Note Chap 2 SP015 (Kinematics of linear motion) Zahasnida

Chapter 02 Physics

CHAPTER 2:
Kinematics of linear motion

(F2F:6.5 hours)

1

Overview: Chapter 02 Physics

Motion

Linear Uniformly Projectile
(1-dimensional) accelerated (2-dimensional)

motion

2

Chapter 02 Physics

Learning Outcome:

2.1 Linear Motion (2.5 hours)

At the end of this chapter, students should be able
to:

 Define

 instantaneous velocity, average velocity and
uniform velocity

 instantaneous acceleration, average
acceleration and uniform acceleration

 Discuss the physical meaning of displacement-
time, velocity-time and acceleration-time graphs.

 Determine the distance travelled, displacement,

velocity and acceleration from appropriate

graphs. 3

Chapter 02 Physics

2.1. Linear motion (1-D)

2.1.1. Distance, d
 scalar quantity.
 is defined as the length of actual path between two points.
 The S.I. unit of distance is metre (m).
 For example :

Q

P

 The length of the path from P to Q is 25 cm.

4

Chapter 02 Physics


2.1.2 Displacement, s

 vector quantity
 is defined as the distance between initial point and final

point in a straight line.
 The S.I. unit of displacement is metre (m).

Example 2.1 :
An object P moves 30 m to the east after that 15 m to the south
and finally moves 40 m to west. Determine the displacement of P
relative to the original position.

5

Solution : Chapter 02 Physics

W N

O 30 m E
 30 m 15 m


P 10 m

S 6

The magnitude of the displacement is given by

OP  152  102  18 m

and its direction is

θ  tan 1 15   56 from west to south
 10 

Chapter 02 Physics

2.1.3 Speed, v

 is defined as the rate of total distance travelled.

 scalar quantity.

 Equation:

speed  total distance travelled
time interval

v  d

Δt

7

 Chapter 02 Physics
2.1.4 Velocity, v

At the end of this lesson, students should be able to:
 Define instantaneous velocity, average velocity and uniform velocity.

 is a vector quantity.

 The S.I. unit for velocity is m s-1.

Average velocity, vav
 is defined as the rate of change of displacement.

 Equation: vav  change of displacement
time interval

vav  s2  s1 Δs
t2  t1 vav  Δt

 Its direction is in the same direction of the change in 8
displacement.

Chapter 02 Physics

Instantaneous velocity, v

 is defined as the instantaneous rate of change of
displacement.

 Equation: v  limit s
t  0 t

v  ds
dt

 An object moves in a uniform velocity when

ds  constant
dt

and the instantaneous velocity equals to the average

velocity at any time. 9

Chapter 02 Physics

s

s1 The gradient of the tangent to the curve at point Q

Q

= the instantaneous velocity at time, t = t1

0 t1 t

 Therefore

Gradient of s-t graph = velocity

10

Chapter 02 Physics


2.1.5 Acceleration, a

At the end of this lesson, students should be able to:
 Define instantaneous acceleration, average acceleration and uniform

acceleration.

 vector quantity

 The S.I. unit for acceleration is m s-2.

Average acceleration, aav
 is defined as the rate of change of velocity.

 Equation:

aav  changeof velocity aav  v2  v1 aav  Δv
time interval t2  t1 Δt

 Its direction is in the same direction of change in velocity.

 The acceleration of an object is uniform when the magnitude

of velocity changes at a constant rate and along fixed

direction. 11

Chapter 02 Physics

Instantaneous acceleration, a

 is defined as the instantaneous rate of change of velocity.

 Equation: limit v
a
t  0 t

a  dv  d 2s
dt dt 2

 An object moves in a uniform acceleration when

dv  constant
dt

and the instantaneous acceleration equals to the average
acceleration at any time.

12

Chapter 02 Physics

Deceleration, a

 is a negative acceleration.

 The object is slowing down meaning the speed of the object
decreases with time.
v

v1 Q

0 The gradient of the tangent to the curve at point Q

 Therefore = the instantaneous acceleration at time, t = t1

t1 t

Gradient of v-t graph = acceleration

13

2.1.6 Graphical methods Chapter 02 Physics

At the end of this lesson, students should be able to:
 Discuss graphs of displacement-time

Displacems ent against time graph (s-t)
s

Gradient = constant Gradient increases
with time
0 t
0t
s (a) Uniform velocity
(b) The velocity increases with time

(c) Q

R Gradient at point R is negative.

P The direction of
velocity is changing.
Gradient at point Q is zero.

