The document discusses key concepts in kinematics including scalar and vector quantities, displacement, velocity, acceleration, free fall acceleration on Earth, displacement-time graphs, velocity-time graphs, and the kinematic equations relating displacement, initial velocity, final velocity, acceleration, and time. It provides examples and sample problems involving calculating displacement, velocity, acceleration, and distance/time using the kinematic equations and interpreting graphs of displacement and velocity over time.

1. Turning Effects of Forces / Moments
Notes
Moment
The moment of a force is the turning effect of a force, or the ability of the force to making
something turn.
Moment of a force (M) about a point O is the product of the force (F) and the
perpendicular distance (D) from the point to the line of action of the force.
Moment = Force x Distance
SI Unit: Newton (N)
The turning effect of a force depends on
- location of applied force
- perpendicular distance between the point of application of the force and the pivot
Type 1,2,3 Levers
Principle of Moments
When a body is in equilibrium, the sum of clockwise moments about the balanced point is
equal to the sum of anticlockwise moments about the same point (pivot).

2. Total clockwise moment = Total anticlockwise moment
When the clockwise moment is not equal to the anticlockwise moment, there is a
resultant moment and the object will rotate in the direction of resultant moment.
If there is no resultant moment, the object is balanced.
Centre of gravity
The centre of gravity (CG) of a body is an imaginery point where the whole weight of the
body seems to act in any orientation.
The CG of a regular object is at the centre.
The CG of an irregular object is determined using a plumb line.
If a body is hanging freely at rest, its CG is always vertically below the pivot, thus the
plumb line method works. It can only be used for flat, irregular objects.
Stability
Stability is a measure of the body's ability to maintain its original position.
3 types of stability:
1. Stable equilibrium
Object will return to original position after slight disturbance.
2. Unstable equilibrium
Object will fall after slight disturbance
3. Neutral equilibrium
Object remains in new position after slight disturbance.
To increase the stability of a body, its base area should be increased, and the height of its centre
of gravity should be decreased.

3. Quick Summary
Example
A light metre rule is allowed to pivot freely at the zero end. The other end is supported by a spring
balance. A weight of 200N is then hung at the 40cm mark. The metre rule stays horizontal. What
is the reading on the spring balance?
Solution
By the principle of moments, taking moments about the pivot

4. Anticlockwise moment = Clockwise moment
F x 1m = 200N x 0.4m
F = 80N
The reading on the spring balance is 80N.
MCQ Questions
1. Which one of the following activities does not apply the turning effect of a force?
a. swinging on a swing
b. sliding down a slide
c. moving up and down on the see-saw
d. rowing a boat
2. Which one of the following quantities is zero when a uniform rod is supported in the middle?
a. mass
b. weight
c. pressure
d. moment
3. When a body is at rest, it obeys the
a. principle of momentum
b. Archimede's principle
c. principle of moments
d. principle of inertia
4. A uniform metre ruler of weight 0.2N balances at the 60-cm mark when a weight W is placed at
the 80-cm mark. What is the value of W?
a. 0.1N
b. 0.15N
c. 0.2N
d. 0.2667N
5. Which one of the following measuring instruments works on the principle of moments?
a. spring balance
b. single pan beam balance
c. micrometer
d. vernier calipers
6. A uniform rod of weight 5N and length 1m is pivoted at a point 20cm from one of its ends. A
weight is hung from the other end so that the rod balances horizontally. What is the value of the
weight?
a. 0N

5. b. 0.05N
c. 5N
d. 7.5N
7. An object will not turn if the applied force on it
a. does not reach its maximum
b. does not produce a moment
c. passes through its centre of mass
d. passes through its centre of gravity
8. Levers are classified into different types according to the position of its
a. fulcrum, load and effort
b. centre of gravity
c. centre of mass
d. moment and load
9. Which one of the following statements does not describe a pair of scissors?
a. its fulcrum lies between the load and the effort
b. it is a lever of type 1
c. it works on the turning effect of a force
d. it does not have a centre of mass
10. Which of the following levers is of type 2?
a. wheelbarrow
b. scissors
c. fishing rod
d. ice tongs
11. The centre of mass of a body
a. has a fixed position
b. depends on the pull of gravity
c. is always outside the body
d. must be in a solid part of the body
12. A drinking glass has a low centre of gravity because
a. it is heavy
b. it is tall
c. it has a broad base
d. its contents are heavy
13. When a body is in neutral equilibrium, any displacement will
a. raise its centre of gravity
b. lower its centre of gravity
c. neither raise nor lower its centre of gravity
d. return the body to its original position

