CHAPTER 2.1 LINEAR MOTION

Chapter 2 Kinematics

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Fifth place - Your text here

1. linear motion can be separated into Horizontal and
Vertical motion.

Learning Outcome:

(a) derive and use 2. Horizontal motion best explain using four fundamental
equations of physical quantities:
motion with
constant Displacement velocity
acceleration;
acceleration time
(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Distance,l is a measure of how far an object has moved.

Learning Outcome: Displacement,s is defined as the distance moved in a
particular direction. (It is the change in position)
(a) derive and use
equations of Example:
motion with To the right = + (+5km)
constant To the left = - (-10 km)
acceleration;
Speed,v is defined as the distance moved per second
(b) sketch and use (or the rate of change of distance)
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Velocity,v is defined as the rate of change of displacement.

Learning Outcome: Average velocity = s

(a) derive and use t
equations of
motion with Instantaneous velocity = t ⎯⎯lim→ 0 s = ds
constant t dt
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

The instantaneous velocity of an object is its velocity at a
particular instant or time.

Learning Outcome:

(a) derive and use Acceleration,a is defined as the rate of change of
equations of velocity.
motion with
constant
acceleration;

(b) sketch and use Average acceleration = v dv
the graphs of dt
displacement t
time, velocity-time
and Instantaneous acceleration =
acceleration-time
for the motion of a
body with constant
acceleration;

You can use velocity-time graphs to
calculate acceleration.

Learning Outcome: 8

(a) derive and use Acceleration =
equations of
motion with 3.8
constant
acceleration; Acceleration = - 2 ms−2

(b) sketch and use The gradient velocity-time graphs gives the acceleration.
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Learning Outcome: QUIZZ

(a) derive and use https://forms.gle/ioBCAZyQnVD6xKrG6
equations of
motion with
constant
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

The shaded area between
the two times represent
the car’s displacement.

Learning Outcome:

(a) derive and use
equations of = =
motion with = න
constant
acceleration;

(b) sketch and use Area under v-t graph + (4.4 − 1.7) × 22.7
the graphs of 1
displacement
time, velocity-time Displacement = 2 × 1.7 × 22.7
and Displacement = 80.59 m
acceleration-time
for the motion of a
body with constant
acceleration;

For velocity-time graphs :
A sloping line shows changing velocity (accelerating)

Learning Outcome: The gradient of displacement-time graph gives the
(velocity)
(a) derive and use
equations of A negative gradient of displacement-time graph
motion with represents velocity in the opposite (negative) direction
constant
acceleration; The gradient of a velocity-time graph gives the
acceleration
(b) sketch and use
the graphs of A negative gradient of a velocity-time graph represents
displacement negative acceleration (slowing down)
time, velocity-time
and The area under a velocity0time graph gives the
acceleration-time displacement
for the motion of a
body with constant
acceleration;

Learning Outcome: QUIZZ

(a) derive and use https://forms.gle/QDoC7TCKya6y1c5z9
equations of
motion with
constant
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Motion with Constant Acceleration

1. From the definition of acceleration, we learn that

Learning Outcome: a = Change in velocity = v − u
Time taken from initial to final velocity t
(a) derive and use

equations of

motion with

constant v = u + at 1
acceleration;

(b) sketch and use where v is final velocity and u is the initial velocity of the
the graphs of object.
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

2. Displacement = average velocity x time.

s = (u + v) t 2
2

Learning Outcome: Replace (1) to ( 2)

(a) derive and use s = ut + 1 at 2
equations of 2
motion with
constant From (1), v-u = at and (2) v + u = 2s
acceleration;
t
(b) sketch and use
the graphs of v2 = u2 + 2as
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Learning Outcome: Equation of linear motion :

(a) derive and use v = u + at
equations of s = ut + ½ at2
motion with v2 = u2 + 2as
constant
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

a =v–u s = average velocity x time
t
= u+v Xt v–u
at = v - u 2 a

at + u = v 2as = (v + u)(v – u)

v = u + at proven 2as = v2 – u2
2as + u2 = v2

s = average velocity x time v2 = 2as + u2

= u+v x t proven
2

= 1 ( u + ( u + at) )xt
2

= 1 (2ut + at2) s = ut + ½ at2 proven
2

Summary of equation: The idea of ‘s u v a t’

Learning Outcome: Formulae Uses when absence of Eq. No

(a) derive and use a=v−u s, displacement 1
equations of t
motion with
constant s = (u + v)  t a, acceleration 2
acceleration; 2

(b) sketch and use s = ut + 1 at 2 v, final velocity 3
the graphs of 2
displacement
time, velocity-time v2 = u2 + 2as t, time 4
and
acceleration-time
for the motion of a
body with constant
acceleration;

Example 1- 1

Starting at time t = 0 seconds, an object accelerates from
12 ms-1 to 27 ms-1 in 4 seconds. Find the value of time, t (to
Learning Outcome: the nearest second), when its total displacement is 110 m.

