NOTA TUTORIAL SP015 CHAPTER 2 20_21

TOPIC 2: KINEMATICS OF LINEAR MOTION limv = s = ds
2.1: LINEAR MOTION
2.2: UNIFORMLY ACCELERATED MOTION t→0 t dt
2.3: PROJECTILE MOTION
▪ vector quantity
2.0 KINEMATICS ▪ SI unit: m s–1

Description of the motion of objects without consideration Average velocity, vav
of what causes the motion.
Rate of Change of Displacement

 = change in displacement
v av
time taken for the change

 = s = s2 − s1
vav t t2 − t1

▪ vector quantity
▪ SI unit: m s–1

How to Measure Displacement

2.1: LINEAR MOTION Distance:
Learning Outcome:
 The total length of travel in moving from one
a) Define location to another = 200m
i. instantaneous velocity, average velocity
and uniform velocity. Displacement:
ii. instantaneous acceleration, average
acceleration and uniform acceleration  straight line distance from the initial position to the
final position of an object = 120 m, in the direction
b) Discuss the physical meaning of displacement- of Northeast
time, velocity-time and acceleration-time graphs.
 equal to the gradient at any point on the curve of a
c) Determine the distance travelled, displacement, displacement – time (s – t) graph.
velocity and acceleration from appropriate graphs.

Subtopic 2.1 (a)

Keypoint

Velocity Acceleration

Instantaneous Velocity Instantaneous Acceleration

Average Velocity Average Acceleration

Uniform Velocity Uniform Acceleration

Instantaneous Velocity, v

▪ velocity at a specified position or instant of time
along the path of motion.

▪ commonly referred as ‘velocity at point A’ or
‘velocity at time t’.

1

• Slope BE (from point B to point E) = average velocity Uniform Acceleration C
• Slope at C = Instantaneous velocity  Uniform acceleration means the acceleration

Uniform Velocity does not depend on time or always constant.
 motion with constant velocity
 acceleration, a = 0 m s–2 a = dv = constant
dt

 velocity changes at a uniform rate.

The displacement increases by equal amounts in
equal times.

 A graph (v – t) plot is a straight line whose
gradient is equal to acceleration.

 Velocity is vector quantity,
→ a change in velocity may thus
involve either or both magnitude & direction.

 An acceleration may due to:
i) change in speed (magnitude),
ii) change in direction or

iii) change in both speed and direction.

Instantaneous Acceleration Quick Test
 acceleration at a particular instant of time.
–1

A car is traveling at 30 km h to the north.
Then it turns to the west without changing its speed. Is
the car accelerating?

Answer: YES!
Reason: there is a change in direction

 Deceleration: object is slowing down
(direction of acceleration is opposite to the
direction of the motion or velocity).

lima = v = dv = d 2s
t →0 t dt dt 2

 change in velocity divided by the time taken
to make the change.

 = change in velocity = v2 − v1 = v
a
time to make the change t2 − t1 t

 vector quantity
 SI unit: m s–2

2

➢ Velocity – time graphs (v – t)
➢ Acceleration – time graphs (a – t)

i. s v a

Car in figure (a) & (d) → accelerating tt t
Car in figure (b) & (c) → decelerating
a) Object is at rest/static/stationary
b) v = 0 m s-1
c) a = 0 m s-2

ii. s v a

tt t

 For linear motion, (+) and (–) signs is used to a) Object is moving in POSITIVE direction
indicate direction of motion, velocity & b) Velocity constant (+ve)
acceleration. c) a = 0 m s-2

 With horizontal direction we may take: iii. s v a

to the right as positive (+) tt t
to the left as negative (–)
 With vertical direction we may take: a) Object is moving in POSITIVE direction
upward as (+); downward as (–) b) Velocity INCREASE (in positive direction)
c) Acceleration constant (+ve): object accelerate
The sign convention you choose is
entirely up to you. It doesn’t matter as in positive direction
long as you keep the same sign
convention for the entire calculation. iv.
s va
Subtopic 2.1 (b)
PHYSICAL MEANING OF KINEMATICS t

`LINEAR GRAPH` tt
3 types of graph:
➢ Displacement – time graphs (s – t) a) Object is moving in POSITIVE direction
b) Velocity DECREASE (in positive direction)

3

c) deceleration constant (-ve): object decelerate Displacement – time graphs ( s – t )
in positive direction

gradient area under graph

v.

