Principles of Linear Motion of Objects - Physics - Explained by Arun Umrao

1
LINEAR MOTION
AN INTRODUCTION
Arun Umrao
www.sites.google.com/view/arunumrao
DRAFT COPY - GPL LICENSING
Arun Umrao
https://sites.google.com/view/arunumrao

2
Contents
0.1 Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
0.1.1 Body May Have Many Types of Energies . . . . . . . . . . . . . . 3
0.2 Linear Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
0.2.1 Straight Path Motion . . . . . . . . . . . . . . . . . . . . . . . . . 4
0.2.2 Displacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
0.2.3 Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
0.2.4 Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
0.2.5 Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
0.2.6 Direction Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
0.2.7 Uniform Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Average Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
Average Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
0.2.8 Non Uniform Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 12
Instantaneous Velocity And Instantaneous Speed . . . . . . . . . . 12
0.2.9 Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
0.2.10 Single Variable Motion . . . . . . . . . . . . . . . . . . . . . . . . . 23
0.2.11 Motion In Cartesian Plane . . . . . . . . . . . . . . . . . . . . . . 24
0.2.12 Derivative of Unit Vector . . . . . . . . . . . . . . . . . . . . . . . 25
0.3 Newton’s Law Of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
0.3.1 Net Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
0.4 Equations of Motion In Horizontal Plane . . . . . . . . . . . . . . . . . . . 29
0.5 Motion In Vertical Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
0.5.1 Eﬀect Of Center of Mass In Vertical Motion . . . . . . . . . . . . . 45
0.5.2 Velocity Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
0.5.3 Linear Velocity Curve . . . . . . . . . . . . . . . . . . . . . . . . . 47
0.5.4 Vertical Throw . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
0.5.5 Vertical Drop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
0.6 Relative Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
0.6.1 Equations of Motions for Relative Motion . . . . . . . . . . . . . . 54
0.7 Motion in River . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
0.7.1 First Type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
0.7.2 Second Type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
0.8 Motion in Lift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
0.8.1 Lift is In Rest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
0.8.2 Lift Moving Upward Direction . . . . . . . . . . . . . . . . . . . . 65
0.8.3 Lift Moving Downward Direction . . . . . . . . . . . . . . . . . . . 66
0.8.4 Motion of Parachute . . . . . . . . . . . . . . . . . . . . . . . . . . 68
0.9 Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
0.9.1 Walking Man . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
0.9.2 Coeﬃcient of Friction . . . . . . . . . . . . . . . . . . . . . . . . . 70
0.9.3 What is R? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
0.9.4 Object at Rest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
0.9.5 Object is About to Move . . . . . . . . . . . . . . . . . . . . . . . 72
0.9.6 Object is Moving . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
0.9.7 Why μk < μs? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
Arun
of Mo
of Mo
V ti
V ti
Umrao
tion In H
tion In H
l Pl
l Pl
https://sites google com/view/arunumrao
ect Of Center of Mass In Vertical Motion . . . .
ect Of Center of Mass In Vertical Motion . . . .

0.1. MOTION 3
0.9.8 Angle of Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
0.9.9 Angle of Repose . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
0.9.10 Law of Dry Friction . . . . . . . . . . . . . . . . . . . . . . . . . . 77
0.9.11 Rolling & Sliding Wheal . . . . . . . . . . . . . . . . . . . . . . . . 78
0.10 Motion of Coupled Objects . . . . . . . . . . . . . . . . . . . . . . . . . . 84
0.10.1 System Is In Rest . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
0.10.2 System is In Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 85
0.11 Impulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
0.12 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
0.12.1 Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
0.12.2 Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
0.12.3 Total Mechanical Energy . . . . . . . . . . . . . . . . . . . . . . . 89
0.12.4 Potential Energy Graph . . . . . . . . . . . . . . . . . . . . . . . . 89
0.13 Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
0.13.1 Conservation of Momentum . . . . . . . . . . . . . . . . . . . . . . 97
0.13.2 Rebounding of particle . . . . . . . . . . . . . . . . . . . . . . . . . 98
0.13.3 Explosion of particle . . . . . . . . . . . . . . . . . . . . . . . . . . 99
0.14 Collision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
0.14.1 Oblique Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
0.14.2 Rules of Elastic Collision . . . . . . . . . . . . . . . . . . . . . . . 104
0.14.3 Perfect Plastic Collision . . . . . . . . . . . . . . . . . . . . . . . . 107
0.14.4 Conservation of Momentum of Varying particle Mass . . . . . . . . 108
0.14.5 Momentum & Newton’s Law of Motion . . . . . . . . . . . . . . . 113
First Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
Second Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
Third Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
0.15 Two Dimensional Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . 115
0.15.1 Scattering Angle After Collision . . . . . . . . . . . . . . . . . . . 116
0.16 Hook’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
0.16.1 Energy Stored in Spring . . . . . . . . . . . . . . . . . . . . . . . . 118
0.1 Motion
There are four types of motions of an object, (i) Linear Motion, (ii) Rotational Motion,
(iii) Angular Motion and (iv) Vibration. All these types of motions are result of forces
applied on the object. The direction and position of force are deterministic factors for
type of motion. To understand this, consider a wheel that is free to roll in a horizontal
surface under the inﬂuence of external force.
omentu
mentuUmrao
m & New
m & New
st Law . . . . . . . . . . . . . . . . . . . . . . .
st Law . . . . . . . . . . . . . . . .
cond Law
cond Law

4
Diametric Axis
CoM
(a)

F
Action Point
Line of Action
(b)
In first part (a) of above figure, diametric axis of wheel, parallel to the horizontal
surface, is shown. A force is applied as shown in second part (b) of above figure. Imaginary
line that is drawn along the direction of force is called Line of Action. In this case, the
line of action passes through the center of mass of the wheel, i.e. along the diametric
axis. Therefore the wheel will slide on the surface rather than roll.

F
Action Point
Line of Action
(a)

F
Action Point
Line of Action
(b)
Now, the direction of force is changed, so that imaginary line, that is drawn along the
direction of force, is shifted off the diametric axis. In this case, the line of action does
not pass through the center of mass of the wheel. Therefore the wheel will roll on the
surface. Linear motion as name suggest is applied to the moving objects when their path
of motion is a straight line for long period or instantaneously linear for a little time. In
this chapter we will study the properties of object moving in linear path.
0.1.1 Body May Have Many Types of Energies
A question, here I am raising that whether a body may have more than one type of
energies or not? The answer is “Yes”, it may have two or more types of energies. But
How? Suppose two friends wake up early morning and go to morning walk. One of them
is lazy type and other is energetic. First is moving silently, whose hands are down and
stationary. His friend is walking energetically with swinging his hands up and down, front
and back, and left and right. Do both friends are burning same amount of body calories?
Answer is ”No”. First friend burns less body calories than his friend. It means second
friend has more body energy than his friend. But How? It is due the swinging of hands by
second friend. It means, second friend posses two types of energies, first by walking and
second by swinging the hands. Now came to real world problem. A bicycle is standing
on its stand and its “rear wheel” is rotating with angular velocity ω. The wheel has only
Arun o
Umrao
ao
Um
https://site
https://sites
htt / i unumrao
runumrao

0.2. LINEAR MOTION 5
rotational energy. But when bicycle moves with speed v, and its “rear wheel” rotates
with same angular speed ω then “rear wheel” has two types of kinetic energies. First
kinetic energy due to its rotation and second kinetic energy due to its linear motion as it
also changes its position (by a distance of d) alongwith change in the position of bicycle.
ω
ω
ω
v
d
0.2 Linear Motion
There are two states of objects. One static and other dynamic. In static state, object
remains in rest position and in dynamic state object continuously changes its position
with respect to origin. In dynamic position, locus of center of mass (CoM) is either curve
or a simple line. If the locus of CoM is straight line then motion is called linear motion. In
linear motion either particle is (i) in constant velocity, (ii) accelerating or (iii) retarding.
Followings are some definitions used in linear motion.
0.2.1 Straight Path Motion
When a particle is moving such way that its locus is simple line then the motion is called
straight path motion. Velocity of particle is simply found by using relation d/t.
0.2.2 Displacement
Displacement is aerial distance between initial journey point and final journey point. In
other words, displacement is represented by a vector whose tail is at the initial journey
point and its head is at final journey point.
N
O
N N
P

d
In above figure, an object moves from one point to other point and finally reached at
P. Now the line-of-sight or aerial distance from O to P is its displacement. In figure (1),
a vehicle is moving in circular path.
A
r parti
r part U
cle is (i)
cle is (i)
e definitions used in linear motion.
e definitions used in linear motion.

