This belongs to the Horizontal Straight Line Motion of Kinematics.

2. TIPS FOR TAKING ONLINE CLASS
FOR YOURSELF
1: Be ready before the starting time
2: Treat an online course like a real course
3: Hold yourself accountable
4: Practice time management
5: Create a regular study space and stay
organized
6: Eliminate distractions
7: Figure out how you learn best
8: Actively participate
9: Leverage your network
INSIDE THE CLASS
1: Open your camera
2: Mute your audio
3: take your pen, copy and other material
sources
3: Don’t add your note on the screen
4: Do ask, write and share after the
permission. It will be provided at the end of
the class
5: Be more disciplined.

3. UDAY KHANAL
Department of Physics
CCRC

4. The essence of SCIENCE: ask an impertinent question,
and you are on the way to a pertinent answer.

5. MECHANICS
• Science of MACHINE
and ART of making
them
OLD
Definition
• Study of motion of a
body either at REST
or in the MOTION
NEW
Definition

6. MECHANICS
Body at rest
Statics Kinetics
Body at motion
Kinematics Dynamics
Study of motion taking into
account the cause of
motion
Study of motion Without taking
into account the cause of
motion
One Dimensional
Two Dimensional
Three Dimensional
Straight Linear and Vertical motion
Projectile motion
Random motion
Can you visualize?

7. One Dimension Two Dimensions
Three Dimensions
x
y
x
x
y
z

8. LET’S VISUALIZE THE MOTIONS

9. ONE DIMENSIONAL
KINEMATICS
Horizontal Straight Line
Motion
Vertical Straight Line
Motion
If forward motion is positive then
backward motion will be negative
and vice versa.
If upward motion is positive then
downward motion will be negative
and vice versa.

10. DISTANCE and DISPLACEMENT
The length of the actual path
followed by a moving body.
Shortest distance between
initial and final position with
direction.
Can distance and displacement be equal?
Initial position
Final position

11. SPEED
Types
• Uniform
Body covers equal distance in equal interval of time.
• Non uniform
Body covers unequal distance in equal interval of
time.
• Average Speed
• Instantaneous Speed
It is valid for POINT of time. If we are interested
to find the the speed at 5 sec then the speed
we found is instantaneous speed.
The time rate of change in DISTANCE for a moving body. It is a scalar
quantiy.
en
taken tak
time
total
given time
in
covered
distance
Speed
Average 
It is valid for PERIOD of time.
A man went POKHARA from KATHMANDU by
bus. In this case the speaking speed is
average speed. However the speed is not
constant throughout the motion.
1
2
1
t
-
t
-
2 x
x

dt
dx
t
x






0
t
lim
Speed
ous
Instantane

12. VELOCITY
Types
• Uniform
Body covers equal displacement in equal interval of
time.
• Non uniform
Body covers unequal displacement in equal interval
of time.
• Average Velocity
• Instantaneous Velocity
It is valid for POINT of time. If we are interested
to find the the velocity at 5 sec then the
velocity we found is instantaneous velocity.
The time rate of change in DISPLACEMENT for a moving body. It is a
Vector quantity.
interval
time
given time
in
covered
nt
displaceme
Velocity
Average 
It is valid for PERIOD of time.
A man went KOTESHWOR from
TINKUNE by bus linearly. In this case
the speaking velocity is average
velocity. However the velocity may not
be constant throughout the motion.
dt
d
t
s
s






0
t
lim
velocity
ous
Instantane
1
2
1
2
t
-
t
-s
s


13. ACCELERATION
Types
• Uniform
Body covers equal velocity in equal interval of time.
• Non uniform
Body covers unequal velocity in equal interval of
time.
• Average acceleration
• Instantaneous acceleration
It is valid for POINT of time. If we are interested
to find the the velocity at 5 sec then the
acceleration we found is instantaneous
acceleration
The time rate of change in VELOCITY for a moving body.
interval
time
given time
in
changed
Velocity
on
Accelerati
Average 
It is valid for PERIOD of time.
1
2
1
2
t
-
t
- v
v

dt
d
t
v
v






0
t
lim
velocity
ous
Instantane acceleration

14. REMEMBER!
Here we only consider constant
acceleration or zero acceleration.
Otherwise it is complex to solve and
we don’t need to discuss at this level.

