1. Rest & Motion

2. 2.1.1
describe using examples how objects can be at rest and in motion simultaneously;
For example, a person inside a train considers himself to be at rest with respect to the fellow passengers or the walls of the train. But when he looks outside, he finds himself to be in motion with respect to the trees outside

3. 2.2.1
describe different types of motion i.e. translatory, rotatory and vibratory motion and distinguish among them;
Translation -- moving in a straight line.
Vibration -- moving back and forth.
Rotation -- moving in a circle.

4. 2.3.1
define the terms speed, velocity and acceleration and write their formulae;
Speed is defined as distance travelled in a unit interval of time
Velocity is defined as displacement travelled in a unit interval of time
Acceleration is a rate of change of velocity

5. 2.3.2 differentiate between distance and displacement, ;
Length of a path distance in a straight line
Any direction specific direction
Scalar quantity vector quantity ;

6. speed and velocity Distance travel in unit time Scalar quantity
Ratio between distance and time
Direction is not involved
Displacement in a unit time
Vector quantity
Ratio between displacement and time
Direction is their

7. 2.4.1 define scalar and vector quantities;
Scalars are quantities that are fully described by a magnitude (or numerical value) alone like mass, speed & time
Vectors are quantities that are fully described by both a magnitude and a direction velocity acceleration & force.

8. 2.4.2 differentiate between scalar and vector quantities;
Mag and unit is needed
Can be represented graphically
Arithmetic addition is possible
Product of scalars, remains a scalar quantity
mass, speed & time
Mag unit and direction is requied
Cannot be represented graphically
Arithmetic addition is not possible
Product of vectors, ca be a scalar or a vector quantity
velocity acceleration & force

9. 2.5.1 define like and unlike parallel forces
Like Parallel forces are the forces that are parallel to each other and have same direction.
Unlike parallel forces are the forces that are parallel but have directions opposite to each other.

10. 2.6.1
describe ‘head to tail’ rule of vector addition of forces vectors;
Vector addition: Resultant vector is vector sum of two vectors
Head to tail rule : vector sum of two vectors a and b can be obtained by joining head of a vector with the tail of b vector. The sum of the two vectors is the vector s joining tail of a to head of b

11. 2.7.1
describe the resolution of force into its perpendicular components;
Components of a Vector:
x-component of vector: its projection on x-axis ax=a cos
y-component of a vector: Its projection on y-axis ay=a sin
Building a vector from its components a =(ax2+ay2); tan  =ay/ax

12. 2.7.2
determine the magnitude and direction of a force from its perpendicular components;
Building a vector from its components a =(ax2+ay2); tan  =ay/ax

13. 2.8.2 plot and interpret distance-time graph with slop
The graph shows
an object which is not moving (at rest).
The straight horizontal line shows that its distance stays the same as time goes by because it is not moving.

14. distance-time graph
The straight line sloping upwards in the graph above shows that the objects distance increases as time goes by.
Slop of the line shows a velocity
The object has velocity because it is moving.
The straight line shows that it is a constant velocity.

15. Velocity-time graph with slop
The straight horizontal line in the graph above shows that the objects velocity does not change as time goes by.
the object is said to have a constant velocity.

16. Velocity-time graph
The straight line in the graph above shows that the velocity of the object has a constant acceleration.
The slope of the line in the graph shows acceleration.
1.  the acceleration is positive because the line slopes upwards
2. how fast the acceleration is. The greater the slope, the faster the acceleration is.

17. 2.8.5 calculate the area under speed-time graph of uniformly accelerated objects to determine the distance;
The total distance travelled by the object can be calculated by measuring the area between the graph and the baseline.
This is called the area under the graph.

18. The area of triangle
A is half base x height
= 0·5 x 10 x 20
= 100.
           = 0·5 x (70 - 30) x 20
= 400.
The area of rectangle B
= (30 - 10) x 20
= 400.
The distance travelled is the total area = A + B + C
= 100 + 400 + 400
= 900 m.