6
A
NM
B
AFO
C
l

s
Figure 1: Displacement path length in circular path.
It started its journey from point A and following path ABC it reaches to point C.
During the journey it covers path length l while its aerial distance from journey point is

d. Path length l is called distance traveled by vehicle while path length
d is called dis-
placement of vehicle. If velocity-time graph is plotted for a moving object then algebraic
sum of areas between graph line and time axis in positive and negative regions is called
displacement of the object.
1
−1
1 2 3
t

v
1
−1
1 2 3
t

v
+A1
−A2
+A3
Here, displacement is
d is algebraic sum of +A1, −A2 and +A3.

d = A1 + A3 + (−A2)
Solved Problem 0.1 The displacement functions of a particle along x and y axes are
x = t2
− 4 and y = t − 2 respectively. Find the displacement at t = 3s. Also ﬁnd the
velocity of the particle about origin when t = 1s.
Solution Distances of the particle after t = 3s along x and y axes are given by
x = 32
− 4 = 5m; y = 3 − 2 = 1m
Now, magnitude of displacement of the particle is given by
d =

x2 + y2 =
√
26m
and its direction from +x-axis is
θ = tan−1

1
5

Velocity of the particle along x and y axes after t = 1s are given by
ẋ = 2t = 2m/s; ẏ = 1m/s
Arun Umrao
Um

d l b f
b A A d A

The magnitude of velocity of the particle is given by
v =

ẋ2 + ẏ2 =
√
5m/s
and its direction from +x-axis is
θ = tan−1

1
2

This is desired result.
0.2.3 Velocity
Velocity of a particle is defined as the rate of change of displacement of a moving object.
It is a vector quantity and it may be positive or negative quantity. It is computed either
in average form or in instanteneous form. So instanteneous velocity of the moving object
is given by:

v =
d
x
dt
1
−1
1 2 3
t
+
v
The velocity-time graph is plotted above. It is assumed that when object moves from
left to right, its velocity is positive and when it moves from right to left, its velocity is
negative, similar to the numberline. From the above figure, during the time t = 0 to
t = 2, velocity curve of the object is in +
v axis, hence object is moving from left to
right direction. As its magnitude is decreasing therefore velocity is decreasing in same
direction. Velocity curve intersects to x-axis at t = 2, it means velocity of the object
is zero. After t = 2, velocity curve enters into −
v axis, so, direction of the velocity is
reversed and now the object starts moving from right to left direction.
1
−1
1 2 3
t
+
v
In the above figure, velocity-time graph is plotted. It is assumed that when object
moves from left to right, its velocity is positive and when it moves from right to left, its
Arun Umrao
a
mrao
iew/arunumrao
3
3
https://sites.google.com/view
e 1
1 2
2
/view

8
velocity is negative, similar to the numberline. During the time t = 0 to t = 1, velocity
curve of the object is in +
v axis. It means object is moving from left to right direction and
its velocity is increasing. During the time t = 1 to t = 2, velocity curve of the object is
in same region, i.e. along the +
v axis, so object is continuously moving from left to right
direction with decreasing velocity. At t = 2, object came to rest. From t = 2 to t = 3,
velocity curve of the object is again in +
v axis, hence object is moving from left to right
direction with increasing velocity. Its velocity decreases but remain in same direction from
t = 3 to t = 3.5. At t = 3.5 object came to rest. After t = 3.5, velocity curve intersects to
x-axis and enters into −
v axis, so, direction of the velocity is reversed and now the object
starts moving from right to left direction with increasing velocity. Negative sign in the
−
v tells about the driection of velocity but larger its magnitude, means larger velocity,
i.e. −35m/s is larger velocity from right to left direction in comparison to −10m/s in
same direction.
Solved Problem 0.2 Enterpret the velocity-time graph given below.
Solution
1
−1
1 2 3
t
+
v
Initially at t = 0, object is moving from left to right (say for positive velocity) and
its velocity is decreasing with time. At time t = 2, its velocity becomes zero. After
this time, velocity-time curve enters in −
v regions, therefore, direction of velocity of the
object is reversed, i.e. object starts moving from right to left direction. The slope of
velocity-time graph represents the acceleration of the object. Slope of velocity-time curve
during 0 ≤ t ≤ 2 is less than the slope of velocity-time curve during 2 t ≤ 4. Hence, the
object moves away with low deceleration (−a
) and returns back with high acceleration
(+a). Here, magnitude of retardation (| − a
|) and acceleration (| + a|) are not equal.
If slope was uniform during the motion between t = 0 to t = 4, then retardation and
acceleration would be equal in magnitude.
0.2.4 Distance
Distance is total path length covered by a moving particle. It does not matter whether
particle reaches to initial place from where it started its journey, or not. Assumed that
length of a moving car is negligible in respect of path length that it coveres during its
journey. A car moves in a curved path as shown below:
Arun Umrao
a
e
−
−1
1
w/a

s

s
A
B
Length of path taht is covered by this car during the journey from A to B is called
its distance while shortest aerial distance measured between A and B is called its dis-
placement. If car covers the journey with constant speed v (say) for a time period t then
distance or path length traveled by the car is
d = v × t
If a graph is plot between distance and time then slope of graph line represents speed
of the object and plot will be a straight line. In velocity-time graph, distance is area
between graph line and time axis.
t
d
t1 t2
d1
d2
t
v
t1 t2
v1
v2
d
If speed of the object changes with time, then velocity-time graph is zig-zag line as
shown in the following ﬁgure.
1
2
1 2 3 4
t
v
(a)
1
2
1 2 3 4
t
v
(b)
The distance covered by the object is area between graph line and time line.
t

d
(a)
t
d
(b)
Arun Umrao
object changes with time, then velocity-time gra
object changes with time, then velocity-time gra

10
If displacement-time graph is plotted in x−y plane then distance-time graph is mod-
ulus of the displacement-time graph. It means all the negative directions or negative
values are converted into positive direction or positive values respectively. This is also
true for velocity-time vs speed-time graph too.
Solved Problem 0.3 A car runs with velocity of 45km/h for 120 minutes. Find distance
covered by it.
Solution The distance covered by any moving vehicle is
d = v × t
The car runs for 120 minutes or 2hrs, hence distance covered by the car is
d = 45 × 2 = 90km
This is required distance.
0.2.5 Speed
Speed of a particle is deﬁned as the rate of change of distance of a moving object. It is a
scalar quantity and it is always a positive quantity. It is computed either in average form
or in instanteneous form. So instanteneous speed of the moving object is given by:
s =
dx
dt
Speed is modulus of the velocity. It means, negative velocity curve becomes positive
velocity curve in speed-time graph.
1
−1
1 2 3
t

v
1
−1
1 2 3
t
s
0.2.6 Direction Plane
A quadrant plan is also called a direction plane. It has four directions, East, West, North
and South. North direction is always put in top side of the paper plane.
Arun Umrao
s =
dt
dt

E
W
N
S
E
W
N
S
N-E
A direction between two principal direction is labeled as both name. In above figure,
direction between East and North is termed as North-East (N-E) direction.
Solved Problem 0.4 A man walks 200m in east and then turn left and walks to 50m. Then
he turns to west and walk 400m. Now find (i) distance he travels and (ii) his displacement
form the initial position.
Solution
N
200m
50m
400m

d Pi
Pf
(i)
N
200m
50m
400m
50m

d Pi
Pf
(ii)
i The distance traveled by man is length of actual path followed by him. So,
s = 200 + 50 + 400 = 650m
ii The displacement is line-of-sight distance between intial and final position of the
man. And it is
d =

502 + 2002 = 206.16m
0.2.7 Uniform Motion
When speed or velocity of a moving particle does not change with time then such motion
of particle is called uniform motion.
Arun
Arun Umrao
Umrao
g
tps://sites
200m
200
htt
P
Pi
i
P
umrao
200m
200
un

d
d P
Pi
P

12
A
B
In a time bound race, a driver (red car) drives a car from start point A to end point
B. As the road is empty, hence driver of the car shall able to maintain its constant
speed through out the race. Here we can say that the motion of car is a uniform motion.
In uniform motion, average speed or average velocity can be computed by using simple
relation v =
d
t
, where d is total distance covered by the car and t is total time taken by
him during the journey.
Average Speed
Average speed is distance covered by moving particle in unit time (1sec).
d A
B
Time = t sec
Figure 2: Speed
From ﬁgure (2), let a vehicle is at point A at any instant time t1 and it reaches at
point B at any instant time t2 with covering distance d.
v =
d
t2 − t1
t
d

v
t1 t2
d1
d2
Average Velocity
Average velocity is displacement covered by moving particle in unit time (1sec).
Arun
Arun
Arun Umrao
Umr
Umra
es google com/view/ar
runumrao
runumrao
n
un
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
un
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
un o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
ttps://site
tps://site
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
ht
htt te
site
te
te
te
te
te
te
te
te
te
te
t
t
t
t
t
t
t
t
t
t
t
t
t
t
i
i
i
i
i
i
i
i
si
si
si
si
si
si
si
si
si
si
si
si
si
si
si
si
si
si
si
si
si
si
si
si
si
si
si
si
si
i
i
i
i
i
i
i
it
t
t
t
t
t
t
t
t
t
t
t
t
te
te
te
te
te
te
te
te
te
te
te
site


xB −
xA

xA

xB
O
A
B
Time = t sec
Figure 3: Speed
From figure (3), let a vehicle is at a point A at any time t1 and it reaches at point
B at any time t2. If position vectors of points A and B are
x1 and
x2 respectively then
average velocity of the particle is

v =

x2 −
x1
t2 − t1
0.2.8 Non Uniform Motion
Δx
A
B
In a time bound race, a driver (red car) drives a car from start point A to end point B.
As the road is filled with other cars, driver of the car can not maintain its constant speed
through out the race. He have to applied break or slowdown when he overtake to other
cars or when there is jam in road crosses. Hence speed of the car is not uniform through
out the race. This is why, all practical motions are non-uniform motions. Assume that
drive can maintain uniform motion for a shot distance between A to C. Now the average
speed for this short length is
Δv =
Δx
Δt
This Δv is called instantaneous speed, i.e. speed for instant of time or speed during very
short interval of time.
Instantaneous Velocity And Instantaneous Speed
If a moving particle covers different displacements or distances with different velocities
or speeds respectively then ratio of small element of displacement or distance to the time
required to cover the small displacement or the small distance is called instantaneous
velocity or instantaneous speed.
Δv =
Δx
Δt
(1)
Arun
A
A
A
A
A
A
A
A Umrao
o
o
o
o
o
ao
a
ao
o
a
a
es.goo
o
e g
s g
s g
g
s