(a) derive and use Solution :
equations of
motion with a = v − u = 27 − 12 = 3.75 ms -1
constant t 4
acceleration;
v2 = u2 + 2as Then v = 122 + 2  3.75110 = 31.13ms-1
(b) sketch and use
the graphs of U sin g, v = u + at
displacement
time, velocity-time t = v-u = 31.13 −12 = 5.1s
and a 3.75
acceleration-time
for the motion of a
body with constant
acceleration;

Example 1- 2

Learning Outcome: A bus travels at straight road with the speed of 24 m/s
decreases uniformly
(a) derive and use to 8 m/s by travelling 40 m. Calculate:
equations of (a) The decelerations of the bus.
motion with (b) The time taken for the deceleration.
constant (c) Total distance travels before the bus stops.
acceleration;

(b) sketch and use (a) v 2 = u2 + 2as
the graphs of 82 = 242 + 2a(40)
displacement a = −6.4 ms-2
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Example 1- 2

Learning Outcome: A bus travels at straight road with the speed of 24 m/s
decreases uniformly
(a) derive and use to 8 m/s by travelling 40 m. Calculate:
equations of (a) The decelerations of the bus.
motion with (b) The time taken for the deceleration.
constant (c) Total distance travels before the bus stops.
acceleration;

(b) sketch and use (b) v = u + at
the graphs of 8 = (24) + (-6.4)t
displacement t = 2.5s
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Example 1- 2

Learning Outcome: A bus travels at straight road with the speed of 24 m/s
decreases uniformly
(a) derive and use to 8 m/s by travelling 40 m. Calculate:
equations of (a) The decelerations of the bus.
motion with (b) The time taken for the deceleration.
constant (c) Total distance travels before the bus stops.
acceleration;

(b) sketch and use (c) U sing v = u + at
the graphs of
displacement 0 = (24) + (−6.4)t
time, velocity-time
and t = 3.75s
acceleration-time
for the motion of a s = ut + 1 at 2 = ((1224x 33..7755)) + 1 (−6.4)(3.75) 2 = 904m5 m
body with constant 2 2
acceleration;

Example :

Learning Outcome: The speed of a car travelling along a straight road decreases uniformly
from 12 ms-1 to 8 ms-1 over 88.0 m. Calculate
(a) derive and use
equations of a) The deceleration of the car
motion with b) The time taken for the speed to decreases from 12 ms-1 to 8 ms-1
constant c) The time taken for the car to come to a stop from the speed of 12
acceleration;
ms-1
(b) sketch and use d) The total distance travelled by the car during this time
the graphs of
displacement Solution (a) : 82 = 122 + 2a(88.0) v = u + at
time, velocity-time u = 12 ms-1 a = - 0.455 ms-2 s = ut + ½ at2
and v = 8 ms-1 v2 = u2 + 2as
acceleration-time s = 88.0 m Deceleration = 0.455 ms-2
for the motion of a a=?
body with constant
acceleration;

Example :

Learning Outcome: The speed of a car travelling along a straight road decreases uniformly
from 12 ms-1 to 8 ms-1 over 88.0 m. Calculate
(a) derive and use
equations of a) The deceleration of the car
motion with b) The time taken for the speed to decreases from 12 ms-1 to 8 ms-1
constant c) The time taken for the car to come to a stop from the speed of 12 ms-1
acceleration; d) The total distance travelled by the car during this time

(b) sketch and use Solution (b) : t=? v = u + at
the graphs of u = 12 ms-1 s = ut + ½ at2
displacement v = 8 ms-1 8 = 12 + (- 0.455)t v2 = u2 + 2as
time, velocity-time s = 88.0 m t = 8.8 s
and a = - 0.455 ms-2
acceleration-time
for the motion of a
body with constant
acceleration;

s

Example : t

The speed of a car travelling along a straight road decre-a8s8es uniformly

from 12 ms-1 to 8 ms-1 over 88.0 m. Calculate

a) The deceleration of the car 44
ms-1
Learning Outcome: b) The time taken for the speed ctoomdeectroeaassetsofprofrmom12thmess-1pteoed88.om8fs1-12
c) The time taken for the car to

d) The total distance travelled by the car during this time

(a) derive and use

equations of

motion with

constant Solution (b) :u = 12 ms-1 v = 8 ms-1 s = 88.0 m a = - 0.455 ms-2
acceleration;

(b) sketch and use t=? v = u + at
the graphs of 88 = 12t + ½ (- 0.455)t2 or
displacement ½ (0.455)t2 - 12t + 88 = 0
time, velocity-time s = ut + ½ at2
and t = 8.8 s or 44 s
acceleration-time v2 = u2 + 2as
for the motion of a
body with constant
acceleration;

