Topic 5 kinematics of particle

CHAPTER 5 KINEMATICS OF PARTICLE
1
Dynamics deals with the accelerated motion of a body.
Dynamics involved:
a) Kinematics is a study of the geometry of the motion.
b) Kinetics is a study of the forces that cause the motion.
Kinematics is the study of the geometry of motion of particles, rigid bodies, etc.,
without considering the forces that cause them to move. It is often referred to as
the "geometry of motion"
There are four (4) kinematic equations,
Since acceleration is uniform
at
u
v 

average
v
t
u
v


2
2
1
at
ut
s 

as
u
v 2
2
2


5.2 CONCEPT OF KINEMATICS OF PARTICLES
5
5.1 INTRODUCTION OF DYNAMICS
s = displacement
a = acceleration
t = time
v = final velocity
u = initial velocity

2
Rectilinear motion is another name for straight-line motion. Position, velocity, and
acceleration of a particle as it moves along a straight line.
The linear motion can be of two types: uniform linear motion with constant velocity
or zero acceleration; non uniform linear motion with variable velocity or non-zero
acceleration.
Where something is located, a position vector r starting from the origin O. It is
specific location of particle P at any instant.
Magnitude of s = Distance from O to P
- The sense (arrowhead direction of r) is defined by algebraic sign on s.
=> +ve = right of origin,
-ve = left of origin
5.3 RECTILINEAR MOTION
5.3.1 Position
1) Single coordinate axis, s
2) Origin, O
3) Position vector r – specific
location of particle P at any instant.
4) Algebraic Scalar s in metres
O
P
r
s
s
Position

3
Displacement – change in its position, vector quantity
If particle moves from P to P’
is +ve if particle’s position is right of its initial position
is -ve if particle’s position is left of its initial position
Speed is a measure of how fast something moves. Speed refers to the distance
traveled by an object during a specific amount of time.
time
ce
dis
Speed
tan

Units for measuring speed: km/h, mi/h (mph), m/s
5.3.1 Speed
5.3.1 Displacement
P’
O P
r
s
s
Δr
Δs
s'
r'
Displacement
Scalar quantity,
specified only by
its magnitude.

4
Average speed is the whole distance covered divided by the total time of travel.
Average speed = total distance travelled/time taken
average
v
t
u
v


Velocity is defined as a speed in a given direction; the rate of change of
displacement of a particle.
Velocity = speed and direction; velocity is a vector.
 
s
m
dt
ds
v 
Constant velocity = constant speed and no change in direction
Average velocity, t
r
vavg



Instantaneous velocity is defined as
dt
dr
vins 

5.3.2 Average Speed
5.3.3 Velocity
P’
O P
s
Δs
v
Velocity

5
Acceleration is the rate of change of velocity of a particle per unit of time. It tells
you how fast (the rate) velocity changes:
Acceleration = change in velocity/time interval
Acceleration is a vector
Define the following terms;
i. kinematics of particle
ii. rectilinear motion
iii. position
iv. displacement
v. average speed
vi. velocity
vii. constant velocity
viii. constant acceleration
5.3.4 Acceleration
Example 1

6
A car accelerates from 4 m/s to 40 m/s in 20 seconds. Determine
i. the acceleration
ii. the average velocity
iii. distance travelled
s
m
v 4
1 
s
m
v 40
2  sec
20

v
i. Acceleration
s
m
t
v
v
a 8
.
1
20
4
40
1
2





ii. Average velocity
s
m
v
v
vavg 22
2
40
4
2
2
1





iii. Distance travelled
m
x
xt
v
s avg 440
20
22 


Example 2

7
A car travelling at 110 km/h decelerates uniformly to rest at rate 4 m/s2 . Calculate
the time and distance taken to stop.
s
m
x
v 56
.
30
3600
1000
110
1 

s
m
v 0
2  2
4
s
m
a 
i. Time taken
t
v
v
a 1
2 

sec
64
.
7
4
56
.
30
0
1
2





a
v
v
t
ii. Distance
as
u
v 2
2
2


 
m
a
v
v
s 74
.
116
4
2
56
.
30
0
2
2
2
2
1
2
2





Example 3

8
A vehicle accelerates from 5 m/s to 45 m/s in 20 seconds. Determine the
acceleration. Also find the average velocity and distance travelled.
.
m
x
vt
d
cetravelle
Dis
v
v
ocity
Averagevel
t
v
v
on
Accelerati s
m
500
20
25
tan
25
2
5
45
2
2
20
5
45
1
2
1
2
2