14
As time element tends to zero, distance element also tends to zero and average velocity
becomes more instantaneous hence from equation (1)
lim
t→0
δv = lim
t→0
δx
δt
dv =
dx
dt
(2)
Equation (2) represent to instantaneous velocity or instantaneous speed of any particle
in diﬀerential form.
Solved Problem 0.5 The position of a bird is given by function x = 2t − t2
. What will be
its velocity after t = 3s.
Solution Path of a bird does not stationary with time but it changes parabolic, hence
the instantaneous velocity of the bird at any time is
dv =
dx
dt
=
d
dt
(2t − t2
) = 2 − 2t
The velocity after t = 3s is
v =

dx
dt

3s
= −4m/s
Negative sign does not mean that the velocity is negative in magnitude but it represents
that after 3s, velocity is decrease upto −4m/s.
Force Velocity
External force when applied to a moving object, its velocity changes. This force
may be continuous or may be an impact. In absence of external forces, velocity of the
object remains constant. External force changes kinetic energy of the object.Change in
kinetic energy results in the change in acceleration of the object. Hence, acceleration
is the result of force and velocity is the result of acceleration and ultimately result of
the force. This external force may be gradual in action or an impact like collision.
Practical example of force and velocity relation is visible in tiny rain drops that fall
with constant speed. In this phenomenon, weight of drop is balanced by the drag force
of air (viscous force by air). Both, rain drops and viscous force do work. Speed of drop
may be increasing or decreasing or constant. Work done by drop and viscous force are
equal to the change in kinetic energy or change in potential energy of the tiny rain drop.
In vertical motion, change in kinetic energy is always equal to the change in potential
energy.
Solved Problem 0.6 A sprinter running 100m race starts at rest. He accelerates his particle
at constant acceleration with magnitude a for 2 second, and then runs at constant speed
until the end. (a) Find the position and speed of the runner at the end of the 2s in terms
of a. (b) If runner takes total time of 10s then ﬁnd the value of acceleration.
Solution
Arun
not m
not m
i i d
i i
Umrao
ean that
ean that
d
ity is decrease upto
ity is decrease upto 4m/s
/ .

a If sprinter accelerates his particle for two second with acceleration of a m/s2
then
distance covered is
da =
1
2
at2
= 2a m
The velocity of sprinter is
v = at = 2a m/s
b Total time taken by sprinter is 10s. This time includes the time required for covering
distance da and 100 − da with constant speed v. The time taken by sprinter to cover the
distance 100 − da is 10 − 2 = 8s. Now
100 − 2a
2a
= 8 ⇒ a =
50
9
m/s2
This is required acceleration.
0.2.9 Acceleration
Acceleration is change in velocity or speed of a moving particle during unit interval of
time.

vB −
vA

vA

vB
O
A
B
Time = t sec
Figure 4: Acceleration of moving car.
From ﬁgure (??), a vehicle has velocity
v1 at time t1 and after some time its velocity
is
v2 at time t2, acceleration of vehicle is

a =

v2 −
v1
t2 − t1
If
a is positive then vehicle is said to be accelerating and ﬁnal velocity
v2 is greater than
that of initial velocity
v1. Similarly if
a is negative then vehicle is retarding. A vehicle
moving with constant velocity then vehicle is neither accelerating or retarding.
Arun
Arun
run
Ar
Ar
Ar
Ar
Ar
Ar
Ar
Ar
Ar
Ar
Ar
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Ar
Ar
Ar
Ar
Ar
Ar
Ar
Ar
Ar
Ar
Ar ao
rao
ra
ra
mr
mr
mr
mr
mr
mr
mr
mr
mr
mr
mr
mr
mr
mr
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
m
mr
mr
mr
mr
mr
mr
mr
mr
mr
mr
mr
mr
mr
Um
Um
Um
mr
num
arun

sites.go
oog

vB
A
A
B
B

16
1
2
1 2 3 4
t
v
(a)
1
2
1 2 3 4
t
a
(b)
Figure 5: Figure (a) is graph of velocity as a function of time. Slope of graph line at any
interval of time is acceleration of the moving object. Figure (b) shows the average speed
in form of shaded region.
Solved Problem 0.7 A car is moving with velocity of 45km/h for 120 minutes. Find
acceleration of the car. Justify your answer.
Solution The acceleration of car is ratio of rate of change in velocity with time. So,
a =
45 − 45
2
= 0km/h2
This answer is true in mathematical mode. Here we considered that car runs in a straight
line with any value of acceleration or deceleration. The term ‘velocity’ is a vector quantity
but the question does not clarify the nature of motion of the car (ie linear or circular).
In practical life it is not possible that the car move with constant speed or velocity for a
long period. Hence the answer may be more accurate if path function of the car is given.
t
d B
A
The graph shows relation between distance and time. If a particle covers unit
distance in small interval of time then we say that its acceleration is very large. If it
takes large interval of time then we say that its acceleration is very small. Particles
with high acceleration has high curve slope and particles with low acceleration has low
curve slope. So, now, particle A has high acceleration.
Solved Problem 0.8 A rock dropped off the cliff and same time a bullet is fired from the
same point vertically downward. Find the acceleration velocity of both rock bullet
when they reached at ground. Given muzzle velocity of gun 800m/s.
A
in ma
in ma U
hematic
hematic
of acceleration or deceleration. The term ‘velocity
of acceleration or deceleration. The term ‘velocity
t l if th t f ti f th (i
t l if th

Solution For the case of rock, its initial velocity is zero and constant acceleration is
g. There are not external forces acting on the rock hence acceleration at the ground will
be g. Now velocity of the rock be
v = ut +
1
2
gt2
== 5t2
m/s
For the case of bullet, initial velocity is 800m/s and it also falls under the only influence
of gravitational acceleration. Hence at the ground its acceleration will be g. Now velocity
of the bullet at the ground is
v = ut +
1
2
gt2
= (800t + 5t2
) m/s
The velocity of bullet is more than to velocity of rock by 800t m/s.
Solved Problem 0.9 The position of a honeybee is given by function x = 2t − t2
+ t3
.
What is its acceleration after time t = 5s.
Solution The path function of the honeybee is f(t) = 2t − t2
+ t3
. The instantaneous
acceleration of a path function is
a =
d2
dt2
f(t) = −2 + 6t
Acceleration after 5s is
a5 =

d2
dt2
f(t)

t=5
= 28m/s2
This is the acceleration of honeybee after 5s.
Solved Problem 0.10 Using integration method find the distance covered by a moving
particle having a velocity function v = 5t2
+ 4 in time t = 2s to t = 6s. Also explain the
significance of the question.
Solution We know that the velocity distance relation is
v =
d
dt
f(x)
Where f(x) is the function of time representing the displacement of the moving particle.
Now integrating above function with respect to time we have
f(x) =

v dt
for the time t = 2s to t = 6s, distance covered by the moving particle is
f(x) =
6
2
(5t2
+ 4) dt =

5
t3
3
+ 4t
6
2
After simplification, we have
d = 359.33m
Arun
5
5s
s is
is Umrao
a
a5
5 =
=

d
d
oog
dt2
2
f
f(
(t
t)
)

= 28
= 28m/s
m/s2
2

18
This is distance covered by moving particle in stipulated time. This question has signif-
icance as Without knowing the initial velocity we can find the distance covered by the
moving particle in time between t = 2s to t = 6s. Generally integration uses the principal
of origin reference to find the distance. Even though we do not know the initial velocity
but using the velocity function we get the time t0 where velocity by function was zero.
Now we can find the required distance by using the method d = difference of distances
covered by moving particle in time 6 − t0 and 2 − t0. In summing word we can
say that definite integration is the best method to solve a motion function for
end limits.
Solved Problem 0.11 In a highway you are driving with speed of 40m/s and suddenly
spots a police car at 650m ahead. The police car is running at a constant speed of 30m/s,
that is also speed limit in that highway. You applied brakes and begin decelerating at
1m/s2
to avoid speeding ticket, applied by police officers if you pass them over the speed
limit. Find that, will you get a speeding ticket?
Solution When you see the police car, you start applying brakes and passes to the
police patrolling car after t s. Now in this time the distance covered by you and police
car should not be exceed to 650m. Now dy + dp = 650 and
uyt −
1
2
at2
+ upt = 650
Substituting the values and simplifying the relation
t2
− 140t + 1300 = 0
The time found is t = 10s or t = 130s. But 130s is very large quantity and in this time,
distance covered by you and police team would be larger that 650m, hence after 10s you
will pass the police car. Now if the speed of your car is more than 30m/s you will surely
get a speeding ticket. Now speed of your car after 10s is
v = 40 − 1 × 10 = 30m/s
Now after 10s your car has speed of 30m/s i.e. within the speed limit, hence you will not
get the speeding ticket. Have a good driving!!!
Solved Problem 0.12 Find the instantaneous velocity of particle at t = 3s when position
vector of the particle is
r = 2tî + t2
ĵ.
Solution Instantaneous velocity of a moving particle is d
r/dt. And

v =
d
dt
(2tî + t2
ĵ) = 2î + 2tĵ
Instantaneous velocity at t = 3s is

vt=3s = 2î + 2 × 3ĵ
= 2î + 6ĵ
Arun
alues a
lues a
Umrao
nd simpli
nd simpl
t
t2
2
−
− 140
140t
t + 1300 = 0
+ 1300 = 0

t î ĵ
r
1 2 1
r1 = 2î + 1ĵ
2 4 4
r2 = 4î + 4ĵ
3 6 9
r3 = 6î + 9ĵ
4 8 16
r4 = 8î + 16ĵ
Table 1: Position table.
2
4
6
8
10
2 4 6
î
ĵ

r
t
=
3
2
4
6
8
2
î
ĵ

v
t
=
3
Magnitude of this instantaneous velocity is v =

22 + 62 =
√
40 units per second.
The direction of instantaneous velocity at this instant of time from positive x-axis is
θ = tan−1
3
t î ĵ
v
1 2 2
v1 = 2î + 2ĵ
2 2 4
v2 = 2î + 4ĵ
3 2 6
v3 = 2î + 6ĵ
4 2 8
v4 = 2î + 8ĵ
Table 2: velocity table.
Solved Problem 0.13 A cyclist is moving in a path represented by a relation vector

r = tî − t2
ĵ. Find the velocity and acceleration of the cyclist.
Arun
run Umrao
Um
unumrao
m
https://sites.google.com/view/aru
ps:/ o
ht
htt ru