Example :

Learning Outcome: The speed of a car travelling along a straight road decreases uniformly
from 12 ms-1 to 8 ms-1 over 88.0 m. Calculate
(a) derive and use
equations of a) The deceleration of the car
motion with b) The time taken for the speed to decreases from 12 ms-1 to 8 ms-1
constant c) The time taken for the car to come to a stop from the speed of 12 ms-1
acceleration; d) The total distance travelled by the car during this time

Solution (c) :

(b) sketch and use u = 12 ms-1 v=0 , t=? v = u + at
the graphs of v = 8 ms-1 s = ut + ½ at2
displacement s = 88.0 m 0 = 12 + (- 0.455)t v2 = u2 + 2as
time, velocity-time t = 26.4 s
and a = - 0.455 ms-2
acceleration-time
for the motion of a
body with constant
acceleration;

Example :

The speed of a car travelling along a straight road decreases uniformly from

12 ms-1 to 8 ms-1 over 88.0 m. Calculate

a) The deceleration of the car

Learning Outcome: b) The time taken for the speed to decreases from 12 ms-1 to 8 ms-1
c) The time taken for the car to come to a stop from the speed of 12 ms-1

(a) derive and use d) The total distance travelled by the car during this time

equations of

motion with

constant

acceleration; Solution (d) :

(b) sketch and use u = 12 ms-1 v=0 v = u + at
the graphs of v = 8 ms-1 , t = 26.4 s
displacement
time, velocity-time s=? s = 12(26.4) + ½ (-0.455)(24.62)
and s = ut + ½ at2
acceleration-time a = - 0.455 ms-2
for the motion of a s = 158 m
body with constant
acceleration; v2 = u2 + 2as

Graphical representation of motion

Learning Outcome:

(a) derive and use Displacement-time graph Velocity-time graph
equations of Acceleration-time graph
motion with
constant
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Distance-time graph

Learning Outcome:

(a) derive and use
equations of
motion with
constant
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Distance-time graph

✓ For a distance-time graph, the distance never decrease

Learning Outcome: ✓ When the object is stationary, the distance-time graph
will be horizontal
(a) derive and use
equations of ✓ The gradient of a distance-time graph is the
motion with instantaneous speed of the object
constant
acceleration; ✓ For straight line with positive gradient, it means that the
object is travelling at uniform speed
(b) sketch and use
the graphs of ✓ There is no straight line with negative gradient (as the
displacement distance never decreases)
time, velocity-time
and ✓ For curve, it means that the object is travelling at non-
acceleration-time uniform speed
for the motion of a
body with constant
acceleration;

Velocity-time graph

Learning Outcome:

(a) derive and use
equations of
motion with
constant
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Learning Outcome: ➢ When the object is stationery, it is a straight horizontal line
at 0.
(a) derive and use
equations of ➢ When the object is undergoing uniform motion, it is a
motion with straight horizontal line at ms−1, where v is the velocity of
constant the object
acceleration;
➢ For straight line with positive gradient, it means that the
(b) sketch and use object is accelerating
the graphs of
displacement ➢ For straight line with negative gradient, it means that the
time, velocity-time object is decelerating
and
acceleration-time ➢ For curves, it means that the acceleration of the object is
for the motion of a changing
body with constant
acceleration; ➢ For area under the graph is the change in the displacement
of the object

Learning Outcome:

(a) derive and use
equations of
motion with
constant
acceleration;

(b) sketch and use

the graphs of

displacement

time, velocity-time

and

acceleration-time

for the motion of a

bacocdeyTlehwreaitthifoicgnou;nrsetasnht ows the relationship between displacement graph, velocity graph and
acceleration graph

Uniform velocity /uniform motion

s Displement time-graph

Learning Outcome: Gradient = constant
= velocity
(a) derive and use
equations of t
motion with
constant v velocity time-graph acceleration time-graph
acceleration;
t a
(b) sketch and use
the graphs of t
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Graphical representation of motion

Learning Outcome: Uniform acceleration v t

(a) derive and use s t
equations of
motion with t
constant
acceleration; a

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Graphical representation of motion

Uniform deceleration (negative acceleration)

sLearning Outcome: v

(a) derive and use tt
equations of
motion with Velocity = slope of displacement-time graph
constant Displacement = area under a velocity-time graph
acceleration;
Acceleration = slope of velocity-time graph
(b) sketch and use a t
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Graphical representation of motion

Stationary object

Learning Outcome: s t
v a
(a) derive and use
equations of t t
motion with
constant
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Example 1 : v

s v t
v t
Learning Outcome: t
t0 v
(a) derive agndrausde ient compare

equations of
motion with
constant
acceleration;

(b) sketch and use
the graphs of
displacement

0time, velocity-time

and
acceleration-time
for the motion of a
body with constant
acceleration;