0t The velocity is zero. 14

Chapter 02 Physics

Velocity versus time graph (v-t) B
C
At the end of this lesson, students should be able to:
 Discuss graphs of velocity-time

v vv

Uniform

Uniform velocity acceleration

0 t1 (a) t2 t0 t1 (b) t2 A t2(c) t

t 0 t1

Area under the v-t graph = displacement

 The gradient at point A is positive – a > 0(speeding up)
 The gradient at point B is zero – a= 0
 The gradient at point C is negative – a < 0(slowing down) 15

Chapter 02 Physics

 From v  ds
Therefore dt

ds  vdt

 ds   vdt

s  t2 vdt
t1

s  sha ded area under the v  t graph

16

Chapter 02 Physics

Example 2.2 :
A toy train moves slowly along a straight track according to the
displacement, s against time, t graph in Figure 2.1.

s (cm)

10

8
6

4

2

Figure 2.1 0 2 4 6 8 10 12 14 t (s) 17
a. Explain qualitatively the motion of the toy train.
b. Sketch a velocity (cm s-1) against time (s) graph.
c. Determine the average velocity for the whole journey.
d. Calculate the instantaneous velocity at t = 11 s.
e. Determine the distance travelled by the toy train.

Chapter 02 Physics

Solution :
a. 0 to 10 s : The train at rest.

10 to 14 s : The train moves in positive direction at a constant
velocity of 1.50 cm s1.

b.
v (cm s1)

1.50

0.68

0 2 4 6 8 10 12 14 t (s)

18

Chapter 02 Physics

Solution : s2  s1
t2  t1
c. vav 

 10  4
14  0

vav  0.429 cm s1

d. v  average velocity from10 s to14 s 19
v  s2  s1
t2  t1
v  10  4
14 10

v  1.50 cm s1

Chapter 02 Physics

Solution :
e. The distance travelled by the train, d is given by

d  area under the graph v-t

 1.5014 10

d  6.0 cm

20

Chapter 02 Physics

Example 2.3 :
A velocity-time (v-t) graph in Figure 2.2 shows the motion of a lift.

v (m s 1)

4
2

0 5 10 15 20 25 30 35 40 45 50 t (s)

-2
-4

Figure 2.2

a. Describe qualitatively the motion of the lift.
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. 21

Chapter 02 Physics

Solution : Lift moves upward from rest with a constant
a. 0 to 5 s : acceleration of 0.4 m s2.

5 to 15 s : The velocity of the lift increases from 2 m s1
to 4 m s1 but the acceleration decreasing to
15 to 20 s : 0.2 m s2.
20 to 25 s :
25 to 30 s : Lift moving with constant velocity of 4 m s1.
30 to 35 s :
Lift decelerates at a constant rate of 0.8 m s2.
35 to 40 s :
Lift at rest or stationary.
40 to 50 s :
Lift moves downward with a constant
acceleration of 0.8 m s2.

Lift moving downward with constant velocity

of 4 m s1.

Lift decelerates at a constant rate of 0.4 m s2

and comes to rest. 22

Chapter 02 Physics

Solution :
b. a (m s2)

0.8
0.6
0.4
0.2

0 5 10 15 20 25 30 35 40 45 50 t (s)

-0.2
-0.4
-0.6
-0.8

23

Chapter 02 Physics

Solution : v (m s 1)
c. i.

4

2 A1 A2 A3
5
0 10 15 20 25 30 A435 40 45 50 t (s)
-2 A5

-4

Total distance  area under the graph of v-t

 A1  A2  A3  A4  A5

Total distance  1 25 1 2  410 1 5 104 1 54 1 15  54

22 2 22

To tal d istance  1 1 5 m

24

Solution : Chapter 02 Physics
c. ii.
Displacement  area under the graph of v - t

 A1  A2  A3  A4  A5

Displacement  1 25 1 2  410 1 5 104 1 5 4 1 15  5 4

22 2 2 2

Displacement  15 m

d. aav  v2  v1
t2  t1

aav  44
40  20

aav  0.4 m s2

25

Chapter 02 Physics

Exercise 2.1 :

1. Figure 2.3 shows a velocity versus time graph for an object
constrained to move along a line. The positive direction is to
the right.

Figure 2.3

a. Describe the motion of the object in 10 s.

b. Sketch a graph of acceleration (m s-2) against time (s) for

the whole journey.

c. Calculate the displacement of the object in 10 s. 26
ANS. : 6 m

Chapter 02 Physics

2. A train pulls out of a station and accelerates steadily for 20 s
until its velocity reaches 8 m s1. It then travels at a constant
velocity for 100 s, then it decelerates steadily to rest in a further
time of 30 s.
a. Sketch a velocity-time graph for the journey.
b. Calculate the acceleration and the distance travelled in
each part of the journey.
c. Calculate the average velocity for the journey.
Physics For Advanced Level, 4th edition, Jim Breithaupt, Nelson
Thornes, pg.15, no. 1.11

ANS. : 0.4 m s2,0 m s2,-0.267 m s2, 80 m, 800 m, 120 m;
6.67 m s1.