20
Solution Instantaneous velocity of a moving particle is d
r/dt. And

v =
d
dt
(tî − t2
ĵ) = î − 2tĵ
It shows that the velocity is intending towards the negative y-axis more rapidly as the
time increases. Similarly instantaneous acceleration is d
v/dt and

a =
d
dt
(î − 2tĵ) = −2ĵ
It shows that acceleration of the particle is 2 units per square second along the negative
y-axis.
−2
−4
−6
−8
−10
2
î
ĵ

vt=0

v
t
=
1

v
t
=
2

v
t
=
3
−2
−4
−6
î
ĵ

a
Solved Problem 0.14 A particle undergoes the transnational motion under the relation
vector
r = 2t3
î − t2
ĵ. Find the acceleration along the x-axis.
Solution Instantaneous acceleration is twice derivative of the position vector and

a =
d2
dt2

2t3
î − t2
ĵ

= 12tî − 2ĵ
This acceleration vector has two components, one along the x-axis and another along
y-axis. The acceleration along the x-axis is the coeﬃcient of î, i.e. 12t units per square
second. Here acceleration increases with time along the x-axis linearly.
Solved Problem 0.15 Find the instantaneous velocity of particle at t = 2s when position
vector of the particle is
r = t2
î − t3
ĵ. What is is velocity along the y-axis at that instant
of time.
Solution Instantaneous velocity is ﬁrst derivative of the position vector and

v =
d
dt

t2
î − t3
ĵ

= 2tî − 3t2
ĵ
Arun
−
Umrao
Um
8
8
ao
s s.
j
j

Velocity of the particle at t = 2s is

vt=2s = 2 × 2î − 3 × 22
ĵ
= 4î − 12ĵ
Velocity along y-axis is 12 unit per second along the direction of negative y-axis.
−2
−4
−6
−8
−10
−12
2 4
î
ĵ

r
t
=
2
−2
−4
−6
−8
−10
−12
2 4
î
ĵ

v
t
=
2
−2
−4
−6
−8
−10
−12
î
ĵ

v
y
t=2
Solved Problem 0.16 A plane starts flying to north with speed of 400km/h from an
airport. After one hour, the pilot of the plane is warned about a cyclone approaching to
him. ATC of the airport instructs pilot to turn to west. Find the displacement of the
plane after two hours from the beginning. Also find the velocity of the plane at this time.
Solution
N
400km
400km

d
Pi
Pf
The speed of plane is 400km/h towards north. After one hour it reaches at a distance
of 400km in the north direction from its original position. When ATC asked pilot of the
plan to turn west due to cyclonic activities, he turns west and flies for another one hour.
Arun Umrao
ˆ
ĵ
j
/

22
As only direction of velocity has been changed, therefore, plan covers a distance of 400km
in west direction. The displacement of plane from the airport is
d =

4002 + 4002 = 400
√
2km
The velocity of the plane is 400km/h towards the west after two hours from the beginning.
Solved Problem 0.17 Explain the meaning of acceleration-time graph as given below:
10m/s2
20m/s2
10s 20s 30s
t
a
Solution The acceleration-time graph as given in the problem gives us following
information.
1. The acceleration-time graph is along the +ve side of the acceleration, hence object
is under acceleration with diﬀerent magnitudes of acceleration.
2. When time is zero, acceleration is 20m/s2
. When time is 30 seconds, acceleration is
zero. So, magnitude of acceleration is maximum initially and decreases with time.
3. Till the acceleration is greater than zero (a 0), object is under acceleration and
its velocity shall be continuously increasing.
4. If acceleration is zero, it does not mean that velocity is zero.
5. Velocity attained during motion from t = 0 to t = 30s, is equal to the area covered
between the acceleration-time graph and x-axis.
Solved Problem 0.18 Using acceleration-time graph as given below, ﬁnd the maximum
velocity attained by the object using integration method. Take initial velocity zero.
10m/s2
20m/s2
10s 20s 30s
t
a
Arun
leratio
leratio Umrao
g p
n with di
n with di
ero acceleration is 20
ero acceleration is 20m/s
/ 2
2
When time is 30 sec
Wh

Solution The given acceleration-time graph is shown below.
10m/s2
20m/s2
10s 20s 30s
t
a
θ
α
dt
t
a
Initially, object is in rest. So, u = 0. Now, take a time period t and t + dt where
acceleration is a. Now, the change in the velocity during this instantaneous time is
dv = a dt
The slope of graph line is
m = tan θ = tan(π − α) = − tan α = −
20
30
= −
2
3
Now, acceleration a at time t
is given by
a = mt + c
It gives
a = −
2
3
t + 20
So dv is
dv =

−
2
3
t + 20

dt
On integration, we have
vmax
0
dv =
30
0

−
2
3
t + 20

dt
On solving it, we have
vmax =

−
2
3
×
t2
2
+ 20t
30
0
It gives vmax = 300 meter per second.
Solved Problem 0.19 A car was stolen by a thief. Police was chasing him. When they
were at a distance of 400 meter, thief saw that police is about to catch him by chasing
him at speed of 15 meter per second. He accelerate his car to escape of police, from his
initial car velocity 10 meter per second. Find the minimum acceleration of car should
thief achieve so that he may never be caught by police.
Solution Initially, velocities of police car and thief car were u1 = 15m/s and u2 =
10m/s respectively. When thief accelerates his car, police car was 400 meter behind at
Arun Umrao

24
time t = 0. Thief shall escape from police if police car is just behind the thief’s car and
thief’s is not overtaken by police car (like collision condition). It means, after time t,
velocities of both, thief and police cars must be equal (say v) and police car just reach
behind the thief’s car. During period of time t, thief car travels x meters and in the same
time, police car travel 400 + x meters.
Δx
u1 u2
(x + 400)m
xm
400m
v
v
Now, according to question, for police car, distance is
400 + x = 15t
For thief’s car, distance is
x = 10t +
1
2
at2
Subtracting second equation from first equation, we have
400 = 15t − 10t −
1
2
at2
Or
10t − at2
= 800
During this time, final velocity of the thief car shall be 15 meter per second. From first
equation of linear motion
15 = 10 + at
This gives at = 5. Substitute this at in equation 10t − at2
= 800, we shall get time
10t − 5t = 800 ⇒ t = 160s
Now, acceleration is a × 160 = 5 or a = 5/160 meter per square second.
0.2.10 Single Variable Motion
Sometime motion of a particle is not linear and homogeneous but it varies from time
to time. It means, equation of motion has degree more than one. If motion of particle
depends only on one parameter then it is said that motion is one variable motion. Take
a case of particle motion, whose displacement depends on only time, t, i.e. position x of
the particle depends on the time t according to relation
x = 2t2
+ 3t (3)
Arun Umrao
equation from first equation, we have
equation from first equation, we have

Here, the motion is not linear, hence its velocity and acceleration is always computed
with help of derivative calculus. Now, velocity of the particle is dx/dt.
dx
dt
= 4t + 3 (4)
Similarly, acceleration of the particle is dv/dt.
d2
x
dt2
= 4 (5)
Using this method, we can solve other problems of one variable motion.
0.2.11 Motion In Cartesian Plane
Motion in two dimensional Cartesian plane is the motion in which x and y component
of the particle changes with time. For example a particle is moving in two dimensional
plane and position of the particle at any instant is
x = a cos t (6)
and
y = a sin t (7)
These form of equations are known as parametric equation of the particle. Velocity of
particle along to x axis is
dx
dt
= −a sin t (8)
and vertical component of velocity of the particle is
dy
dt
= a cos t (9)
Now velocity of the particle is
v =

dy
dt
2
+

dx
dt
2
(10)
On substituting the values and solving the relation
v = a (11)
This is the velocity of the particle. Direction of the velocity with +xaxis is
tan θ =

dy
dt

dx
dt

(12)
Similarly acceleration of particle along to x axis is
d2
x
dt2
= −a cos t (13)
Arun
ations
ations
axis is
axis isUmrao
are know
are know
dx
dx
= a sin
i t
t

