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O
Level Physics
Unit
2: Kinematics



1.

Scalar and Vector quantities

Type of Quantity

Scalar

Vector

By definition

Has magnitude only

Has magnitude and direction.

Examples

Distance
Speed
Time
Mass

Displacement
Velocity
Acceleration
Weight

Displacement is distance in a
specified direction.

Note: When asked to find
a) a scalar quantity, only the
magnitude is required (e.g. 70m).
b) a vector quantity, the magnitude and direction are required. (e.g.
75m, 20° from the North.



2.

Kinematic Quantities

Symbol

SI Unit

Time

t

s

Distance/Displacement

s

m

Speed/Velocity
Final velocity
Initial velocity

v

m/s
m/s
m/s

v

u

Acceleration

a

m/s^{2}





Speed and Velocity

3.


Speed

Velocity

Definition

Distance travelled per unit time.

Change in displacement per unit time.

Type of quantity

Scalar

Vector

SI Unit

m/s

Formula


Note

The magnitude of the speed and
velocity of an object will differ if there is a change in direction.




4.

Two moving objects
are said to have the same velocity when they have the same speed and move in
the same direction. In a linear motion, velocity can be positive or negative
to specify the direction of motion denoted as positive and negative. A
positive velocity describes motion in the denoted positive direction; and negative
velocity in the denoted negative direction (or back towards starting point)
Note: In negative
velocity (e.g. 6m/s), the negative sign = motion opposite to the positive
direction denoted. However, its speed is 6m/s as speed is a scalar and does
not involve direction.


Acceleration

5.


Acceleration

Definition

Change in velocity per unit time.

SI unit

m/s^{2}

Type of Quantity

Vector

Formula

Acceleration =
a
=


Note: The formula may be changed to the following:
a) a =
b) v = u + a(∆t)
c) ∆t =



6.

Acceleration
occurs when there is a change in an object’s velocity (change in speed and/or
direction).



7.

Positive
acceleration:

Object accelerates in direction of velocity.
Hence, velocity is increasing.

Negative
acceleration:

Object accelerates in direction opposite to
velocity.
Hence, velocity is decreasing.



For
a linear motion, the direction of the acceleration can be specified by
denoting as positive and negative.
Note:
When objects travel in a nonlinear motion, it undergoes positive
acceleration.




Graphical Analysis of Motion



8.


Displacementtime (dispt) graph

Velocitytime (velt) Graph

Deductions

·
The gradient of the dt graph is equivalent
to the speed OR velocity. [See note]
·
A st line with positive grad ⇒
uniform velocity; st line with negative grad ⇒ uniform velocity in opposite
direction.
·
Curve ⇒ nonuniform velocity
(positive acceleration)
·
Grad of the tangent at a point ⇒ instantaneous speed OR
velocity. [See note]
Note: The grads of distt ⇒ speed; of dspmtt ⇒ velocity. Distance
& speed are scalars, displacement & velocity are vectors. Note that
dspmtt graph involves direction unlike distt.

·
Grad of velt graph ⇒ acceleration. Positive grad ⇒ positive acc; negative grad ⇒ negative acc/deceleration
(moving in opp direction)
·
St
line ⇒
uniform acc; curve ⇒ nonuniform acc (positive)
·
Grad
of the tangent at a point ⇒ instantaneous acc.
·
Area
under the graph ⇒ distance travelled OR
displacement [See note].

Deductions of graphs





9.

Summary Graphical Analysis of Motion

Displacementtime (dispt) graph

Velocitytime (velt) Graph

At rest



Uniform velocity



*Uniform velocity in opposite
direction

*Note: Not applicable for distancetime
graph.

*Note: Not applicable for speedtime
graph.

Uniform acceleration
(just acceleration for dispt)



Uniform deceleration
(just deceleration for dispt)



Increasing acceleration



Decreasing acceleration



Increasing deceleration



Decreasing deceleration





Acceleration of Free Fall

10.

During free fall when there is no air resistance,
all objects falling under gravity experiences constant acceleration
(acceleration due to gravity). The acceleration due to gravity, g, for a body
close to Earth’s surface is a constant 10m/s^{2}. Every one second,
the object’s speed will increase by 10m/s.
Acceleration due to gravity where air resistance is
negligible is independent of mass and surface area. All objects undergo the
same constant acceleration.

g =

is equivalent to

10^{ }=

;

W=mg

is equivalent to

W=m(10)

, since the constant g is known.



11.

Free fall Downward motion
Dspmt





Vel/ms^{1}

g/ms^{2}


0m


0s




0

10

At the moment when object was dropped,
it was at rest.







10m


1s


10

10

Object accelerates due to gravity at
10m/s^{2}. In 1s, it travels 10m.