A missile is fired vertically with an initial velocity of 300 m/s. It is acted on by
gravity. Calculate;
i. The height it reaches
ii. The time taken to go up and down again.
s
m
v 300
1 
s
m
v 0
2  2
81
.
9
s
m
a 

i. The height it reaches
as
v
v 2
2
1
2
2 

Example 5
Example 4

9
 s
81
.
9
2
300
0 2
2



 
 
m
s 15
.
4587
81
.
9
2
300
2




ii. Time taken
t
v
v
a 1
2 

sec
58
.
30
81
.
9
300
0
1
2






a
v
v
t
A graph can be established describing the relationship with any two of the
variables, a, v, s, t
Using the kinematics equations a = dv/dt, v = ds/dt, a ds = v dv
To determine the particle’s velocity as a function of time, the v-t Graph, use v =
ds/dt. Velocity as any instant is determined by measuring the slope of the s-t
graph
v
dt
ds

Slope of s-t graph = velocity
In figure 5.1 are examples of the displacement-time graphs you will encounter.
5.4 GRAPHICAL METHOD OF RECTILINEAR MOTION
5.4.1 The s-t Graph

10
Time (t)
Time (t)
Displacement
(s)
Displacement
(s)
An object stationary
over a period of time.
The gradient is zero,
so the object has zero
velocity.
Displacement
(s)
Time (t)
An object moving at a constant
velocity. The displacement is
increasing as time goes on. The
gradient, however, stays constant,
so the velocity is constant. Here the
gradient is positive, so the object is
moving in the direction we have
defined as positive.
An object moving at a constant
acceleration. Both the displacement
and the velocity (gradient of the
graph) increase with time. The
gradient is increasing with time,
thus the velocity is increasing with
time and the object is accelerating.

11
Velocity
(m/s)
Time (t) (sec)
Constant
deceleration
Constant
acceleration
Constant velocity
The gradient of the velocity-time gradient gives a value of the changing rate in
velocity, which is the acceleration of the object. The area below the velocity-time
graph gives a value of the object's displacement.
On a velocity-time graph a horizontal (flat) line indicates the object is travelling at
a constant velocity.
A straight diagonal line indicates the object's velocity is changing.
In the graph on the left, the line sloping upwards shows the object is accelerating
and the line sloping downwards in this case towards v = 0, shows it
is decelerating.
A lift is accelerated from rest to 15 m/s at a rate of 5m/s2 . It then moves at
constant velocity fro 8 seconds and then decelerates to rest at 1.4 m/s2 . Draw the
velocity-time graph and deduce the distance travelled during the journey. Also
deduce the average velocity for the journey.
5.4.1 The v-t Graph
Velocity – Time Graphs
Example 6

12
Time taken for accelerate
t
v
v
a 1
2 

sec
3
5
0
15
1
2





a
v
v
t
Time taken for decelerate
sec
71
.
10
4
.
1
15
0
1
2






a
v
v
t
Distance travelled (area under graph)
     
m
x
x
x
s
75
.
222
25
.
80
120
5
.
22
15
71
.
10
2
1
15
8
15
3
2
1







Velocity
(m/s)
Time (t) (sec)
a=5m/s
2
a=-1.4m/s2
t2
t1 8
15

13
Average velocity
s
m
vavg 26
.
10
71
.
21
75
.
222


A car start from rest and accelerates uniformly for 7 seconds and reaches a
velocity of 5 m/s at the end of the acceleration. Its velocity is maintained for 12
seconds and then it decelerates and finally stop in 4 seconds.
i. Draw the graph of velocity versus time
ii. Calculate distance travelled for the journey
iii. Calculate the average velocity
iv. Calculate the acceleration over first part of the journey
.
i. Graph of velocity versus time
Example 7
Velocity
(m/s)
Time (t) (sec)
A
23
7 19
5
B C