6. MCQ Answers
1. b
2. d
3. c
4. a
5. b
6. d
7. b
8. a
9. d
10. a
11. a
12. c
13. c
Structured Questions
1. A uniform metre rule AB is supported at its centre of gravity by a knife edge. A force of
5N is applied at a point which is 30cm from end A of the rule. Calculate the force which
must be applied to point B to restore equilibrium.
[2.0N]
2. A boy of weight 600N sits on the see-saw as shown at a distance of 1.5m from the pivot.
What is the force F required at the other end to balance the see-saw?
[450N]
3. A very light rod 40cm long is pivoted at the centre. A weight of 50N is placed at one end.
Where is the place to put a weight of 200N in order that the rod is in equilibrium?
[5cm from the centre]
4. A very light rod 20cm long has weights of 60N and 40N at its ends. About which point can
the rod balance horizontally?

7. [8cm from the 60N weight]
5. A uniform rod 1m long has masses of 100g and 40g at its ends. If it balances 30cm from
one end, what is the weight of the rod?
[0.1N]
6. The figure shows a uniform metre rule pivoted at the 50cm mark. 125g and 200g weights
hang from the rule as shown.
a. Calculate where you would hang a 25g mass in order to balance the rule horizontally
b. State, without calculation, how the rule with the two masses hanging as shown in the
figure could be balanced without using any extra mass.
[40cm from the pivot on the side of the 200g mass]
https://sites.google.com/site/urbangeekclassroomsg/chemistry-
classroom/stoichiometry#TOC-Molar-Volume-and-Molar-Mass

8. Kinematics: Speed, Velocity, and
Acceleration
Notes
Scalar vs Vector quantities
• Scalar quantities:described by a magnitude only.
 eg. distance, mass, length, temperature
• Vector quantities: quantities described by a magnitude and direction
 eg. displacement, weight, acceleration, force, momentum
Some terms
• Displacement: The distance measured along a straight line in a stated direction w.r.t. the
• original point (vector).
• Velocity: Rate of change of displacement
• Acceleration: Rate of change of velocity
 Note: Negative Acceleration = Retardation
Acceleration of free-fall
• The acceleration of free-fall near the surface of the Earth is constant and is
approximately 10m/s2. It is derived from the gravitational force felt by objects near
the Earth surface and independent of the mass of any object.
• Speed of a free-falling body (experiencing no other forces other than gravity)
increases by 10m/s every second or when the body is thrown up, it decreases by
10m/s every second.
• The higher the speed of an object, the greater the air resistance.
• Terminal Velocity: When an object is moving at constant velocity, acceleration is
0.
• As an object falls, it picks up speed, increasing air resistance. Eventually, air
resistance becomes large enough to balance the force of gravity where the
acceleration of the object is 0, reaching constant velocity.

9. Displacement-Time Graphs
• Used to show displacement over time.
• Horizontal line: Body at rest.
• Straight line with positive gradient: Uniform velocity.
• Straight line with negative gradient: Uniform velocity in the opposite direction.
• Curve: Non – uniform velocity.
• The gradient of the tangent of this graph gives the instantaneous velocity of the object.
Velocity-Time Graphs
• Used to show velocity over time.
• Such a graph can be used to find:
 Velocity
 Acceleration: Gradient
 Distance travelled: Area under the graph
Ticker Tape
Download the presentation here
chapter2-speed_velocity
The Equations
They are called the 'suvat' equations because the quantities s, u, v, a and t are used in
the equations, with four of the symbols used in each equation.
= displacement (measured in metres)
= initial velocity (measured in metres per second, ms -1)

10. = final velocity (also measured in ms-1)
= acceleration (measured in metres per second per second, ms -2)
= time (measured in seconds, s)
Below are the equations:
Note
• It is important to bear in mind that these equations can only be used for CONSTANT
ACCELERATION ONLY. When acceleration is not constant, these equations do not
work. For variable acceleration, either graphical methods or calculus would be needed.
• Furthermore, these equations can only be used for motion in a straight line or one-
dimensional motion.
• Thus these equations are known as the equations of rectilinear motion.
• Rectilinear motion is one-dimensional motion with uniform acceleration.
MCQ Questions
1. Which of the following is a vector?
a. area
b. volume
c. density
d. force
2. The displacement of an object from a fixed point is the distance moved by the object
a. in a particular interval of time
b. in a particular direction
c. at a constant speed
d. at a constant velocity
3. A car accelerates from rest at 5ms-2 for 0.5 minute. The final velocity of the car is
a. 150ms-1
b. 5.5ms-1
c. 10ms-1
d. 2.5ms-1

