9.8)(4)2 = 78.4 M
THE FULL RANGE = 2 (78.4) = 156.8M
• IN GEOMETRY AND ALGEBRA: BY MEASURING THE AREA UNDER
CURVE OF VELOCITY WE COULD FIND THE DISTANCE AS THE
FOLLOWING:
• ½(B)(H) = ½(8)(39.2) = 156.8 ( IF THERE ARE NO AIR RESISTANCE)
STUDENTS` DOCUMENTS
Israa Elsayed, Waldy Reyes, Ashraful Chowdhury
December 4, 2014’
#81
A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction between the
block and the surface is 0.25. The block is connected by a massless string to the second
block with a mass of 300 g.
The string over a light frictionless pulley as shown above. The system is released from
rest.
a. Draw clearly labeled free-body diagrams for each of the 500 g and the 300g masses.
Include all forces and draw them to relative scale. Draw the expected direction of
acceleration next to each free-body diagram.
b. Use Newton’s Second Law to write an equation for the 500 g mass.
c. Use Newton’s Second Law to write an equation for the 300 g mass.
d. Find the acceleration of the system by simultaneously solving the system of two
equations.
e. What is the tension force in the string?
Given:
m1= 500g → 500
1000
m1 = 0.5kg
m2 = 300g → 300
1000
m2 = 0.3kg
g= 9.8m/s2
a= ?
𝝁=
0.25
Formulas:
a= (m2 - 𝝁m1)
(m2 + m1)
g
FT= (g - a ) m2
Part A:
Part B:
FT-FFT=m1a
Part C:
-FT+m2g=m2a
Part D:
a= (m2 - 𝝁m1) g
(m2 + m1)
a= 0.3 - (0.25) 0.5 ( 9.8)
0.3 + 0.5
a= 0.3 - 0.125 (9.8)
0.8
a= 0.175 (9.8)
0.8
a= 2.14 m/s2
Part
FT1=
FT1=
FT1=
FT1=
E:
m1a + 𝝁m1g
(0.5x2.14)+ (0.25x0.5x 9.8)
( 1.07 ) + (1.225)
2.3N
FT2=
FT2=
FT2=
FT2=
(g - a ) m2
( 9.8 - 2.14) (0.3)
( 7.66) ( 0.3 )
2.3N
Siddhartha
Das
STUDENTS DOCUMENTS`
12-18-14
The Pulley System
The pulley in my opinion is one of the greatest
inventions because it used everywhere in our world.
For example, such as cranes, elevators, flagpoles, zip
lines, motors, bicycle rings/chains, clotheslines, water
well bucket/rope, and rock climbing devices. The
pulley system has been used for over 2000 years!
Despite all the advancement in technology we use
this simple system for many things.
Who invented it and What is it?
•The pulley system was first invented by Archimedes
back in 287 B.C.
•It is a wheel with a grooved rim around which a cord
passes. It acts to change the direction of a force
applied to the cord and is chiefly used (typically in
combination) to raise heavy weights. For example,
How does an elevator work?
•An elevator uses a pulley machine as well,
without the use of pulleys, an elevator will
have require a large motor to pull it up.
•The keys part of an elevator is,
• One or more cars (metal boxes) that
rise up and down.
• Counterweights that balance the cars.
• An electric motor that hoists the cars
up and down, including a braking
system.
• A system of strong metal cables and
pulleys running between the cars and
the motors.
• Various safety systems to protect the
passengers if a cable breaks.
•It is supposed to have a motor, two pulleys,
cables, and a counter part. Here is an
example
EVELISE RIVERA (via Google Docs)
Mr. Salama Physics class
A roller coaster car has a mass of 500 kg and is traveling through a vertical loop of
radius 20 m
a roller coaster car has a mass of 500 kg and is traveling through a vertical loop of radius
20 m with a speed of 20m/s
What is the apparent weight at the bottom of the loop?