14
ii. Distance travelled
Total distance = area A + area B + area C
   
   
 
m
x
x
x
s
5
.
87
10
60
5
.
17
5
19
23
2
1
5
7
19
7
5
2
1









iii. the average velocity
s
m
vavg 8
.
3
23
5
.
87


iv. acceleration over first part of the journey
acceleration = first slope of graph
2
71
.
0
0
7
0
5
s
m
a 



A car moves with a uniform velocity of 20 m/s for 25 seconds and then accelerates
uniformly at 4m/s2 to a velocity of 30 m/s. After that it continues with that velocity of
30 m/s before it decelerates and finally stop in 20 seconds. The total time the car
takes to stop is 60 seconds
i. Draw the graph of velocity versus time
ii. Calculate the deceleration
iii. Calculate the time taken for constant acceleration
Example 8

15
i. graph of velocity versus time
ii. Decelerate
2
1
2
5
.
1
20
30
0
s
m
t
v
v
a 





iii. Time taken for constant acceleration
t
v
v
a 1
2 

sec
2
5
20
30
1
2





a
v
v
t
Velocity
(m/s)
Time (t) (sec)
a=5m/s2
20 sec
20
15
30
15
25
15

16
Non uniform acceleration means that the acceleration of an object is
not constant, but either increasing or decreasing over a passage of time, as
shown in the diagram above. It refers to variation in the rate of change in velocity.
This variation can be expressed either in terms of time (t) or position (s).
Instantaneous acceleration at time t is found by taking smaller and smaller values
of Δt and corresponding smaller and smaller values of Δv,
dt
dv
dt
s
d
a
dt
ds
v



2
2
We obtain expression for velocity as below


 dt
t
a
v )
(
We obtain expression for position as below
1.4 NON-UNIFORM ACCELERATION
5.3.1 VELOCITY AND ACCELERATION IN EXPRESSED IN
TERM OF TIME (t)

17


 dt
t
v
s )
(
A particle moves along a straight line such that its position is define by s = (2t3 –
2t2 + 8)m. Determine the average velocity, average speed and acceleration of the
particle when t = 3s.
s = (2t3 – 2t2 + 8)
when t=0 sec, s= 8
t=3 sec, s=2(33)-2(32) + 8 = 44
i) V average=
s
m
t
s
Vaverage /
12
3
8
44






ii) Speed average
when t=0 sec, s= 8
t=1 sec, s=2(13)-2(12) + 8 = 8
t=2 sec, s=2(23)-2(22) + 8 = 16
t=3 sec, s=2(33)-2(32) + 8 = 44
s
m
taken
time
travel
Total
Speedaverage /
12
3
36
_
_



iii) Velocity
Example 8

18
t
t
dt
ds
v 4
6 2



when t=3 sec
s
m
v 42
)
3
(
4
)
3
(
6 2



iv) Acceleration
4
12 

 t
dt
dv
a
when t=3 sec
2
32
4
)
3
(
12
s
m
a 


The position of particle which moves along a straight line is defined by the relation
s=3t3-5t2-25t+30, where s is expressed in meter and t in seconds. Determine:
a) The time at which the velocity will be zero
b) The position and distance traveled by the particle at that time
c) The acceleration of the particle at that time
d) The distance traveled by the particle from t=4s to t=6s
Example 8