27

Chapter 02 Physics

Learning Outcome:

2.2 Uniformly accelerated motion (2 hour)
At the end of this chapter, students should be able
to:
 Apply equations of motion with uniform

acceleration:

v  u  at

s  ut  1 at2
2

v2  u2  2as

28

Chapter 02 Physics

2.2. Uniformly accelerated motion

At the end of this lesson, students should be able to:
 Apply equations of motion with uniform acceleration.

 From a  vu
t
(1)
v  u  at

where v : final velocity
u : initial velocity
a : uniform (constant) acceleration
t : time

29

Chapter 02 Physics

 From equation (1),
velocity

v

u

Figure 2.4 t time

0
 From the graph,

The displacement after time, s = shaded area under the
graph

= the area of trapezium

 Hence, s  1 u  vt (2)

2 30

Chapter 02 Physics

 By substituting eq. (1) into eq. (2) thus (3)

s  1 u  u  att

2

s  ut  1 at2
2

 From eq. (1), v u  at
 From eq. (2),
v  u  2s multiply

t

v  uv  u   2s at 

t

v2  u2  2as (4)

31

Chapter 02 Physics

 Notes:
 equations (1) – (4) can be used if the motion in a straight
line with constant acceleration.

 For a body moving at constant velocity, ( a = 0) the
equations (1) and (4) become

vu

Therefore the equations (2) and (3) can be written as

s  vt constant velocity

32

Chapter 02 Physics

Example 2.4 :

A plane lands on a runway at velocity 50 m s1 and decelerates at
constant rate. The plane travels 1.0 km before stops. Calculate

a. the deceleration of the plane.

b. the time taken for the plane to stop.
u  50 m s1 a  ?
Solution : v0

a. Use v2  u 2  2as s  1000 m

t?

0  502  2a1000

a  1.25 m s2

Hence the deceleration of the plane is 1.25 m s2.

33

Chapter 02 Physics

Solution :
b. By using the equation of linear motion,

v  u  at

0  50  1.25t

t  40 s

OR

s  ut  1 at2
2

1000  50t  1 1.25t2

0.625t 2  50t 1000  0 2

t  40 s

34

Chapter 02 Physics

Example 2.5 :

A bus travelling steadily at 30 m s1 along a straight road passes a
stationary car which, 5 s later, begins to move with a uniform
acceleration of 2 m s2 in the same direction as the bus.
Determine

a. the time taken for the car to acquire the same velocity as the
bus,

b. the distance travelled by the car when it is level with the bus.

Solution : vb  30 m s1  constant; uc  0; ac  2 ms 2

a. Given vc  vb  30 m s1 35
Use vc  uc  actc

30  0  2tc

tc  15 s

Chapter 02 Physics

b.

b vb  30 m s1 b vb b vb

c uc  0 ac  2 m s2 c

tb  0 s tb  5 s tb  t

From the diagram, sc  sb

tb  t; tc  t  5 Therefore

0 u1ctc212t sc  sb sc  vbt

actc 2  vbtb sc  3039.4

 52  30t sc  1183 m

2 t  39.4 s

36

Chapter 02 Physics

Example 2.6 :

A particle moves along horizontal line according to the equation

s  t 3  2t

Where s is displacement in meters and t is time in seconds.
At time, t =2 s, determine
a. the displacement of the particle,

b. Its velocity, and

c. Its acceleration.

Solution :

a. t =2 s ; s  t 3  2t

 23  22

s  12 m

37

Chapter 02 Physics

Solution :
b. Instantaneous velocity at t = 2 s,

Use v  ds
dt

 d t3  2t 
dt

v  3t 2  2

Thus v  32 2  2

v  14 m s1

38

Chapter 02 Physics

Solution :
c. Instantaneous acceleration at t = 2 s,

Use a  dv
dt

 d 3t 2  2
dt

a  6t

Hence a  62 

a  12 m s2

39

Chapter 02 Physics

Exercise 2.2 :

1. A speedboat moving at 30.0 m s-1 approaches stationary
buoy marker 100 m ahead. The pilot slows the boat with a
constant acceleration of -3.50 m s-2 by reducing the throttle.

a. How long does it take the boat to reach the buoy?

b. What is the velocity of the boat when it reaches the buoy?