19. 2.9.1 derive equations of motion for a body moving with a uniform acceleration in a straight line
FIRST EQUATION OF MOTION Vf = Vi + at 
Consider a body initial moving with velocity "Vi". After certain interval of time "t", its velocity becomes "Vf". Now Change in velocity = Vf - Vi  OR  Due to change in velocity, an acceleration "a" is produced in the body. Acceleration is given by
a = (Vf – Vi)/t at = Vf – Vi at + Vi =Vf OR

20. SECOND EQUATION OF MOTION S = Vit + 1/2at2 
Consider a car moving on a straight road with an initial velocity equal to ‘Vi’. After an interval of time ‘t’ its velocity becomes ‘Vf’. Now first we will determine the average velocity of body.
Average velocity = (Initial velocity + final velocity)/2 OR Vav = (Vi + Vf)/2 but Vf = Vi + at
Putting the value of Vf
Vav = (Vi + Vi + at)/2
Vav = (2Vi + at)/2
    Vav = 2Vi/2 + at/2
    Vav = Vi + at/2
     Vav = Vi + 1/2at (i)
we know that S = Vav x t
Putting the value of ‘Vav’
S = [Vi + 1/2at] t

21. THIRD EQUATION OF MOTION OR  2aS = Vf2 – Vi2 
Consider a body moving initially with velocity ‘Vi’. After certain interval of time its velocity becomes ‘Vf’. Due to change in velocity, acceleration ‘a’ is produced in the body. Let the body travels a distance of ‘s’ meters.
 According to first equation of motion : Vf = Vi + at     
OR  Vf – Vi = at
                      (Vf – Vi)/a = t (i)
Average velocity of body is given by:
Vav = (Initial velocity + Final velocity)/2
                 Vav = (Vi + Vf)/  (ii)
we know that :          S = Vav x t  (iii)
Putting the value of Vav and t from equation (i) and (ii) in equation (iii)
S = { (Vf + Vi)/2} { (Vf – Vi)/a}
2aS = (Vf + Vi) (Vf – Vi)
According to [ (a+b)(a-b)=a2-b2]

22. 2.9.2 solve problems related to uniformly accelerated motion using appropriate equations;
Starting from rest, an object accelerates at a rate of 12 m/s2. What is the velocity of the object at the end of 3 seconds?
Vi = 0 (Starting from rest) a = 12 m/s2
Vf = ? t = 3.0 seconds
a = (Vf - Vi)/t
 12  = (Vf - 0)/3.0
(Cross Multiply and Solve)
Vf = 36 m/s

23. An object at 100. m/s accelerates at a rate of 120. m/s2
An object at 100. m/s accelerates at a rate of 120. m/s2. What will the velocity of this object be after .100 seconds?
a = (Vf - Vi)/t
120  = (Vf - 100)/100  
Cross Multiply
120 = (Vf - 100)
Vf = 112 m/s

24. Given:a = +3.2 m/s2 t = 32.8 s vi = 0 m/s Find: d = ??
An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.
Given:a = +3.2 m/s2 t = 32.8 s
vi = 0 m/s Find: d = ??
d = vi*t + 0.5*a*t2
d = (0)*(32.8)+ 0.5*(3.20)*(32.8)2
d = 1720 m

25. A car starts from rest and accelerates uniformly over a time of 5
A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car
Given: d = 110 m t = 5.21 s
vi = 0 m/s Find: a = ??
d = vi*t + 0.5*a*t2
110 = (0)*(5.21)+ 0.5*(a)*(5.21)2
110 m = (13.57)*a
a = (110 m)/(13.57)
a = 8.10 m/ s2

26. A race car accelerates uniformly from 18. 5 m/s to 46. 1 m/s in 2
A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.
Given:vi = 18.5 m/s vf = 46.1 m/s t = 2.47 s Find: d = ?? a = ??
a = (vf -vi)/t a = ( )/(2.47)
a = 11.2 m/s2
d = vi*t + 0.5*a*t2
d = (18.5)*(2.47)+ 0.5*(11.2)*(2.47)2
d =
d = 79.8 m