30m


2s


20

10

Object accelerates due to gravity at
10m/s^{2}. In 1s, it travels 20m




















Object
at rest. Velocity and acceleration are both zero.

50m


3s






0

0





Area under velocity/speed  time
graph= displacement/distance dropped respectively.
















12.

Free fall Upward and downward motion.









Pstn

Time

Vel/ms^{1}

g/ms^{2}

Dspmt/m










A

0s

20

10

0




C




















B

1s

10

10

20



B


D



















C

2s

0

Nonzero

30
















A




E




D

3s

10

10

20























E

4s

20

10

0





















F



F

5s

30

10

30








Area under velocity/speedtime graph =
displacement/distance respectively.



Air resistance

13.

Air resistance is a frictional force which opposes
motion. It causes the acceleration of an object in free fall to be lower than
the acceleration of free fall.
(g of object falling with air resistance < 10m/s^{2})
An object falling under gravity with air resistance
experiences decreasing acceleration (air resistance increases with increasing
velocity) until terminal velocity is reached.



14.

Air resistance is dependent on
a)
Velocity: Air resistance increases with
velocity till terminal velocity.
b)
Surface area: Air resistance is directly
proportional to the surface area of an object.



15.

Terminal velocity is the maximum constant velocity
that can be reached where a = 0 and which occurs when (weight = air
resistance).
Terminal velocity is dependent on:
a)
Mass of object: A heavier object will
experience a higher terminal velocity than a lighter object
b)
Surface area of object: An object with a
larger SA will experience lower terminal velocity than one with a smaller SA.



16.

Decreasing acceleration due to increasing velocity
hence increasing air resistance.


t/s

R and W

v/ms^{1}

a/ms^{2}

dspmt/m

0

R = W

0

0

0

1

R_{1} < W

8

8

8

2

R_{2} < W
R_{1} < R_{2}

14

6

22

3

R_{3} < W
R_{1} < R_{2} < R_{3}

18

4

30

4

R = W

20

0

32

Note: W is a constant, R increases with velocity.

Questions involving
calculations Unit 2: Dynamics


Speed and Velocity

1.

A car travels 7km north and then 3km west in 10mins.
Calculate its
a) average
speed; b) average velocity.


a) Ave
speed =
=
60km/h

b) By
Pythagoras’ Theorem,



AC
Displacement

=
≈

7.6158km

Ave velocity =
= 45.7km/h (3sf)
Bearing of C
from A =
= 336.8° (1dp)
Ave velocity is
45.7km/h, 336.8° from the North.


Acceleration

2.

The figure shows the displacementtime graph of an
object.
a)
Fill in the table below and sketch the
corresponding velocitytime graph.
Time

Displacement

Velocity

0 to 4s



4 to 8s



8 to 16s



b)
Calculate
i)
the average speed of the object
ii)
the average velocity of the object
c)
If the object decelerates uniformly to rest in
8s at the end of the 16s, calculate the distance moved in this period.







3.

A motorist, who saw the traffic light he was
approaching turn red, was travelling at 15m/s. His reaction time is 0.4s.
a) Given
that the maximum deceleration of the car is 3.75m/s^{2}, calculate
the distance travelled by the car before it comes to a complete stop.
Given that the motorist is 40m away from the
junction which he stopped exactly at, calculate
b)
the car’s deceleration.
c)
the time taken for the car to stop, using your
answer in (b).



Acceleration of Free Fall

4.

A brick falls
from the top of the building and takes 4.0s to reach the ground. Calculate
a)
the speed of the brick when it reaches the
ground;
b)
the height from which it falls.
a)
g =
10 =
v = 10 x 4
= 40m/s
b) Height
= distance travelled
=
average speed x time
=
(0 + 40)(4.0)
=80m

5.

A book was dropped from a window 9m above the
ground. Calculate
a)
the time taken for the book to hit the ground
b)
the speed of the book when it hits the ground.



a) g
=
10 =
Final speed = 10t
Distance = average speed x
time
9 = (10t) x t
5t^{2} = 9
t = 1.34s

Alternative mtd for (a)
g =
10 =
Final speed = 10t
Area under vt graph = distance = = 5t^{2}
5t^{2} = 9
t
= 1.34s

b)
Final speed = 10t
= 13.4m/s




6.

A
man takes off from a spring board as shown in the diagram. He jumps up into
the air to reach the highest point of his jump before falling downwards.
(Take air resistance
as negligible).
a) It
is given that that the man took 0.7s to reach the highest point of his jump.
Calculate his speed when he leaves the spring board.
b) He
takes another 1.02 before entering the water. Calculate his speed when he
enters the water.
c) Find
the height of the springboard above the water.