15. Let’s figure out average and instantaneous (say velocity ) by two
more examples:
Departing with
10m/s from KTM
Reaching with 15m/s to
POKHARA
Constant acceleration
Therefore, average velocity =
2
v
u 
Kirtipur Ratnapark
If displacement covered is given by a function
1
2 2

 t
s
0
4t
dt
ds
Since, 


v
Instantaneous Velocity at 5 sec = ?
Now, instantaneous velocity at 5 sec is given by:
Vinst at 5 sec = 4 x 5 = 20 m/s

16. EQUATIONS OF KINEMATICS
2
2
2
2
1
t
2
t
t
a
u
s
as
u
v
a
u
v






Here s, u, v and a are Vectors, t is a scalar. These equations can be
derived from the same condition. So, as per problems we can use a
suitable equation.
REMEMBER!
These equations are valid only for constant acceleration. They are not
valid for variable acceleration.

17. DISTANCE TRAVELLED IN NTH (or in last one )
SECOND
1sec 1sec
1sec
1sec
s3 s4
a const
S4
th or distance
travelled in last one
second
)
1
2
(
2
1
s
-
s
s
Here,
th
th
n
3
4
4




n
a
u
S
General formula is
……………++++++++++=============***************
rest
…………………………………s4…………………………...
……s1........
….…….s2……....….…….
………………………s3……………….…….
HW: Prove the relation

18. TIME GRAPH

19. DISPLACEMENT TIME GRAPH

20. VELOCITY TIME GRAPH

21. ACCELERATION TIME GRAPH

22. HOW TO UNDERSTAND CONSTANT ACCELERATION?
2 m/s
……
6 m/s
4 m/s
1 sec 1 sec 1 sec 1 sec
rest
velocity
t
U
V
a


since, a = 2m/s2 a = 2m/s2 a = 2m/s2 and so on .. .. ..
CONCLUSIONS:
Thus, we see that at constant acceleration velocity is increasing.
At zero acceleration velocity is constant.
Zero acceleration means either the body at rest or moves with uniform velocity.

23. NUMERICALS FOR HORIZONTAL STRAIGHT LINE MOTION
HOMEWORK
1: A particle is moving with a velocity of V= (3 + 6t + 9t2) cm/s. Find out,
a: the acc. of the particle at t = 3 sec.
b: the displacement of the particle in the interval t = 5 sec, to
t=8 sec.
Ans; a: 60 cm/s2
b: 1287 cm
2: The motion of a particle along a straight line is described by the function
x = (2t – 3)2 where x is in the meters and t is in seconds.
a: find the position, velocity, and acc. at t = 2 sec.
b: find the velocity of the particle at origin.
Ans: a: 1m, 4m/s, 8m/s2
b: 0 m/s

24. FOR STUDENTS HELP!
To check and discuss the answers at informal time, please feel free to join on
the following face book page and the group.
PAGE: https://www.facebook.com/101UKPhysics/
Group: https://www.facebook.com/groups/394635377344127/

25. Problem 1
A student walks on a straight road from his home to CCRC 1km away with a
average speed of 1m/s. Taking the admit card, he immediately turns and run
back home with a average speed of 2 m/s. What is the magnitude of average
velocity and the average speed for the whole path.

26. Problem 1
A student walks on a straight road from his home to CCRC 1km away with a
average speed of 1m/s. Taking the admit card, he immediately turns and run
back home with a average speed of 2 m/s. What is the magnitude of average
velocity and the average speed for the whole path.
Hints:
……………………………….
1 m/s
1 km away
2 m/s
taken
time
Total
nt
displaceme
Total
velocity
average
:
a 
m/s
0
2
1000
1
1000
)]
1000
(
1000
[





taken
time
Total
distance
Total
speed
average
:
b 
s
m /
33
.
1
2
1000
1
1000
1000
1000





27. Problem 2
A Motor cyclist travels from Ratnapark to Bhaktpur with an average speed of
60km/hr and returns back to Ratnapark with an average speed of 40km/hr. What
is the average speed of the cyclist for the total time of travel ?