19
a) The time at which the velocity will be zero
30
25
5
3 2
3



 t
t
t
s
25
10
9 2



 t
t
dt
dx
v
2
.
1
sec
31
.
2
0
25
10
9 2







t
t
t
t
b) The position and distance traveled by the particle at that time
When t = 2.31 sec
30
25
5
3 2
3



 t
t
t
s
      m
s 45
.
17
30
31
.
2
25
31
.
2
5
31
.
2
3
2
3






c) The acceleration of the particle at that time
10
18 

 t
dt
dv
a
  2
58
.
31
10
31
.
2
18
s
m
a 


d) The distance traveled by the particle from t=4s to t=6s.
When t = 4 sec
      m
s 42
30
4
25
4
5
4
3
2
3





      m
s 348
30
6
25
6
5
6
3
2
3





Distance traveled= 348 – 42 =306 m

20
Integrate v dv , assuming that initially v= vo at s= so

 
s
s
c
v
v
ds
a
vdv
0
0
 
0
2
2
2 s
s
a
v
v c
o 


Integrate ac=dv/dt., assuming that initially v= vo when at t=0

 
t
c
v
v
dt
a
dv
0
0
at
v
v o 

5.3.2 Velocity as a Function of Position
5.3.3 Velocity as a Function of Position

21
The car moves in a straight line such that for a short time its velocity is defined by
v = (9t2 + 6t) m/s where t is in sec. Determine it position and acceleration when t =
5s. When t = 0, s = 0.
i. Position.
Noting that s = 0 when t = 0, we have
 
t
t
dt
ds
v 6
9 2



 
 
2
3
0
2
3
0
0
2
0
3
3
3
3
6
9
t
t
s
t
t
s
dt
t
t
ds
t
s
t
s





 

When t = 5s,
    m
s 450
75
375
5
3
5
3
2
3





ii. acceleration
 
6
18
6
9 2





t
t
t
dt
d
dt
dv
a
When t = 5s,
 
2
96
6
5
18
6
18
s
m
t
a





Example 8

22
Rotational Displacement, 
Consider a disk that rotates from A to B:
Angular displacement o
1 rev = 3600 = 2 rad
Measured in revolutions, degrees, or radians.
θ must be expressed in the units of radians for this expression to be valid

r
s 

r
v 
t

 
5.4 RELATIONSHIP BETWEEN LINEAR AND ANGULAR
MOTION
v
r – radius
θ – angle rotated in radian
v – linear velocity
ω - angular velocity
a - linear acceleration
α – angular acceleration
N - revolution per second
s - length of arc AB
- length of arc AB
- linear velocity
- angular velocity

23
N

 2

t
1
2 




A rope is wrapped many times around a wheel of radius 0.6 m. How many
revolutions of the wheel are required to raise a bucket to a height of 25 m?
rad
r
s
7
.
41
6
.
0
25




rad
rev 
2
1 
rev
rev
64
.
6
2
1
7
.
41 









Example 1
h = 25 m
r
- angular acceleration

24
A motorcycle tire has a radius of 35 cm. If the wheel makes 500 revolutions, how
far will the motorcycle have traveled?
rad
rev
rev 6
.
3141
1
2
500 









 
m
rad
r
s 35
.
0
6
.
3141

 
m
s 6
.
1099

Angular velocity ( ω ), is the rate of change in angular displacement.
sec
rad
t

 
Angular velocity can also be given as the frequency of revolution, f (rev/sec or
rpm):
ω = 2 f . Angular frequency f (rev/sec).
Example 2
5.4.1 Angular Velocity

25
A wheel rotates 280º in 6 seconds. Calculate the following.
i) The angle turned in radians?
ii) The angular velocity in rad/s.
Solution:
rad
rev o

2
360
1 

rad
rad
x
o
89
.
4
360
2
280
280 


A rope is wrapped many times around a drum with diameter 60 cm. Calculate the
angular velocity of the drum are required to lift the bucket to 18 m in 10 second?
Example 4
Example 4
h = 18 m
r

26
Solution:
rad
r
s
60
3
.
0
18




sec
6
10
60 rad
t





Angular acceleration ( ) is the rate of change in angular velocity.
2
1
2
sec
rad
t





The angular acceleration can also be found from the change in frequency, as
follows:
 