26
and vertical component of acceleration of the particle is
d2
y
dt2
= −a sin t (14)
Now acceleration f of the particle is
f =

d2y
dt2
2
+

d2x
dt2
2
(15)
On substituting the values and solving the relation
f = a (16)
tan α =

d2
y
dt2

d2x
dt2

(17)
This is the method of finding of velocity and acceleration of the particle in two dimensional
motion.
0.2.12 Derivative of Unit Vector
Assume a unit vector r̂ whose horizontal and vertical components are |r̂| cos θ and |r̂| sin θ
respectively. Now unit vector can be written as
r̂ = cos θî + sin θĵ (18)
Here θ is angular direction of unit vector. Now differentiating equation (18) with respect
to t
d
dt
r̂ =
d
dt
cos θ î +
d
dt
sin θ ĵ
x
y
r̂
θ
|r̂| cos θ
|r̂|
sin
θ
n̂
α
|n̂| sin α
|n̂|
cos
α
O
A
B
Figure 6: Differentiation of unit vector.
On differentiation
dr̂
dt
= − sin θ î
dθ
dt
+ cos θ ĵ
dθ
dt
A
unit ve
unit veU
ctor can
ctor can
ˆ
r̂ cos θ
θˆ
î
i + sin
i θ
θˆ
ĵ
j

0.3. NEWTON’S LAW OF MOTION 27
dr̂
dt
=

− sin θ î + cos θ ĵ
dθ
dt
(19)
Component − sin θ î + cos θ ĵ tells us that sin θ is measured along the −î i.e. along the
−x axis and cos θ is measured along the y axis. If a unit vector n̂ is drawn for these
components and it make angle α with −x axis, then its horizontal component is cos α. This
cos α must be equal to the sin θ and sin α must be equal to the cos θ so that − sin θ î+cosθ ĵ
is represented by n̂. Now
cos α = sin θ ⇒ sin
π
2
− α

= sin θ ⇒ α =
π
2
− θ (20)
Similarly for vertical component
sin α = cos θ ⇒ cos
π
2
− α

= cos θ ⇒ α =
π
2
− θ (21)
This means ∠AOB is
∠AOB = π −
π
2
− θ − θ =
π
2
(22)
This means n̂ is normal vector on the unit vector r̂. And − sin θ î + cos θ ĵ is n̂. Now
result is
dr̂
dt
= n̂
dθ
dt
(23)
This is the required answer.
0.3 Newton’s Law Of Motion
The basic concept of motion in horizontal plane are defined by Sir Isaac Newton. Newton’s
three principles of motion are
1. If a particle is in motion or in rest, there is no change in its state until, unless an
external force is applied.
2. Change in energy or momentum of the particle, which is in rest or in motion is
directly proportional to the quantum of force applied. Assume that mass of the
particle is m and external force F1
changes its energy. The force F is given by
F = ma
Here a is linear acceleration. If impact time is very small then
F = m(v − u)
Where u and v are initial and final velocity of the particle. Always remember that
velocity and acceleration of a particle is dependent on the current forces not the
history of the forces. When a force is removed, velocity becomes constant and
acceleration becomes zero instantaneously in the direction of force. For example,
when a particle is dropped from a window of a accelerating train in x-axis (say),
then just after drop, the acceleration of the particle along x-axis becomes zero due
to force in x-axis is removed and only gravitational acceleration acts on the particle.
1
If force exerted in such way that change in velocity of a particle taken place in finite time then force
is equal to ma. But if impact time is very small then force is equal to the change in momentum, i.e.
(mv2 − mv1).
n’s Law O
Law O
f ti i h i t l l d fi d b Si I
ti i h i t

28
3. There is always a reaction of an action.
u1 u2
(a)
v2
v1
(b)
Figure 7: Figure (a), two bodies just before collision. Figure (b), the colliding bodies just
after collision.
Solved Problem 0.20 A wave travels in a medium according to x = A sin(ωt). Find its
velocity and acceleration.
Solution The propagation equation of wave is
x = A sin(ωt)
Velocity of a wave which covers a distance x int time t is
dx
dt
. So
dx
dt
= A cos(ωt) ω
v = Aω cos(ωt) (24)
Now again diﬀerentiate the equation (24) with respect to time
dv
dt
= Aω (− sin(ωt) ω)
a = −Aω2
sin(ωt) (25)
This is the acceleration of wave motion.
0.3.1 Net Force
If two forces
F1 and
F2 are acting on a particle and passing from same point2
then
resultant or net force is
F =

F2
1 + F2
2 + 2F1F2 cos θ
Where θ is the angle between two forces. This net force tries to change the inertia of the
particle. If particle is accelerating with constant acceleration ‘a’ then
F = ma
If particle is moving without acceleration then F = 0 and in this case net force on the
particle is zero. It is also cleared that if a force applied on a moving particle then it
continuously accelerates.
2
If both forces are not passing through same point then they form force couple and tried to twist or
rotate the both about its axis.
Arun Umrao
v
v =
= Aω
Aω cos(
cos(ωt
ωt)
)

0.3. NEWTON’S LAW OF MOTION 29
Solved Problem 0.21 A balloon is tied to a block. The mass of the block is 2kg. The
tension of the string is 30N. Due to the wind, the string has an angle θ relative to vertical
direction as θ = sin−1 3
5
. Ignoring other forces ﬁnd (a) the components of force on the
block by the string. (b) the component of acceleration of block and (c) magnitude of the
force of the wind on the balloon if mass of the balloon is zero and the force of the wind
on the balloon is in the x−direction.
Solution
Fw
T
mg
θ
T T
mg
Fw
Fx
Fy T
mg
Fw Fx
Fy
Angle between string and vertical is given as θ = sin−1 3
5
that gives sine cosine
values from right angle triangle rule.
sin θ =
3
5
cos θ =
4
5
a Angle made between sting and vertical is θ. From symmetry of the angle, horizontal
vertical components of the force on block are
Fx = T sin θ
Fy = T cos θ
b Components of the acceleration can be obtained by using the relation F = ma for
both components of the forces acting on the string.
ax =
T
m
sin θ = 9m/s2
ay =
T
m
cos θ = 12m/s2
c From the ﬁgure, force by wind on the balloon is Fx and its value is
Fa = T sin θ = 30 ×
3
5
= 18N
These are the required answers.
Umrao
c
cos
os θ
θ =
=
m/
5
5

30
0.4 Equations of Motion In Horizontal Plane
There are three parameter of motion, distance, velocity and time. Three parameters are
inter-related as
v = u + at, (26a)
s = ut +
1
2
at2
, (26b)
v2
= u2
+ 2as, (26c)
Equation (26a) is called velocity time relation. Equation (26b) is called position time re-
lation. Equation (26c) is called position velocity relation. Acceleration is not independent
parameter of motion as its relation with velocity is a = (v − u)/t.
Proof of Ist Equation of Motion At any instant, initial velocity of a particle is u and
after time t its velocity changes to v u. From deﬁnition of acceleration
a =
v − u
t
or
at = v − u
or
v = u + at (27)
If ﬁnal velocity v u then particle is retarding then equation (27) becomes
v = u − at (28)
Proof of IInd Equation of Motion
Ist Method At any instant initial velocity of a particle is u and after time t its velocity
changes to v u. Average velocity of particle for this interval of time is u +
v − u
t
(initial
velocity + average velocity during time interval t).
u
u +
v − u
2
v
Figure 8:
Distance covered by particle during time interval t is
s = Average velocity × t
or
s =

u +
(v − u)
2

× t (29)
Arun
u
u the
theU
n particl
n particl
v
v = u at
t

0.4. EQUATIONS OF MOTION IN HORIZONTAL PLANE 31
or
s =

ut +
(v − u)
2t
t2

or
s = ut +
1
2
at2
(30)
Here (v − u) = at. If particle is decelerating then equation (30) becomes
s = ut −
1
2
at2
(31)
IInd Method Let a particle has initial velocity u and after time interval t its velocity
is vt. After attaining velocity vt, particle covers a small distance ds in time interval dt
then,
ds = vtdt
Where vt is any velocity within velocity range (from u to v). Writing value of vt from
equation (27)
ds = (u + at)dt
Integrating above equation from t = 0 to t = t we get
s
0
ds =
t
0
(u + at)dt
Above equation becomes
s = ut +
1
2
at2
(32)
If particle is retarding then equation (32) becomes
s = ut −
1
2
at2
(33)
Proof of IIIrd Equation of Motion Writing value of t from equation (27) into equation
(32) we get
s = u
(v − u)
a
+
1
2
a

(v − u)
a
2
Solving this equation we get
2as = 2(uv − u2
) + v2
+ u2
− 2uv
or
2as = v2
− u2
or
v2
= u2
+ 2as (34)
If particle is retarding then
v2
= u2
− 2as (35)
Arun Umrao


32
Solved Problem 0.22 An object is moving in a horizontal plane such that time is function
of velocity. Acceleration of object is a. Now find the rate of change of acceleration.
Solution Assume an object is moving with velocity v at any time t and its position
is x at that instant. Now
t = f(v) (36)
Differentiating above relation with respect to time t to get acceleration
1 = f
(v)
dv
dt
(37)
That gives
a =
1
f(v)
(38)
To get the rate of change of acceleration, above relation is again differentiated with respect
to t
da
dt
= −
f
(v)dv
dt
(f(v))
2 (39)
Substitute value of f
(v) from equation (38)
da
dt
= −
f
(v)a
1
a