11. 4. When the brakes of a bicycle were applied, the bicycle was brought to rest from 4ms -
1
in 2 minutes. What is the acceleration of the bicycle?
a. -1/30 ms-2
b. 1/30 ms-2
c. -2ms-2
d. 2ms-2
5. A free falling object is said to be in linear motion. This is because the object is falling
a. due to its weight
b. at constant velocity
c. at constant acceleration
d. in one direction
6. An object has been falling freely from rest for 3 s. The maximum velocity of the object
is
a. 30ms-1
b. 3.3ms-1
c. 10ms-1
d. 13ms-1
7. A body is moving in a circle at a constant speed. Which of the following statements
about the body is true?
a. There is no acceleration
b. There is no force acting on it
c. There is a force acting at a tangent to the circle
d. There is a force acting towards the centre of the circle
e. There is a force acting away from the centre of the circle
8. What must change when a body is accelerating?
a. the force acting on the body
b. mass of the body
c. speed of the body
d. velocity of the body
MCQ Answers
1. d
2. b
3. a
4. a
5. c
6. a
7. d
8. d

12. Structured Questions Worked Solutions
1. A car accelerates uniformly from a speed of 20ms-1 to a speed of 25ms-1 in 2s.
Calculate
a. the average speed for this period of 2s
b. the distance travelled during this period
c. the acceleration
Solution
1a. 22.5 m/s
1b. 45 m
1c. 2.5 m/s2
2.
Describe the motion of the lorry over the following sections of graph along...
a. PQ
b. QR
c. RS
Solution
2a. moving with decreasing speed
2b. stationary/zero speed
2c. moving with increasing speed

13. 3.
a. How far has the object travelled during the first 5 seconds?
b. What is the acceleration of the object
c. For how long does the object move at uniform velocity?
d. What is the average speed of the object during the first 15 seconds?
Solution
3a. 25 m
3b. 2 m/s2
3c. 0
3d. 11.7 m/s2
4. In an experiment, a student measured the accelerations of a steel ball and a
feather falling freely. Will the two accelerations be the same or different? Give a
reason for your answer.
Solution
The two accelerations are different. The air resistance slowed the downward motion of
the falling feather much more than that of the falling steel ball.
5. The figure below shows the velocity of a bus moving along a straight road over
a period of time.

14. a. What does the portion of the graph between O and A indicate?
b. What can you say about the motion of the bus between B and C?
c. What is the deceleration of the bus between C and D?
d. What is the total distance travelled by the bus in 100 s?
e. What is the average velocity of the bus?
Solution
5c. 1 m/s2
5d. 2000 m
5e. 20 m/s
6. Two cyclists, A and B, start a race. A accelerates for the first 5 s, until his
velocity reaches 12ms-1, after which he travels with constant velocity. B
accelerates for the first 10 s, until his velocity reaches 15ms-1, after which he
travels with constant velocity.
a. Sketch the velocity-time graphs for the two cyclists
b. Calculate the distance travelled by both cyclists in the first 10 s
c. Who is in the lead after 10 s?
Solution
6b. 90 m; 75 m
7. A car travels along a straight road. The speedometer reading after every 5 s is
tabulated below
Time/s 0 5 10 15 20 25 30 35 40
Velocity 0 10 20 30 30 30 30 15 0

15. m/s
a. Draw a velocity-time graph to show the variation of velocity with time
b. Describe the motion of the car
c. How far from the starting point is the car after 20 s
d. What is the total distance travelled by the car
e. What is the average velocity of the car for the whole journey
Solution
7c. 373 m
7d. 825 m
7e. 20.6 m/s
8. Find the average velocity of a car which travels 360km in 6 hours in
a. km/h
b. m/s
Solution
8a. 60 km/h
8b. 16.7 m/s
9. Find the average velocity of an athlete who runs 1500m in 4 minutes in
a. m/s
b. km/h
Solution
9a. 6.25 m/s
9b. 22.5 km/h
10. The graph below shows the speed-time graph for a child on a swing

16. a. Write down
i. the maximum speed
ii. the time at which the maximum speed occurs
bi. On the graph, mark with 'Z' the point where the magnitude of acceleration of the
child is maximum.
bii. Mark with 'M' one point at which the acceleration is zero
c. Estimate the distance travelled by the child in 1.2s
d. Describe briefly the changes in acceleration during the period shown on the
graph
Solution
10ai. 6.0 m/s
10aii. 0.60 s
10b.