2
sec
2 rad
t
f



 since  
f

 2

The bucket is lifted from rest until the angular velocity of the wheel is 30 rad/s after
a time of 8 sec. Calculate the average angular acceleration?
Solution:
8
0
30
1
2 



t



2
sec
75
.
3 rad


5.4.2 Angular Acceleration
Example 4

27
From the definition of angular displacement,

r
s 















t
r
t
r
t
s
v


Since
t





Thus,
sec
m
r
v 

From the velocity relationship we have,





















t
r
t
r
t
v
a

.
Since
t





Thus, 2
sec
m
r
a 

5.4.3 Angular and Linear Velocity
Linear speed = angular velocity x radius
5.4.4 Angular and Linear Acceleration
Linear acceleration = angular acceleration x radius

28
t
v
v
s 




 

2
2
1
t







 

2
2
1 


at
v
v 
 1
2 t


 
 1
2
2
1
2
1
at
t
v
s 
 2
1
2
1
t
t 

 

as
v
v 2
2
1
2
2 
 

 2
2
1
2
2 

Define the following terms;
i. kinematics of particle
ii. rectilinear motion
iii. position
iv. displacement
v. average speed
vi. velocity
vii. constant velocity
viii. constant acceleration
REVIEW QUESTIONS
5.4.5 Comparison: Linear and Angular Parameter
EXERCISE 2.1

29
a) Define the following terms:-
i. Constant velocity
ii. Constant acceleration
b) An object being thrown up from the top of 100 meters building from the ground with a
velocity 20 m/s. Determine:-
i. The velocity of an object exactly when it hits the ground
ii. Time taken from it starts until it reaches the ground
A missile was shot with an upward velocity of 90 m/s from the top of a 50 meters
tower. Determine the maximum height of the missile from the ground.
a) A truck travelling along a straight road at 20 km/h and the speed increases its speed 120
km/h in 15 s. Calculate the distance travelled. (The answer must be in SI unit)
a) A ball in Figure 3(b) is thrown vertically upward with a speed of 15 m/s. Determine
the time of flight when it returns to its original position.
EXERCISE 2.1
EXERCISE 2.1
EXERCISE 2.1
EXERCISE 2.1

30
A rocket travel upward at 75m/s. When it is 40m from the ground, the engine fails.
Determine max height sB reached by the rocket and its speed just before it hits the
ground
.
Maximum Height. Rocket traveling upward,
EXERCISE 2.1

31
vA = +75m/s when t = 0. s = sB when vB = 0 at max ht. For entire motion,
acceleration aC = -9.81m/s2 (negative since it act opposite sense to positive
velocity or positive displacement)
)
(
2
2
2
A
B
C
A
B s
s
a
v
v 


sB = 327 m







s
m
s
m
v
s
s
a
v
v
C
B
C
C
B
C
/
1
/
80
/
1
.
80
)
(
2
2
2
2
The negative root was chosen since the rocket is moving downward
a) A train starts from rest at a station with constant acceleration of 2.5 m/s2
until it
achieves velocity of 70 km/h. Then, the train decelerate until it stops in 10 s. Calculate:-
i. Distance travelled by the train
ii. Deceleration of the train
b) A car starts from rest with a constant acceleration of 2.3 m/s2
for 16 seconds. Then, the
car decelerates uniformly at 3 m/s2
and finally stops.
i. Sketch a graph of velocity(v) against time(t) for the journey
ii. Determine the velocity for the car
iii. Calculate the time taken when it uniformly decelerates
EXERCISE 2.1
EXERCISE 2.1

32
A car is travelling along a straight road. The car takes 120 sec to travel between
two sets of traffic light which are 2145 m apart. The car starts from rest at the first
set of traffic lights and moves with constant acceleration for 30 sec until its speed
is 22m/s. The car maintains this speed for T seconds. The car then moves
constant deceleration, coming to rest at the second set of traffic lights.
i. Draw, velocity-time graph
ii. Calculate the value of T
a) A car starts from rest and accelerates uniformly for 70 seconds and reaches a velocity
of 80 m/s at the end of the acceleration. Its velocity is maintained for a while and then it
stops within 65 seconds with constant deceleration. The total distance travelled by the
car is 12.2 km.
i. Draw a Velocity-Time graph
ii. Determine the acceleration of the car
iii. Calculate the time taken for the journey
iv. Determine the deceleration of the car
EXERCISE 2.1
EXERCISE 2.1