2 (40)
On simpliﬁcation answer is
da
dt
= −f
(v)a3
(41)
Equation (41) is desired answer.
Solved Problem 0.23 The displacement vector is given by s = 3tî + 4t2
ĵ − 4k̂. Find the
velocity and the acceleration vector and its magnitude at time t = 2s.
Solution The velocity of an object is

v =
ds
dt
=
d
dt
(3tî + 4t2
ĵ − 4k̂)
= 3î + 8tĵ
Velocity at t = 2s is

v = 3î + 16ĵ
Velocity magnitude, in unit per second is
v =

32 + 162 =
√
265 = 16.28
Arun Umrao
dt

/
a
a

Similarly acceleration of and object is

a =
dv
dt
=
d
dt
(3î + 8tĵ)
= 8ĵ
Acceleration at t = 2s is

a = 8ĵ
Direction of acceleration is along the y−axis. Magnitude of acceleration in unit per square
second is
a = 8
These are the required answers.
Solved Problem 0.24 A cyclist is cycling in horizontal plane with velocity of 20m/s. He
saw an inclined plane of inclined angle 30o
. When cyclist head-on the foot of incline
plane, he stops applying force on the cycle paddle. Cyclist moves in inclined plane upto
a distance s. Find the distance that the cycle cover in the inclined plane.
Solution
u
30o
g
gh
gv
(a)
s
(b)
A cyclist moves in horizontal plane with velocity 20m/s. When cyclist starts to climb
inclined plane, it halts the applying force. Cyclist will run for a distance s in inclined
plane and ﬁnally came to rest. Here motion is neither in horizontal plane nor in vertical
plane. The retarding force that reduces the velocity of the cyclist is the mass force of
the cyclist working opposite to the direction of motion along the surface of the inclined
plane. Using third equation of linear motion for inclined plane
v2
= u2
− 2g sin(30o
) × s
Simplifying and substituting the value of u, we have
s =
u2
2g sin(30o)
=
202
2 × 10 × 0.5
= 40m
The distance covered by cyclist is 40m.
Arun Umrao
U
U
tes
es
p t
it
sit
:/
://
://s
/s gle
gl
g
g
gh
h
le
e
g
gv
ru
ru

34
Solved Problem 0.25 The position function of a particle undergoing linear motion is
s = t2
− 9t + 3. Find the velocity of the particle when t = 2s. Also find the average
velocity between 2 ≤ t ≤ 5s. If time is measured by a faulty clock and there is a probability
of error of ±0.1s, in each measurement, then find the error in the measured velocity when
t = 2s.
Solution The distance function is not linear function, hence velocity is changing non-
linearly with time. Therefore, instantaneous velocity of the particle is ds/dt. So,
v =
ds
dt
=
d
dt
(t2
− 9t + 3)
It gives, velocity of the particle at any time t
v = 2t − 9
Now, velocity at time t = 2s is
v|t=2s = 2 × 2 − 9 = −5m/s
The average velocity is given by
vavg =
s|t=5s − s|t=2s
5s − 2s
On substituting the values of instantaneous velocities at t = 2s and t = 5s, and simplifying
the expression, we have
vavg =
−17 + 11
3
= −
6
3
= −2m/s
Error in velocity measurement with time is obtained by getting time derivative of the
velocity.
dv
dt
= |2|
So,
dv = 2 × dt
Now, the error in velocity measurement is dv = 2 × 0.1 = ±0.2m/s.
Solved Problem 0.26 The distance function of a particle undergoing linear motion is
s = t3
−6 in meter. Find the acceleration of the particle when t = 2s. If time is measured
by a faulty clock and there is a probability of error of ±0.15s, in each measurement, then
find the error in the measured distance when t = 2s. Meter scale, which is used here has
least count of 1cm.
Solution The distance function is not linear function, hence velocity is changing non-
linear with time. Therefore, instantaneous velocity of the particle is ds/dt. So,
v =
ds
dt
=
d
dt
(t3
− 6)
Arun
e value
e value Umrao
of instan
of instan
have
have

It gives, velocity of the particle at any time t
v = 3t2
Acceleration of the particle is obtained by getting time derivative of the velocity.
a =
dv
dt
= 6t
So a = 6 × 2 or a = 12m/s2
. Error in distance measurement with time, is obtained by
getting time derivative of the distance function.
ds
dt
= 3t2
So,
ds = 3t2
× dt
Substituting the values of t and dt, we get
ds = 3 × 22
× 0.15 = ±1.8m
Meter scale used in measurement of distance has least count of 1cm. The error is found
in one decimal place of the meter, i.e. multiple of 10cm. Hence the error is ±1.80m.
Solved Problem 0.27 A motor driver driving vehicle with speed of um/s and accelerate
for time t and then retarded for time t with same acceleration or retardation a. If (s)he
covers a distance d then ﬁnd the time for which (s)he accelerate or retarded.
Solution Let the distance covered by vehicle in time t is s1 when vehicle is accelerated
then
s1 = u t +
1
2
a t2
Now the velocity of vehicle, after time t after acceleration is v = u + a t. The distance
covered by the vehicle in time t during retardation is s2
s2 = v t −
1
2
a t2
Now according to question
d = s1 + s2
d = u t + v t (42)
Substituting the value of v in equation (42)
d = u t + (u + a t)t
d = 2 u t + a t2
The time t can be calculated by solving above equation for t
t =
−2 u ±
√
4 u2 + 4a d
2 a
(43)
27 A Umrao
motor dr
n retarded for time t w th same acceleration or re
d then ﬁnd the time for which (s)he accelerate or

36
t =
−u ±
√
u2 + a d
a
(44)
There are two values of times in which the vehicle can cover same distance d but negative
time has no meaning hence time t is
t =
−u +
√
u2 + a d
a
(45)
Solved Problem 0.28 A motor driver driving vehicle with speed of um/s and (s)he see a
man at distance d. (S)he use brakes and stop the vehicle just near the man. Find the
retardation of the vehicle.
Solution The distance covered by vehicle before halt is d. If retardation of vehicle is
a and ﬁnal velocity is v = 0 then using third equation of linear motion
v2
= u2
− 2 a d
On simplifying the equation result will be
u2
= 2 a d
a =
u2
2 d
(46)
This is the retardation of vehicle to halt it just before the man.
Solved Problem 0.29 Two motor drivers are driving vehicles in same direction with initial
velocities u m/s and 2u m/s respectively and acceleration of vehicles are a m/s2
and
a/2 m/s2
respectively. If vehicle 1 travels a distance d from beginning then ﬁnd how far
behind the vehicle 2 is.
Solution Let both vehicle travels for time t in which vehicle 1 travels a distance d.
Now time for this distance is
d = 2u t +
1
2
a
2
t2
Solving above equation the time t is
t =
2
√
4 u2 + a d − 4 u
a
The distance covered by vehicle 2 is s then
s = u t +
1
2
a t2
Substituting the values of t in above equation
s = u

2
√
4 u2 + a d − 4 u
a

+
1
2
a

2
√
4 u2 + a d − 4 u
a
2
Arun Umrao
tion of vehicle to halt it just before the man.
tion of vehicle to halt it just before the man.

0.5. MOTION IN VERTICAL PLANE 37
Or
s =

2
√
4 u2 + a d − 4 u
a

u +
1
2
a

2
√
4 u2 + a d − 4 u
a

Or
s =

2
√
4 u2 + a d − 4 u
a

4 u2 + a d − u

Or
s =
1
a

2(4u2
+ ad) − 2u

4u2 + ad − 4u

4u2 + ad + 4u2

Or
s =
1
a

12u2
+ 2ad − 6u

4u2 + ad

Now the distance between them after time t is d − s hence vehicle 2 is lagging behind by
1
a

−12u2
− ad + 6u

4u2 + ad

(47)
0.5 Motion In Vertical Plane
A particle moving in horizontal plane has no significant effect of Earth’s gravity. Velocity
of particle changes when external force is applied. This force alters the acceleration of the
particle. In vertical direction, particle falls or jumps under the effect of gravitational force
of the Earth. Gravitational acceleration is equal to 9.8m/s2
in SI unit. Three equations
of vertical motion for a falling particle are
v = u + gt, (48a)
h = ut +
1
2
gt2
, (48b)
v2
= u2
+ 2gh. (48c)
Here g is gravitational acceleration and h is vertical height. If particle is thrown vertically
upward under gravitational force of the Earth then equation of vertical motion become
v = u − gt (49a)
h = ut −
1
2
gt2
(49b)
v2
= u2
− 2gh (49c)
Velocity Position Gradient
Acceleration can also be represented in terms of velocity and velocity gradient.
Assume a function y which depends on time t. First derivative of this function is given
by
dy
dt
=
d
dt
y
Arun
when e
when Umrao
xternal f
xternal f
direction, particle falls or jumps under the effect o
direction, particle falls or jumps under the effect o
itational acceleration is equal to 9
itational acceleration is equal to 9 8
8m/s
/ 2
2
in SI un
i SI

38
Now again derivating above relation, we have
d
dt
dy
dt
=
d
dt

dy
dt

Or
d2
y
dt2
=
d
dt

dy
dy
×
dy
dt

=
dy
dt

d
dy
×
dy
dt

Substituting dy/dt = v, we have
d2
y
dt2
= v

d
dy
v

This relation is useful when acceleration, velocity and position relation is to be ob-
tained.
Vertical motion in constant and variable gravity
Three equations, described above for vertical motion are useful when gravity does
not varies when object moves from Earth’s surface to a height of h. If so, we can not
use relations described above. For variable gravity we have to used center force law.
Suppose at any height y from the earth center, mass is placed. Now from Newton’s
gravitational force law
F = −G
Mem
y2
(50)
Negative sign shows that force on the mass acting downward that is opposite to the
direction of motion of mass. Now if gravitational acceleration at the height y from earth
center is g
then F = mg
and
mg
= −G
Mem
y2
R2
e
R2
e
(51)
This gives
g
= −
gR2
e
y2
(52)
Where g is the gravitational acceleration of earth at surface. Now g
depends on the
position of point from Earth’s center. So
g
=
d2
y
dt2
(53)
And
d2
y
dt2
= −
gR2
e
y2
(54)
bject m
U
moves fro
ribed above. For variable gravity we have to use
g y
h h t