27. A feather is dropped on the moon from a height of 1. 40 meters
A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon
Given:vi = 0 m/s d = 1.40 m a = 1.67 m/s2
Find: t = ??
d = vi*t + 0.5*a*t2
1.40 = (0)*(t)+ 0.5*(1.67)*(t)2
1.40 = 0+ (0.835)*(t)2
(1.40)/(0.835) = t2
1.68  = t2
t = 1.29 s

28. Graphs Slide:
4. 1
During the course of a laboratory experiment, a team of students
obtained the graph on the right representing the force exerted as
a function of the acceleration of a cart.
Calculate the mass of the cart.
A) 3.0 kg
B) 2.0 kg
C) 1.0 kg
D) 0.50 kg
E) 0.25 kg
Note that the slope
represents mass.
Click
Click
Click

29. 4. 2 Reminder The area under the curve is the distance travelled.
Graphs Slide:
4. 2
Reminder
The area under the curve
is the distance travelled.
200 m
250 m
350 m
200 m
+ 100 m
Answer: 350 m – 200 m = 150 m
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30. 4. 3 5 min = 5 x 60 s = 300 s Graphs Slide:
Illustrated below is the velocity versus time graph of a particle.
NOTE: The area under the curve represents distance.
50 000
How far has the particle travelled in 5 minutes?
A) m
B) m
C) m
D) m
Convert to seconds
5 min = 5 x 60 s = 300 s
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31. 4. 4 Graphs Slide: Consider the position versus time graph below.
Which one of the following statements best describes the motion illustrated by the above graph.
A) Increasing velocity, constant velocity, increasing velocity
B) Increasing velocity, zero velocity, increasing velocity
C) Constant velocity, constant velocity, constant velocity
D) Constant velocity, zero velocity, constant velocity
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32. Zero acceleration
Graphs Slide:
4. 5
The velocity-time graph on the left represents the motion of a
car during a 6 s interval of time.
Positive
acceleration
Negative acceleration A) B) C) D)
Which of these graphs represents
the acceleration
of the car?
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33. 4. 6 Fastest velocity here thus maximum KE at this point. Click
Graphs Slide:
4. 6
Fastest velocity here thus
maximum KE at this point.
This point shows that the car
has zero KE which means it
has zero velocity at point I
which is incorrect.
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34. 4. 7 Graphs Slide: Velocity = 750 m/25 s = 30 m/s
Area = velocity = 50 m/s
Velocity = 600 m/25 s = 24 m/s
Velocity = 25 m/s
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35. The slope represents acceleration
Slide:
4. 8
This means the second segment has a negative acceleration (slope).
This means the first segment has to be a straight line (a=0).
The slope represents acceleration
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36. 4. 9 Graphs Slide: Click Since the distance between
dots is increasing, the object
is accelerating.
Zero velocity
Constant
reverse
velocity
Constant
forward
velocity
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37. Note that the slope represents acceleration.
Graphs Slide:
4. 10
Carlo Martini owns a Ferrari. Carlo performed a speed test on his car and plotted the graph below.
Note that the slope represents acceleration.
Answer
Knowing that Carlo’s Ferrari has a mass of 1288 kg, calculate the net force of the car based on the above graph.
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38. Graphs Slide:
4. 13
Both a car and a truck drive off at the same time and in the same direction. The graphs below illustrate their movement.
Slope = Average velocity = 10 m/s
Area to x-axis = distance
Determine how far apart the two vehicles are after 30 seconds.
Step-1 Distance of car = area under the curve = 275 m
Step-2 Distance of car = VAt = (10 m/s)(30 s) = 300 m
Step-3 Distance apart = 300 m – 275 m = 25 m
Answer
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