33
DISEMBER 2016
b) A train starts from rest at a station with constant acceleration of 2.5 m/s2
until it
achieves velocity of 70 km/h. Then, the train decelerate until it stops in 10 s.
Determine:-
i. Distance travelled by the train
ii. Deceleration of the train
JUN 2016
b) A car starts from rest and accelerates uniformly for 70 seconds and reaches a velocity
of 80 m/s at the end of the acceleration. Its velocity is maintained for a while and then it
stops within 65 seconds with constant deceleration. The total distance travelled by the
car is 12.2 km.
i. Draw a Velocity-Time graph
ii. Determine the acceleration of the car
iii. Calculate the time taken for the journey
c) The motion of a particle is defined by the relation x = 6t4
– 2t3
– 12t2
+ 3t + 3, where x
and t are expressed in meters and seconds respectively. Determine the time, the position
and the velocity when a = 0
EXERCISE 2.1
EXERCISE 2.1
EXERCISE 2.5

34
A particle moves along a straight line such that its position is defined by
s=2t3+3t2-12t-10 m. Determine the velocity, average velocity and average speed of
the particle when t = 3s.
c) A vehicle moves in a straight line such that for a short time its velocity is defined by v
= (0.9t2
+ 0.6t) m/s where t is in second. When t = 3s, determine:-
i. Displacement(s)
ii. Acceleration(s)
b) The motion of a particle is defined by the relation x = 1.5t4
– 30t2
+ 5t + 10, where x
and t are expressed in meters and seconds, respectively. When t = 4s, determine:-
i. the position
ii. the velocity
iii. the acceleration of the particle
EXERCISE 2.1
EXERCISE 2.1
EXERCISE 2.1

35
JUN 2017
c) A car starts from rest with a constant acceleration of 1.25 m/s2
until it achieves 50
km/h. The velocity maintained as far as 1.2 km. When the brake is applied it stops with
10 seconds with constant deceleration.
i. Calculate the total time taken during the journey
ii. Determine the total distance during the journey
The motion of particle is define by the relation x =t3-9t2+24t-6, where x is
expressed in meter and seconds. Determine the position, velocity and acceleration
when t=5s.
A particle moves along a horizontal path with a velocity of v = (3t2 – 6t) m/s. if it is
initially located at the origin O, determine the distance traveled in 3.5s and the
particle’s average velocity and speed during the time interval.
EXERCISE 2.1
EXERCISE 2.1
EXERCISE 2.1

36
Coordinate System. Assuming positive motion to the right, measured from the
origin, O
Distance traveled. Since v = f(t), the position as a function of time may be found
integrating v = ds/dt with t = 0, s = 0.
 
 m
t
t
s
tdt
dt
t
ds
dt
t
t
vdt
ds
s t
t
2
3
0 0
0
2
2
3
6
3
6
3







 

 
 m
t
t
s
tdt
dt
t
ds
dt
t
t
vdt
ds
s t
t
2
3
0 0
0
2
2
3
6
3
6
3







 

0 ≤ t < 2 s -> -ve velocity -> the particle is moving to the left, t > 2a -> +ve velocity -
> the particle is moving to the right
0
0


t
s
m
s
s
t
0
.
4
2




37
m
s
s
t
125
.
6
5
.
3


The distance traveled in 3.5s is
sT = 4.0 + 4.0 + 6.125 = 14.125m
Velocity. The displacement from t = 0 to t = 3.5s is Δs = 6.125 – 0 = 6.125m
And so the average velocity is






 s
m
t
s
aavg /
75
.
1
0
5
.
3
125
.
6
Average speed,
  s
m
t
s
v T
avg
sp /
04
.
4
0
5
.
3
125
.
14






Topic 5 kinematics of particle

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