sv a Velocity – time graphs ( v – t )

t area under graph

gradient

t t Acceleration – time graphs ( a – t )

a) Object is moving in NEGATIVE/OPPOSITE i. Displacement – time graphs
direction

b) Velocity constant (in negative direction)
c) a = 0 m s-2

vi.
s va

t
t

t Instantaneous velocity,
v = ds
a) Object is moving in NEGATIVE/OPPOSITE
direction. dt
= gradient of ( s – t ) graph
b) Velocity INCREASE (in negative direction)
c) Acceleration constant (-ve): object accelerate in ii. Velocity – time ( v – t ) graphs

negative direction

vii.
s va

t

t
t

a) Object is moving in NEGATIVE/OPPOSITE
direction.

b) Velocity DECREASE (in negative direction).
c) Deceleration constant (+ve): object decelerate in

negative direction

Subtopic 2.1 (c) Shaded AREA under the (v - t) graph =
Displacement of the object
DETERMINE THE DISTANCE,
DISPLACEMENT, VELOCITY AND
ACCELERATION FROM APPROPRIATE

GRAPH:

4

Example 2.1.1: c.
A toy train moves slowly along a straight track
according to the displacement, s against time, t graph vav = s2 − s1
in Figure 2.1. t2 − t1

s (cm) vav = 10 − 0
14 − 0
10
vav = 0.714 cm s−1
8
6 d.

4 v = average velocity from10 s to 14 s
2
v = s2 − s1
0 2 4 6 8 10 12 14 t (s) t2 − t1

Figure 2.1 v = 10 − 4
14 −10

v = 1.50 cm s−1

e. The distance travelled by the train is 10 m

a. Explain qualitatively the motion of the toy train. Example 2.1.2:
b. Sketch a velocity (cm s-1) against time (s) graph. −1
c. Determine the average velocity for the whole v (m s )

journey. 4
d. Calculate the instantaneous velocity at t = 12 s.
e. Determine the distance travelled by the toy train.

Solution: : The train moves at a constant 2
a. velocity of 0.68 cm s−1.
0 to 6 s 0
6 to 10 s : The train stops. 5 10 15 20 25 30 35 40 45 50 t (s)
10 to 14 s : The train moves in the same
-2
b. direction at a constant velocity of
1.50 cm s−1. -4
−1
Figure 2.2
v (cm s )
A velocity-time (v-t) graph in Figure 2.2 shows the
1.50 motion of a lift.
a. Describe qualitatively the motion of the lift.
0.68 b. Sketch a graph of acceleration (m s−2) against time
(s).
c. Determine the total distance travelled by the lift and
its

displacement.
d. Calculate the average acceleration between 20 s to
40 s.

0 6 8 10 12 14 t (s) 5
24

Solution: Lift moves upward from rest with a c. i. −1
a. constant acceleration of 0.4 m s−2. v (m s )
0 to 5 s : The velocity of the lift increases
5 to 15 s : from 2 m s−1 to 4 m s−1 but the 4
acceleration decreasing to 0.2 m s2.
15 to 20 s : Lift moving with constant velocity of 2
20 to 25 s : 4 m s−1. A2 A3
25 to 30 s :
Lift decelerates at a constant rate of 0 A1
0.8 m s−2. 5 10 15 20 25 30 35 40 45 50 t (s)

Lift at rest or stationary. -2 A4
A5
30 to 35 s : Lift moves downward with a constant
35 to 40 s : acceleration of 0.8 m s−2. -4
40 to 50 s : Lift moving downward with constant
velocity of 4 m s−1. Total distance = area under the graph of v-t
Lift decelerates at a constant rate of
0.4 m s−2 and comes to rest. = A1 + A2 + A3 + A4 + A5

b. Total distance = 1 (2)(5)+ 1 (2 + 4)(10)+ 1 (5 +10)(4)+ 1 (5)(4)+ 1 (15 + 5)(4)

22 2 22

Total distance = 115 m

−2 ii.

a (m s ) Displaceme nt = area under the graph of v-t

0.8 = A1 + A2 + A3 + A4 + A5
0.6
0.4 Displacement = 1 (2)(5)+ 1 (2 + 4)(10)+ 1 (5 +10)(4)+ 1 (5)(− 4)+ 1 (15 + 5)(− 4)
00..22
22 2 2 2
0
- 0.2 Displaceme nt = 15 m
-0.4
-0.6 d.
-0.8
t (s) aav = v2 − v1
5 10 15 20 25 30 35 40 45 50 t2 − t1

aav = −4−4
40 − 20

aav = −0.4 m s−2

Test your concept
1. Can you accelerate a body without speeding

up or slowing down? Is it possible?
2. A car is traveling at 30 km h–1 to the north.