𝜮F= ma
Fn-mg=ma
+mg +mg
Fn= mg+ma
Fn=m(g+a)
FN=m(g+v2/r)
Fn=500(9.8+202/20)
=500(29.8)
14,900N
a roller coaster car has a mass of 500 kg and is
traveling through a vertical loop of radius 20 m with a
speed of 20m/s
What is the apparent weight at the bottom of the
loop?
𝜮𝑭 = 𝒎𝒂
FN-mg=-ma
FN=mg-ma
FN=m(g-a)
FN-m(g-V2/r)
FN=500(9.8-202/20)
500(9.8-20)
500(10.2) =5,100N
WHAT ALGEBRAIC SKILLS DO STUDENTS
LEARN IN PHYSICS CLASS ?
• SOLVING FOR VARIABLES:
• * KE = ½ MV2 SOLVE FOR M SO THAT FROM KINETIC
ENERGY WE WILL FIND THE MASS OF THE OBJECT TO BE
M= 2KE/V2.
OR THEY WILL BE ABLE TO SOLVE IT FOR V v  2 KE
M
• *FN=M(G-V2/R) SOLVE IT FOR V v   Fn  g r
m

• SCIENTIFIC NOTATION:
A TRUCK TRAVELS AT A CONSTANT SPEED OF 45 M/S. HOW MUCH
WORK DID THE TRUCK ENGINE DO DURING A 2 HOUR PERIOD IF IT
SUPPLIED A FORCE OF 25 KN OF FORCE.
THE ANSWER: V=45M/S, T=2HR=7200S, D=VT, W=FD, W=?
D= 324,000M, F=25KN= 25(1000)=25,000N
W= (45M/S)(7200S) (25KN)(324000)= 8,100,000,000 J CONVERT
TO SCIENTIFIC NOTATION
W= 8.1X10^9J
• GRAPH AND ANALYZE:
HOOK`S LAWA
Mass
(grams)
50
100
150
200
Displacement
(cm)
2
4
6
8
As stated above the relationship depicted on the graph is W = kx
where k is the spring constant. Therefore, the spring constant is the
slope of the line.
STUDENTS SCORES IN PHYSICS AND ALGEBRA
AFTER 12 WEEKS OF LEARNING PHYSICS.
PHYSICS PERFORMANCE
• PHYSICS STUDENTS HAVE LEARNED VARIETY SKILLS OF ALGEBRA FROM
BASIC OPERATIONS SUCH AS ADDITION, SUBTRACTION, MULTIPLICATION,
DIVISION, DECIMAL, FRACTION AND PERCENTAGE TO BIG OPERATIONS
SUCH AS CRITICAL THINKING AND OPEN FREE RESPONSE QUESTIONS.
I HAVE HANDED A SURVEY TO STUDENTS TITLED “DO YOU THINK THAT
LEARNING PHYSICS HAS ENHANCED YOUR ALGEBRA SKILLS?”
THE RESULT OUT OF 67 STUDENTS WAS AS THE FOLLOW:
•
Number of
students who
studied physics
year 2014 1nd
2015
67
Yes
No
Partially
41 (61%)
7 (10%)
19 (28%)
NON PHYSICS STUDENTS` PERFORMANCE IN
ALGEBRA
Grade
A
B
C
D
F
Average
performance
Change
M.P. 1
1
3
5
5
8
66.67%
3.28%
M.P. 2
2
2
6
4
7
69.95%
Physics Students performance in Algebra
Grade
A
B
C
D
F
Average
performance
Change
M.P. 1
0
4
9
7
4
69.29%
7.12%
M.P. 2
3
9
7
2
3
76.41%
REFERENCE
• HTTP://WWW.SUPERSTRINGTHEORY.COM/MATH/MATH1.HTML
• HTTP://WWW.PHYSICSCLASSROOM.COM/CLASS/1DKIN/LESSON-4/DETERMINING-THESLOPE-ON-A-V-T-GRAPH
• HTTP://WONDEROPOLIS.ORG/WONDER/WHO-INVENTED-MATH/
• HTTPS://WWW.NJCTL.ORG/COURSES/SCIENCE/ALGEBRA-BASED-PHYSICS/
For more information,
visit our website: www.deltadvanced.org
or email : [email protected]
Ahmed Salama