17. 10c. 3.6 m
10d. The acceleration increases from zero initially to a maximum and decreases to zero.
The body then decelerates with deceleration increasing to a maximum and decreases to
zero.
11. A body is accelerated uniformly from rest and in the first 8.0s of its motion it
travels 20m. Calculate
a. the average speed of this period of 8 s
b. the speed at the end of this period
c. the acceleration
Solution
11a. 2.5 m/s
11b. 5 m/s
11c. 5/8 m/s2
12. A car of length 6.0m accelerates from rest along a straight level road as shown.
The car takes 2.0s to pass the point P.
10.0s later the car has just passed point Q.
The car takes 0.40s to pass point Q.
Calculate

18. ai. the average speed of the car as it passes P
aii. the average speed of the car as it passes Q
aiii. the average acceleration of the car between P and Q
bi. Estimate the distance between P and Q
bii. What assumption did you make when you estimated the distance between P
and Q?
Solution
12ai. 3 m/s
12aii. 15 m/s
12aiii. 1.2 m/s2
12bi. 90 m
12bii. The car accelerates uniformly between P and Q. The acceleration remains
constant.
13a. What is meant by the period of a simple pendulum?
13b. The period of a simple pendulum 1 m long is approximately 2 s. State
accurately how you would determine the period of such a pendulum as accurately
as possible, using a stopclock accurate to within 0.1 s.
Solution
13a. The period of a simple pendulum is the time taken for one complete oscillation
made by the pendulum.
13b.
The pendulum as shown above is slowly brought to point X and then released.
Simultaneously, the stopclock is started. For every time the bob returns to X, it is
considered one oscillation. Count to 30 oscillations and read the time taken. The whole
experiment is repeated three times. Then average the three readings. The period is
found by taking average time over 30.
14. Students, investigating motion down an inclined plane, measure the speed of a

19. steel ball at one second intervals after the ball starts to roll from rest down one
such plane:
time in s 0.00 1.00 2.00 3.00
speed in m/s 0.00 0.60 1.20 1.80
a. calculate the average acceleration over the first 3.00 s
b. calculate the average speed over the first 3.00 s
c. what was the distance travelled by the ball in the first 3.00 s?
d. how do the numbers in the table show that the acceleration was constant?
Solution
14a. average acceleration = (final speed - initial speed) / time = (1.80 - 0.00) / 3.00 =
0.60 m/s2
14b. average speed = 1/2 x (u + v) = 1/2 x (0.00 + 1.80) = 0.90 m/s
14c. distance travelled = average speed x time = 0.90 x 3 = 2.7 m
14d. The speed increases by a constant value of 0.60 m/s every one second, hence the
acceleration can be seen to be constant.
15. A metal box, attached to a small parachute, is dropped from a helicopter.
a. Explain in terms of the forces acting, why
i. its velocity increased immediately after being dropped
ii. it reached a uniform velocity after a short time
b. The total force opposing the motion of the box and parachute at a particular
instant during its fall is 30N. The combined mass of the box and parachute is
5.0kg. Calculate the resultant downward force on the box and parachute. (g = 10
N/Kg)
Briefly describe the motion of the box and parachute at this time
c. At the end of this fall the parachute is caught on a tall tree. The box is then cut
loose and falls from rest to the ground. The time of fall is 2.4 s. Calculate
i. the velocity with which the box strikes the ground
ii. the average velocity during its fall
iii. the distance fallen (g = 10 m/s2)
Solutions

20. 15ai. This is because at the moment the box left the helicopter, the force of gravity pulled
it downwards, causing an increase in velocity
15aii. The presence of the parachute increased the air resistance which acted as a
constant retarding force on the box. Once this force was equal to the weight of the box,
velocity became constant.
15b. resultant downward force = (5 x 10) - 30 = 20 N
From Newton's Second Law of Motion, the box and parachute at this time will accelerate
at a rate of 4 m/s2
15ci. given t = 2.4 s, u = 0, g = 10 m/s2,
v = u + at
v = 10(2.4) = 24 m/s
15cii. average velocity = (u + v)/2 = (0 + 24)/2 = 12 m/s
15ciii. distance = 12 x 2.4 = 28.8m