Then it turns to the west without changing its
speed. Is the car accelerating?
3. How would you draw a displacement time
graph for a stationary object?
4. What would the gradient of a distance time
graph represent?
5. What does the area a speed - time graph
represent?

6

2.2 UNIFORMLY ACCELERATED MOTION v = 22.62 ms−1

Learning Outcomes: Convert to km h –1

At the end of this chapter, students should be 22.62 km
able to: v = 1000
1 hour
• Apply equations of motion with uniform 60(60)
acceleration:
v = 81.43 km hour−1
v = u + at
s = ut + 1 at2 (b) Assume: time elapsed, t

2 v = u + at
v2 = u2 + 2as 22.62 = 27.78 + (−6.5)t

Kinematics Equation for uniform acceleration − 6.5t = −5.16
Assume a car has uniform acceleration & consider the t = 5.16
motion between X and Y: 6.5
t = 0.794 s

u = initial velocity ( velocity on passing X ) Example 2.2.2
v = final velocity ( velocity on passing Y ) A park ranger driving on a back country road
a = acceleration suddenly sees a deer ‘frozen’ in his headlights. The
s = displacement ( in moving from X to Y ) ranger, who is driving at 11.4 m s−1 immediately
t = time taken ( to move from X to Y ) applies the brakes and slows with an acceleration of
3.8 m s−2.
Example 2.2.1
The driver of a pickup truck going 100 km h–1 applies a) If the deer is 20.0 m from the ranger’s vehicle
the brakes, giving the truck a uniform deceleration of when the brakes are applied, how close does
6.50 m s–2 while it travels 20.0m. the ranger come to hitting the deer?
(a) What is the speed of the truck in kilometers per
b) How much time is needed for the ranger’s
hour at the end of this distance? vehicle to stop?
(b) How much time has elapsed?
Solution:

Solution: sGiven: u = 11.4 m s−1; a = − 3.80s m s−2
Given:
1st find the distance traveled before stopped
a = – 6.50 m s–2 (deceleration)
s = 20.0 m • −1 −2
u = 100 km hour –1
Given: u = 11.4 m s ; a = − 3.80 m s
= 100(1000)m = 27.78 m s−1 st
60(60)s • 1 find the distance traveled before stopped

(a) Final velocity, v =? From:
v2 = u2 + 2as
v2 = u2 + 2as
v2 = (27.78)2 + 2(−6.5)(20)
(0)2 = (11.4)2 + 2(−3.8)(s)

7

s = − 129.96 (c) Using:
− 7.6
v = u + at
= 17.1 m a = v−u

The distance between the stopped vehicle & deer: t
x = 20 −17.1 = 2.9 m = 0−4

(b) Time needed to stop =? 1
a = −4 m s−2
From:

v = u + at

0 = 11.4 + (−3.8)t (d) Displacement, s = area under the graph

t = 3.0 s s = 1 (3 + 6)(4)
2
Example 2.2.3 s = 18 m
A toy car moves with an acceleration of 2 m s–2 from
(e) Acceleration – time graph
rest for 2.0 s. It then moves with constant velocity for
–2
another 3.0 s. It finally comes to rest after another 1.0
a (m s )
s.

a) Sketch a velocity-time graph to shown the
motion of the toy car.

b) What is the velocity of the toy car after first 2
seconds?

c) Calculate the deceleration of the car.
d) What is the total displacement of the car for

the whole journey?
e) Sketch the acceleration-time graph for the

motion of toy car.

Solution: Deceleration 2 56 t (s)
(a) 02
v decreases from
v (m s–1)
–1 –1

4 m s to 0 m s in 1s

–4
4

t (s) 2.3 PROJECTILE MOTION

02 56 Learning Outcomes:
(b) Using:
At the end of this chapter, students should be
v = u + at able to:
= 0 + 2(2)
a) Describe projectile motion launched at an
v = 4 m s−1 angle,  as well as special cases when  = 0
and  = 90 (free fall).

b) Solve problems related to projectile
motion.