Example 2 : v

a v

Learning Outcome: t t t
t
(a) derive and use a v t
equations of or
t t
maotion with
a
constant or
acceleration; t

(b) sketch and use
the graphs of
displacement

timae, velocity-time

and
acceleration-time
for the motion of a
body with constant
acceleration;

Example 3: Gradient = 0
Velocity = 0
s

Learning Outcome: Gradient negative Direction
Velocity negative change
(a) derive and use
equations of t v
motion with
constant t t

Garcacdeileernattipono;sitive
Velocity positve

(b) sketch and use
the graphs of

displacaement

time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Calculation :

Learning Outcome: v = ds a = dv
dt dt
(a) derive and use
equations of = gradient of (s-t) graph = gradient of (v-t) graph
motion with
constant s = v dt
acceleration;
= area under (v-t) graph
(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Acceleration due to gravity (ignoring friction)

Learning Outcome: 1.Objects in gravitation field experience a downwards force
- their weight. If unbalanced, this produces acceleration
(a) derive and use downwards.
equations of
motion with 2.There is an acceleration free fall or acceleration due to
constant gravity when the object moves at the same acceleration.
acceleration;

(b) sketch and use 3. Acceleration free fall is represented by the symbol "g" as
the graphs of assume constant 9.81m s-2.
displacement
time, velocity-time 4. The value of g not depends on the displacement or location
and of the object.
acceleration-time
for the motion of a
body with constant
acceleration;

Study the following figure:

Learning Outcome:

(a) derive and use
equations of
motion with
constant
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Learning Outcome: Case 1 Case 2 Case 3

(a) derive and use An object travels An object travels An object travels
equations of upwards downwards below the reference
motion with
constant level
acceleration;
Displacement = +s Displacement = +s Displacement = -s
(b) sketch and use
the graphs of Velocity = +v Velocity =-v Velocity = -v
displacement
time, velocity-time Acceleration = -g Acceleration = -g Acceleration = -g
and
acceleration-time
for the motion of a
body with constant
acceleration;

• Useful information for calculation:

Learning Outcome:

(a) derive and use
equations of
motion with
constant
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Information Analysis

The velocity decreases 1. Calculation of maximum
to zero when the object displacement:
reaches the highest H = u2

Learning Outcome: point. 2g

(a) derive and use 2. Calculate total time for the objects to
equations of
motion with travel.
constant
acceleration; Using the formula: s = ut + 1 at2
2
(b) sketch and use Time taken for the s is H and a is g
the graphs of object to reach the 1
displacement highest point is the H = ut − 2 gt 2
time, velocity-time
and same as the time it Total time for traveling is 2 x t.
acceleration-time
for the motion of a takes to drop back to its 2. Calculate velocity before reaching the
body with constant
acceleration; initial point. ground.

v = u – gt

u=0
v = -gt

a = v-u a = v-u = 10-0 =10ms-2
t t1

= 30-0 a = v-u = 30-20 =10ms-2
3 t1

=10ms-2 a = v-u = 40-30 =10ms-2
t1
Learning Outcome:

(a) derive and use
equations of
motion with
constant
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

s = ut + ½ at2 s = ut + ½ at2
=(20)(1)+½(-10)(1)2 =(20)(2)+½(-10)(2)2
= 15 = 20

vLearning Outcome: v = u + at
= 20 + (-10)(2)
(a) derive and use
equations of s =0
motion with
constant t2 t

0 tacceleration; 1

(b) sketch and use 0 t1 t2 t
the graphs of
displacement
time, velocity-time
and

va=ccuele+raatiton-time
=for2t0he+m(o-t1io0n)(o1f )a

bacocd=eyl1ew0raitthiocno;nstant

s = ut + ½ at2 s = ut + ½ at2
=(20)(1)+½(-10)(1)2 =(20)(2)+½(-10)(2)2
= 15 = 20

vLearning Outcome: v = u + at
= 20 + (-10)(2)
(a) derive and use s
equations of 0 =0
motion with
constant t2 v = u + at

0 tacceleration; 1 t = 0 + (-10)(1)
= -10

(b) sketch and use t1 t2 t
the graphs of
displacement s = ut + ½ at2
time, velocity-time
and =(0)(1)+½(-10)(1)2

va=ccuele+raatiton-time = -5
=for2t0he+m(o-t1io0n)(o1f )a d = 20-5 = 15

abcocd=eyl1ew0raitthiocno;nstant

Learning Outcome:

(a) derive and use
equations of
motion with
constant
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Learning Outcome:

(a) derive and use
equations of
motion with
constant
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Example :

Learning Outcome:

(a) derive and use
equations of
motion with
constant
acceleration;

(b) sketch and use A stone thrown upward with initial velocity 30 ms-1.
the graphs of Calculate the time interval between 25 m from the surface.
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;