This is the equation of motion of mass in varying gravity. In terms of velocity above
equation can be written as
v
dv
dy
= −
gR2
e
y2
(55)
Both equations are very important in real projectile motion.
Solved Problem 0.30 The instantaneous height of an object moving in vertical direction
is given by h = −12t2
+ 36t + 86 where h is in meter and t is in second. Find (a) the
velocity of object when t = 0, (b) Maximum height attained by the object and time when
it occurs, and (c) its final velocity when h = 0.
Solution The height function is given by h = −12t2
+ 36t + 86. The velocity function
will be
dh
dt
= −24t + 36
a the velocity of object when t = 0 is

dh
dt

t=0
= −24 × 0 + 36 = 36m/s
b The initial velocity is 36m/s and final velocity at the maximum height will be zero.
Now the time when object reached at the highest point is when v = 0
0 = −24t + 36
It gives t = 1.5s. This means the moving object will attain its maximum height after
1.5s. Now the maximum height is
h = −12 × 1.52
+ 36 × 1.5 + 86 = 113m
c When h = 0 then
−12t2
− 36t + 86 = 0
And it gives roots of time t = −1.59s and t = 4.57s. Only t = 4.57s is acceptable due to
positive time position. Now velocity at h = 0 is

dh
dt

t=4.57
= −24 × 4.57 + 36 = −73.6m/s
Arun Umrao
locity is 36
locity is 36m
m/s
/s and final velocity at the maximum
and final velocity at the maximum
n object reached at the highest point is when
n object reached at the highest point is when v

40
−100
0
100
−3 −2 −1 0 1 2 3 4 5 6
h
v
Figure 9: Height and velocity are graphed as function of time ‘t’.
Solved Problem 0.31 In deep and dark well, a piece of rock is dropped. Sound of hit is
heard after 6s. Find the depth of well.
Solution
h
Figure 10: Well-rock-sound model.
We know that an object falling under the gravity of the earth. The rock particle hits
well ﬂoor with after time t1 hence the depth of well is
h = ut1 +
1
2
gt2
1
The same depth is covered by the sound with a speed of 332m/s in rest of time (6 − t1)s,
hence
h = 332 × (6 − t1)
Solving above two relations, we have
332 × (6 − t1) =
1
2
gt2
1
as initial rock particle dropping velocity is zero. Now solving for time t1
t1 = 5.54s
Arun Um
mrao
m o
mrao
m
sites.g
sites.g
g
s
sites g
t w/arun
w/arun
w/arun
a mr
h
oog
oog

Now depth of well is
h = 332 × (6 − 5.54) ≈ 153m
Solved Problem 0.32 A man drops two balls from heights 25m and 30m. Find the
velocities of both balls when they hit ground. Also ﬁnd the velocity-to-gravity ratio.
Solution
25m
30m
u1
v1
u2
v2
Figure 11:
The two drop balls fall under the gravitational force only. Initially both balls are
dropped from rest, hence the velocity at the ground of the two towers are
v2
1 = u2
+ 2gh1 = 2 × 10 × 25
= 500
v1 =
√
500m/s
and
v2
2 = u2
+ 2gh2 = 2 × 10 × 30
= 500
v2 =
√
600m/s
Now the velocity-to-gravity ratio of the two bodies are
v1
g
=
√
500
10
=
√
5s
v2
g
=
√
600
10
=
√
6s
balls f
balls f Umrao
all under
all under
hence the velocity at the ground of the two towe
hence the velocity at the ground of the two towe

42
Solved Problem 0.33 Mr Bablu is at the top of the roof and Mr Ram is at ground. Both
throw balls with initial velocity 12m/s towards each other. Find the distance where both
balls will pass to each other. Given, height of building is 24m.
Solution
12
m
s
12
m
s
24m
h1
h2
Figure 12:
Let after time t, both balls will meet to each-other. So the height covered by the
ground ball is
h1 = 12 × t −
1
2
gt2
and the depth covered by top ball is
h2 = 12 × t +
1
2
gt2
Adding both equations,
h1 + h2 = 24 × t
Total height of tower is 24m so
t = 1s
Now the height covered by both balls are
h1 = 12 × 1 −
1
2
× 10 × 12
= 7m
h2 = 12 × 1 +
1
2
× 10 × 12
= 17m
This is required results.
Solved Problem 0.34 The gravity of moon is one-sixth of the earth. A particle is dropped
in both planets from same height. Find Moon-to-Earth ratio of times taken by both
bodies to reach surface.
Arun Umrao
h
h1
1 = 12
= 12 ×
× t
t −
−
1
vi
2
2
g
gt
t2
2

Solution Let both objects dropped from a height of h from rest. The time taken by
object in moon and earth are t1 and t2 respectively. Now
h = ut1 +
1
2
g1t2
1
h = ut2 +
1
2
g2t2
2
Equating both relations
1
2
g1t2
1 =
1
2
g2t2
2
On simplifying
t1
t2
=

1
6
This is the required relation.
Solved Problem 0.35 A ball starts rolling in inclined plane at its highest position as shown
in ﬁgure (13). Find the velocity at the end of inclined plane. Again ﬁnd the velocity of
ball if it falls freely in vertical plane. Analyze and explain your answer.
Solution
vp
vd
vR
a
b
θ
g
g cos θ
Figure 13:
Ball starts rolling from the highest position of the inclined plain. Ball rolls along the
surface of plain. Gravity along the surface of the plain is g cos θ. Now the velocity of the
ball at the lowest point of the plain is
v2
p = u2
p + 2gphp
The value of hp, i.e. length of plain is
hp =
b
sin θ
Now the velocity of ball at the bottom of inclined plain is
v2
p = 2g cos θ ×
b
sin θ
vp =

2gb cotθ
Arun
n
n Umrao
mra
mr
mr
sites.g
s θ
θ

44
If ball falls freely vertically from the height a from the rest. Now velocity of the ball at
the lowest point is
vd =

2ga
a and b are mutually related as
a = b cot θ
and
vd =

2gb cotθ
It shows that velocities of ball at lowest point when it moves along the surface of the
inclined plain or when it falls freely vertically downward from the top of the inclined
plain are same. But, ball travels longer when it moves along the surface of the inclined
plain.
Solved Problem 0.36 Two persons start walking from one corner of a rectangular field
along the perimeter. If the velocities of persons are 3m/s and 2m/s respectively, then
find the point where they would meet each other. Dimensions of field are 120m × 40m.
Solution Perimeter of the rectangle is 320m. Assume after time t they will meet with
each other. The distance covered by both persons after time t is (2 + 3)t = 5t.
120m
40m
2m/s
3m/s
8m
This distance should be equal to the perimeter of the rectangular filed. It gives
t = 320/5 = 64s
The distance covered by the first and second persons are 192m and 128m respectively.
Now the coordinates in rectangle where both will be met is (120, 8).
120m
40m
3m/s
2m/s
88m
If the direction of velocities of mutually interchanged then point where both persons
will meet is (88, 40).
A un
3m Umrao
U
/s
s://sites.google.com/view/arunum
ht
m
s:
s
t //

Solved Problem 0.37 A rocket, initially at rest and close to the ground is accelerate
vertically upward with a acceleration a for a time t = 0 to t = t1. After that fuel is
exhausted. Find the maximum height upto which rocket can reach. If the rocket’s net
acceleration is equal to the gravity then ﬁnd the total time upto which the rocket is in
the air. Ignore other factors and forces.
Solution Initially rocket is in rest and accelerates vertically for time t = 0 to t = t1
with an acceleration a. Height of rocket during the forced motion from the time t = 0 to
t = t1
h1 = 0 +
1
2
at2
1 =
1
2
at2
1
Now velocity attained by the rocket during this motion is
v = u + at1 = at1
Now the fuel of rocket is exhausted and it moves vertically as motion under the gravity
with initial speed of v. The height attained by the rocket is obtained from the third
equation of linear motion in vertical plane. Final velocity is zero.
h2 =
v2
2g
=
a2
t2
1
2g
Now, total height attained by the rocket is
h1 + h2 =
a2
t2
1
2g
+
at2
1
2
Time, t2, taken by the rocket to reach its maximum height, is given by ﬁrst equation of
linear motion in vertical plane. So,
v
= u
− gt2
Here, v
= 0 and u
= at1. So,
t2 =
at1
g
Now, time taken by rocket in covering the height h1 + h2 under the free fall condition
with initial velocity zero. So,
a2
t2
1
2g
+
1
2
at2
1 = 0 +
1
2
gt2
3
It gives
t3 =

a2t2
1
g2
+
at2
1
g
Total time of rocket upto which it is in air is given by sum of t1, t2 and t3. So,
T = t1 + t2 + t3
Or
T = t1 +
at1
g
+

a2t2
1
g2
+
at2
1
g
Arun Umraoh
h
2g
g 2

46
If a = g then
T = 2t1 + t1
√
2 = (2 +
√
2)t1
0.5.1 Effect Of Center of Mass In Vertical Motion
Center of mass is point inside or outside of the particle where gravitational acceleration
or gravitational force acts. If object is hang freely in air then vertical axis of the mass
system or mass particle passes through the center of mass. Discussion of center of mass
is important as total energy of mass particle changes according to the position of center
of mass. For example kinetic or potential energy of small bodies or point mass3
depends
on the position of center of mass. Kinetic energy of mass m is
KE =
1
2
mv2
(56)
Where v is velocity of point mass m. Potential energy of mass is given by
PE = mgh (57)
Where h is height of the center of mass. Now suppose a system in which particle mass is
not point mass but distributed in large volume. For example a ball of mass m1 is floating
in the liquid of mass m2 filled in a cylinder. Liquid is filled in cylinder up to a height of
h and area of cross section is A. The center of mass of this cylindrical water column is
at height of h/2 and total mass can be assumed to be concentrated at this height. Now
potential energy of the liquid column is
PEw = m2g
h
2
(58)
m2
m1
(a) (b) (c)
Figure 14: In figure (a), center of mass of water column and ball is shown separately. In
figure (b), ball is at the bottom of the liquid column and their common center of mass is
between the center of masses of liquid column and ball. Similarly in figure (c) center of
masses of liquid column, ball and common center of mass is shown.
3
Point mass is an assumption in which it is assumed that whole mass of the particle is concentrated
at its center of mass and particle shape or mass distribution through-out the particle has no effect in the
system.
Arun
s sectio
s sectioUmrao
n is
n is A
A.
.
nd total mass can be assumed to be concentrated
nd total mass can be assumed to be concentrated
th li id l i
li id l i