8

PROJECTILE MOTION - the object moves upward or downward it also
A motion where object travels at uniform velocity in moving horizontally.
horizontal direction; at the same time undergoing
acceleration in downward direction under the Relationship Between Horizontal & Vertical
influence of gravity. Component

2 dimensional motion: Physical Horizontal Vertical Magnitude
▪ Horizontal (x-component) Quantiti (x- (y-
▪ Vertical (y-component)
es component) componen
t)

Displace sx sy s= s2 + sy2
ment, s x

Initial ux uy u= u2 + u 2
velocity, x y

u

Final vx vy v= v2 + vy2
Velocity x

,v

Acceler ax = 0 ag = −g -
ation, a

Direction,    = tan−1 vy

vx

Kinematics Equations

Linear Horizontal Vertical

(x-component) (y-component)

v = u + at vx = ux vy = uy − gt
sx = uxt
s = ut + 1 at2 sy = uyt − 1 gt 2
2 vx = ux 2

Consider an object thrown with a velocity u at an v2 = u2 + 2as vy2 = uy2 − 2gsy
angle θ° relative to the horizontal.

✓ At maximum height, vy = 0
✓ ux = u cos 
✓ uy = u sin 

S (Horizontal sy ( vertical
x displacement )

displacement

)



–2 9

a =0ms
x

a =–g
y

i. Projectile Motion Case 1 iii. Projectile Motion Case 3
(special case,  = 0)

 = 0

Initial point

 Explanation:
▪ An object thrown horizontally from a certain
s = range, R
height with a velocity, u at angle  = 0.
x ▪ Also called as zero launch angle / horizontal

s =0 projectile.
▪ ux = u
y ▪ uy = 0
▪ sy = −ve value (final point is below/lower
Explanation:
▪ An object thrown with a velocity, u at angle initial point)

 relative to the horizontal. iv. Projectile Motion Case 4
▪ ux = u cos 
▪ uy = u sin 
ii. Projectile Motion Case 2


Initial point

Explanation: Explanation:
▪ An object thrown from a certain height with a ▪ An object thrown from a certain height with a

velocity, u at angle  above the horizontal velocity, u at angle  below the horizontal
line.
line. ▪ ux = u cos 
▪ ux = u cos 
▪ uy = u sin  10
▪ sy = −ve value (final point is below/lower

initial point)

▪ uy = −u sin  From u = 30 m s–1 & θ = 35˚, resolved u into x & y
▪ sy = −ve value (final point is below/lower comp.

initial point) ux = u cos 35° = 30 cos 35°= 24.6 m s–1
uy = u sin 35° = 30 sin 35° = 17.2 m s–1
v. Projectile Motion Case 5
(special case,  = 90) (a) Maximum height (comp – y);
At maximum height, vy = 0
Using:
vy2 = uy2 – 2gsy
(0)2 = (17.2 )2 – 2 (9.81) (sy)
sy = 295.84
19.62
= 15.1m

(b) Range =?
Max. horizontal displacement (comp – x)
R = Sx = ux t;

Explanation: to find R, must know value of time of flight, T.
▪ Object in motion solely under the influence of
→ Time of flight, T = 2 tup
gravity
▪ Also called as free fall motion. At max. height, vy = 0
▪ Does not involved x-component. From: vy = uy – gt
(0) = (17.2) – 9.81(tup)
(use equation for y-comp only) tup = 17.2 = 1.753s
▪ uy = +ve (thrown upward) 9.81
▪ uy = −ve (thrown downward)
▪ uy = 0 (dropped/released) Total flight time, T = 2 tup = 2(1.753) = 3.506 s
Procedure for Solving Projectile Motion Problems Range, R = Sx(max) = ux (T)
1. Separate the motion into the x (horizontal)
= 24.6 (3.506)
part and y (vertical) part. = 86.25 m
2. Consider each part separately using the
Example 2.3.2
appropriate equations for x and y motion. A ball is projected from a height of 25.0 m above the
ground. It is thrown with an initial horizontal velocity
Example 2.3.1 of 8.25 m s–1.
A cannonball is fired with an initial velocity of 30.0 m
s–1 at an angle of 35° to the horizontal. a) How long is the ball in flight before striking
(a) What is the maximum height reached by the ball? the ground?
(b) What is its range?
Solution: b) How far from the building does the ball strike
Given: u = 30.0 m s–1 ; θ = 35° ; ay = g = 9.81 m s–2 the ground?

c) What is the velocity of the ball just before it
strikes the ground?

u= 30 m s–1 H

θ=35°

11

R

Solution:

Consider x and y separately: v is 23.66 m s–1 at an angle 69.59° below
+x axis.
x-motion y-motion
Example 2.3.3
sx = ? sy = – 25.0 m A hockey player hits a “slap shot “in practice (with no
ux = 8.25 m/s uy = 0 goalie present) when he is 15.0 m directly in front of
the net. The net is 1.20 m high and the puck is initially
ax = 0 ay = – 9.81 m/s 2 hit at an angle of 5° above the ice with a speed of 35.0
m s–1.

a) Make a sketch of the situation using x - y
coordinates, assuming that the puck is at the
origin at the time it is hit. Be sure to locate the
net in the sketch and show its height.

b) Determine if the puck makes it into the net.

Solution:

(a) From: Sy = uyt – ½ g t 2 s =
(– 25) = (0) t – ½ (9.81) (t2) y

t = 2(25) = 2.26 s s = 15 m
9.81 x

(b) (a)
For the puck goes into the net,
sx = ux(t ) → Sy ≤ 1.20 m
initially: θ = 5° ; u = 35.0 m s–1 ; Sx = 15 m
= (8.25)(2.26)
Resolved u into x & y component:
= 18.6 m ux = u cos 5 = 34.9 m s–1
uy = u sin 5 = 3.05 m s–1
(c) In horizontal, -- constant velocity motion.
In order to find Sy, we must calculate t using
x- component of the velocity is unchanged
vx = ux = 8.25 m s–1 Sx ;

In vertical, g acts on the object, so velocity From : Sx = ux t
t = sx = 15.0 = 0.43s
changed with time. ux 34.9

From: From : Sy = uy (t) – ½ g t 2
= (3.05)(0.43) – ½ (9.81)(0.43) 2
vy = uy – g t = 1.31 – 0.906
= (0) – 9.81(2.26) = 0.40 m
= – 22.17 m s–1 12

i. Velocity, v = vx2 + vy2

= (8.25)2 + (−22.17)2

= 23.66 ms−1 vx
v
ii. Direction of v: 

tan = vy = 22.17
vx 8.25
  = 69.59

vy

sy ≤ 1.2 m → puck goes into the net Solution:

Example 2.3.4 a. At the maximum height, H, vy = 0 and u = uy
A book is dropped 150 m from the ground. Determine = 10.0 m s−1 thus
a. the time taken for the book reaches the ground.
b. the velocity of the book when it reaches the ground. v 2 = u 2 − 2gsy
(given g = 9.81 m s-2) y y

0 = (10.0)2 − 2(9.81)H

H = 5.10 m

Solution: b. From point A to C, the vertical displacement,

a. sy= 0 m thus

The vertical displacement is Sy = −150 m sy = uyt − 1 gt 2
2
Hence

sy = uyt − 1 gt 2 0 = (10.0)t − 1 (9.81)t2
2
2

−150 = 0 − 1 (9.81)t 2 t = 2.03 s

2 c. From point A to D, the vertical displacement,
sy= −30.0 m thus
t = 5.53 s
b. The book’s velocity is given by

vy = uy − gt sy = uyt − 1 gt 2
2
vy = 0 − (9.81)(5.53)

vy = −54.2 m s−1 − 30.0 = (10.0)t − 1 (9.81)t2

Or 2

vy2 = uy2 − 2gsy 4.91t 2 −10.0t − 30.0 = 0

vy2 = 0 − 2(9.81)(−150 ) t = 3.69 s or −1.66 s Time don’t
have negative
vy = 54.2 m s−1 value.

d. Time taken from A to D is t = 3.69 s thus

Example 2.3.5 vy = uy − gt

A ball is thrown from the top of a building is given an vy = (10.0)− (9.81)(3.69)
initial velocity of 10.0 m s−1 straight upward. The
vy = −26.2 m s−1
building is 30.0 m high and the ball just misses the
OR
edge of the roof on its way down, as shown in figure From A to D, sy = −30.0 m

2.7. Calculate vy2 = uy2 − 2gsy

a. the maximum height of the stone from point A. vy2 = (10.0)2 − 2(9.81)(− 30.0)

b. the time taken from point A to C. vy = 26.2 m s−1
Therefore, the ball’s velocity at D is
c. the time taken from point A to D.
vy = −26.2 m s−1
d. the velocity of the ball when it reaches point D.

(Given g = 9.81 m s−2) B

−1 C

u =10.0 m s

A

30.0 m

D
13