No. 23,pg. 51,Physics for scientists and engineers with
modern physics, Serway & Jewett,6th edition.

ANS. : 4.53 s; 14.1 m s1

2. An unmarked police car travelling a constant 95 km h-1 is
passed by a speeder traveling 140 km h-1. Precisely 1.00 s

after the speeder passes, the policemen steps on the
accelerator; if the police car’s acceleration is 2.00 m s-2, how

much time passes before the police car overtakes the

speeder (assumed moving at constant speed)?

No. 44, pg. 41,Physics for scientists and engineers with

modern physics, Douglas C. Giancoli,3rd edition.

ANS. : 14.4 s 40

Chapter 02 Physics

3. A car traveling 90 km h-1 is 100 m behind a truck traveling 75
km h-1. Assuming both vehicles moving at constant velocity,
calculate the time taken for the car to reach the truck.
No. 15, pg. 39,Physics for scientists and engineers with
modern physics, Douglas C. Giancoli,3rd edition.

ANS. : 24 s
4. A car driver, travelling in his car at a constant velocity of 8

m s-1, sees a dog walking across the road 30 m ahead. The
driver’s reaction time is 0.2 s, and the brakes are capable of
producing a deceleration of 1.2 m s-2. Calculate the distance
from where the car stops to where the dog is crossing,
assuming the driver reacts and brakes as quickly as
possible.
ANS. : 1.73 m

41

Chapter 02 Physics

Learning Outcome:

2.3 Projectile motion (2 hours)
At the end of this chapter, students should be able

to:
 Describe projectile motion launched at an angle,

 as well as special cases when  =0 and 

=90 (free fall).
 Solve problems on projectile motion.

42

2.3. Projectile motion Chapter 02 Physics

At the end of this lesson, students should be able to:

 Describe projectile motion launched at an angle,  as well as special
cases when  =0 and  =90 (free fall).

 consists of two components:

 vertical component (y-comp.)

 motion under constant acceleration, ay= g
 horizontal component (x-comp.)

 motion with constant velocity thus ax= 0
y

v1y v1 B v

P 1 sy=H v2y Qvv222x

v1x C
t2
uy u t1 sx= R x

A  43

Figure 2.5 ux

Chapter 02 Physics

 The x-component of velocity along AC (horizontal) at any
point is constant,

ux  u cos θ

 The y-component (vertical) of velocity varies from one point
to another point along AC.

u y  u sin θ

44

Chapter 02 Physics

 Velocities at points P and Q.

Velocity Point P Point Q

x-comp. v1x  ux  u cosθ v2x  ux  u cosθ

y-comp. v1y  uy  gt1 v2 y  uy  gt2

       magnitude v1  v1x 2  v1y 2 v2  v2x 2  v2 y 2

direction θ1  tan1 v1y  θ2  tan 1  v2 y 
v1x v2 x

Table 2.1

45

Chapter 02 Physics

2.3.1 Maximum height, H

 The ball reaches the highest point at point B at velocity, v

where

 x-component of the velocity, vx  v  ux  u cosθ

 y-component of the velocity, vy  0

 y-component of the displacement, sy  H

 Use v 2  u 2  2gsy
y y

0  u sin  2  2 gH

H  u 2 sin 2 

2g

46

Chapter 02 Physics

2.3.2 Time taken to reach maximum height, t’

 At maximum height, H

 Time, t = t’ and vy= 0

 Use v y  u y  gt

0  u sin    g t '

t'  u sin 

g

2.3.3 Flight time, t (from point A to point C)

t  2t'

t  2u sin θ
g

47

Chapter 02 Physics

2.3.4 Horizontal range, R and value of R maximum

 Since the x-component for velocity along AC is constant hence

ux  vx  u cos

 From: s x  u xt and s x  R

R  u cos  t 

R  uu2co2ssin2cuosgsin  
R 
g 

48

Chapter 02 Physics
 From : sin 2  2 sin  c o s 

R  u 2 sin 2

g

 The R is maximum when  = 45 and sin 2 = 1

Rmax  u2
g

49

Chapter 02 Physics

Example 2.7 :

A tennis ball is thrown upward from the top of a building with
velocity 15 m s-1 at an angle 30 to the horizontal. The height of
the building is 40 m. Calculate

a. the maximum height of the ball from the ground.

b. the magnitude of the velocity of the ball just before it strikes the

ground. (given g = 9.81 m s-2) ux  u cos 30
Solution : u = 15 m s1  15 cos 30
 13.0 m s1
30

ax  0 H ? uy  u sin 30
ay  g  15 sin 30
h  40 m  7.50 m s1

v  ? 50