28. Problem 2
A Motor cyclist travels from Ratnapark to Bhaktpur with an average speed of
60km/hr and returns back to Ratnapark with an average speed of 40km/hr. What
is the average speed of the cyclist for the total time of travel ?
This problems is same like the problem 1. This type of problems can be
solved directly by a short cut method as:
2
1
2
1
2
v
v
v
v

Try to solve, you will get the ans 48 km/hr.

29. Problem 3
A body starts from rest and moves with uniform acceleration of 10m/s2 for 5
seconds. During the next 10 seconds, it moves with uniform velocity. It is
brought to rest in 5 seconds by applying uniform retardation. What is the
average velocity of the body for the whole path?

30. Problem 3
A body starts from rest and moves with uniform acceleration of 10m/s2 for 5
seconds. During the next 10 seconds, it moves with uniform velocity. It is
brought to rest in 5 seconds by applying uniform retardation. What is the
average velocity of the body for the whole path?
Hints:
....................................
.
+++++++++++++++
================
rest
5 sec
10 m/s2
10 sec
Constant velocity
So, a= 0 m/s2
5 sec
retardation
rest
S1, t1
S2, t2 S3, t3
3
2
1
3
2
1 s
s
s
velocity
average
here,
t
t
t 




Try to solve, you will get the answer 37.5 m/s

31. Problem 4
A car travelling east at a constant speed of 20 m/s begins to accelerate at
3.8 m/s2 when it is 200m behind a truck currently moving east at a constant
speed of 42 m/s. How long will it take for the car to catch up with the truck?
How far will the truck and car travel during this time.
...200m…... …………………………………….
….…………t………………………………………
………………..t………………...
……………
Both coming with a
constant velocities
……………………
………………
a=3.8 m/s2
Constant velocity
origin
Struck = 42t + 0
Scar = 20t + ½ 3.8 t2
I……..……………..d car = dtruck…….…….…..…..I
dcar = 0 + 20t + 1/2at2 dt = 200 + St
E
N
Catch up
A B
C

33. Thus, we can write; 20 t + ½ x 3.8 t2 = 200 + 42 t :quarditic equation
Try to solve you will get the answer t = - 5.99 sec and t = 17.57 sec.
It is noted that the positive value of the ans is accepted.
Putting this value of time, you will finally get the distances travelled by car
as 937.9m and truck as 737.9m respectively during that time.

34. Problem 5
A car travelling with a speed of 15 m/s is braked and it slows down with
uniform retardation. It covers a distance of 88 m as its velocity reduce to 5
m/s . If the car continues to slow down with the same rate after what further
distance will it be brought to rest? [ 11 m]

35. Short Question asked on NEB from This Part ( HORIZONTAL STRAIGHT
LINE MOTION.
1: Can an object have an eastward velocity experiencing a westward
acceleration ?
2: If the displacement of a body is proportional to the square of time, state
whether the body is moving with uniform velocity or uniform acceleration.
3: Can a body have a constant speed but changing velocity?
4: : If the displacement of a body is proportional to the square of time, state
the nature of the motion of the body.

36. For exam HORIZONTAL STRAIGHT LINE motion is fundamentally important for
the concept.
It doesn’t mean that it is less important, it is highly important for building your
mind smart to solve the problems of MECHANICS.
REMEMBER!

37. DID YOU ENJOYE THE CLASS?
Leave your valuable suggestions so that I will be better for you all in
the next class. Your suggestions are highly appreciated.
NO?
Yes?

38. NEXT CLASS
VERTICAL STRAIGHT LINE
MOTION
VERY IMPORTANT FOR THE EXAM

39. STAY HOME,STAY SAFE!

41. THANK YOU !