A small ball of mass m1, let to ﬂoat in water, center of mass of this system is shifted
upward. New position of the center of mass is
h
=
m1h + m2
h
2
m1 + m2
(59)
Here value of h
is greater than that of h/2, this means the potential energy of this system
is increased. New potential energy of the system is
PEs = (m1 + m2)g × h
= (m1 + m2)g ×

m1h + m2
h
2

m1 + m2
(60)
Or
PEs = g

m1h + m2
h
2

(61)
This equation shows that the total potential energy of the system is sum of the potential
energies of each individual bodies. Change in final potential energy is from the work done
by external force to put the ball at surface of water by doing some work. This concept is
valuable for finding the energy of a system in which center of mass of system constituents
are changing their position. A mass system in which heavy particle is moving in vertical
liquid column, resultant center of mass changes with time and total potential energy
varies correspondingly with instantaneous center of mass of the mass system.
Solved Problem 0.38 A vessel is filled with water upto height of 2m. The area of cross-
section of water vessel is 20cm2
. Find the center of mass of the system. A ball is let to
float in water column. Mass of the ball is 400gm. Find the change in potential energy of
the system. Explain the reason behind this change in the energy. If ball is made of iron
and settle down at the bottom of vessel then what is change in the potential energy of
the system and how it differ with previous result. If this potential energy is lesser than
the previous one then where is the energy loss gone?
Solution Assume base of vessel is the zero line from where we have to calculate the
potential energies. Mass of the water column is
m = 20 × 10−4
× 2 × 103
kg
On simplification m = 4kg. The center of mass of this system is at 1m from the base of
the vessel. The potential energy of the system is
PE1 = mgh = 4 × 10 × 1 = 40.00 Jule
If ball of 0.4kg is placed at the surface of water and it is floating at the height of 2m.
Now center of mass of the system with respect to the bottom of the vessel is
CPnew =
4 × 1 + 0.4 × 2
4.40
= 1.09 m
Now new potential energy of the system is
PE2 = 4.4 × 10 × 1.09 = 47.96 Joule
Arun
.38 A U
vessel is
essel is 20cm . Find the center of mass of the sys

48
Now change in the potential energy is
PE = PE2 − PE1 = 47.96 − 40.00 = 7.96 Joule
This enhancement of energy is due to the work done by external force in placing it on the
surface of water in vessel. If ball is made of iron, then it would not float but settle down
at the bottom. Now new center of mass of the system with respect to the bottom of the
vessel is
CPnew =
4 × 1 + 0.4 × 0
4.40
= 0.91 m
Now new potential energy of the system is
PE3 = 4.4 × 10 × 0.90909 ≈ 40.00 Joule
Now there is no change in the energy as it is equal to the initial potential energy of the
system. This is because, potential energy of the ball that also increases potential energy
of the system is converted into the kinetic energy of the ball when it start moving down
and finally settle down at the bottom or at the base line. Ball has to do work to overcome
the drag force or viscous force of water during downward motion. The extra energy of
the system 7.96J, dissipated in form of heat during downward motion of the ball.
0.5.2 Velocity Curve
The velocity curves are drawn between time and instantaneous velocity of an object at
that time. The velocity behavior of an object is dependent on, direction of velocity, force
acting on the object and collisions if any.
0.5.3 Linear Velocity Curve
An object moving under constant acceleration, then its velocity increases continuously
with time and velocity curve is
t
v
(a)
t
v
−u
u = 0
t
(b)
t
v
u
(c)
In the curve 0.5.3 (a), when time is zero, velocity is zero, i.e. initial velocity of the
object is zero. Velocity of the object continuously increases with time. In the figure
0.5.3 (b), when time is zero, velocity is −u. Velocity is zero at time t
. After t t
,
object accelerates and velocity continuously increases. In this case, first velocity of object
decreases, came in rest and then object accelerated. In the figure 0.5.3 (c), when time
is zero, object velocity is u, i.e. velocity of the object is neither negative nor zero when
time measurement is started.
ocity b
ocity b Umrao
ehavior o
ehavior o
t and collisions if any.
t and collisions if any.

0.5.4 Vertical Throw
When an object is throws vertically upward, initially its velocity is maximum and it
reduces gradually to zero when object reached upto its maximum height. Now, under the
gravitational force, object starts moving downward, i.e. in opposite direction to initial
velocity. Its velocity increases continuously until it reached to the earth. Taking the
direction of upward velocity as positive velocity, +u, and downward velocity as negative
velocity, −u. The velocity curve of this system is shown below.
u
u = 0
−u
t
v
If object-Earth collision is elastic then after collision, object will rebound with velocity
u. The velocity curve shall be as shown below:
u
u = 0
−u
u
u = 0
−u
t
v
If object earth collision is not elastic then velocity after ﬁrst rebound shall not be
equal to +u but it shall be +v +u and in this case, height of object in ﬁrst rebound
shall be less than the height of object in upward throw.
Arun Umrao
http
tt
tt
0
0 /si
s
su
u =
= 0
0
v

50
u
u = 0
−u
v
v = 0
−v
t
v
0.5.5 Vertical Drop
When an object is dropped vertically from a height of h, initially its velocity is zero and it
increases gradually to maximum when object reached to the ground. When object collides
with ground it rebounds vertically upward. and direction of its velocity is changed. While
magnitude remains same. Again, object moves vertically upward direction and its velocity
reduces gradually due to negative gravitational attraction. At its maximum height its
velocity becomes zero. The velocity curve of this system is shown below.
u
u = 0
−u
t
v
After ﬁrst rebound, object reached its maximum height and stars moving downward
again. The velocity curve shall be as shown below:
u
u = 0
−u
u
u = 0
−u
t
v
If object Earth collision is not elastic then velocity after ﬁrst rebound shall not be
Arun Umrao
a
://site w un

equal to +u but it shall be −v +u (in magnitude) and in this case, height of object
after first rebound shall be less than the initial height of object at first fall.
u
u = 0
−v
v
v = 0
−w
t
v
Solved Problem 0.39 Find the maximum height of an object whose displacement function
is h(t) = 2t3
− 4t + 2 in vertical direction. Also find the velocity of the particle at t = 2s.
Solution The given displacement function is h(t) = 2t3
− 4t + 2. Object, that is
moving in vertical plane in accordance with this displacement function, when achieves its
maximum height, its velocity becomes zero, i.e. v = dh/dt = 0. So,
v =
dh(t)
dt
= 6t2
− 4 = 0
It gives, t =

2/3 = ±0.81. Now, the maximum height is at either t = +0.81s or at
t = −0.81s.
h(+0.81) = 2 × (0.81)3
− 4 × 0.81 + 2 = −0.18m
and
h(−0.81) = 2 × (−0.81)3
− 4 × −0.81 + 2 = 4.18m
So maximum height is 4.18m. Velocity of particle at t = 2s is
v = 6t2
− 4 = 20m/s
Solved Problem 0.40 The head top rotor of a helicopter provides maximum thrust of
60kN. Mass of the helicopter is 1500kg. Now find (i) maximum uplift acceleration, (ii)
maximum translational acceleration when it carry 500kg payload.
Solution The head top rotor of a helicopter rotates about the vertical axis as shown
in the first part of following figure. When it moves ahead, then rotor is slightly tilted
downward as shown in the second part of the following figure.
Arun Umrao
v
v
dt
dt

52
x
y
mg
F
F
x
y
mg
F Fv
Fh
mg
θ
i The maximum upward acceleration is found when helicopter lifts empty. Now in
this case
F − mg = mav
Substituting the values, we have
60000 − 1500g = 1500av
On solvign it, we have av = 30m/s2
.
ii Mass of loaded helicopter is M = 1500 + 500 = 2000kg. Helicopter is moving
forward, therefore head top rotor is tilted by an angle θ from the vertical as shown in
second part of the above ﬁgure. Resoluting the thrust F vertically and horizontally.
Resoluted forces are Fv = F cos θ and Fh = F sin θ. When helicopter achieves suﬃcient
height, Fv is balanced by Mg. So
60000 cosθ = 2000 × 10 = 20000
It gives cos θ = 1/3. Now, translational acceleration is
Mah = Fh = 60000 × sin θ = 60000 ×

1 − cos2 θ = 60000 ×

8/9
It gives ah = 28.28m/s2
.
0.6 Relative Velocity
When an observer observes motion of a particle with respect to ones frame of reference
(rest or moving) then the velocity of observing particle is called relative velocity of that
particle w.r.t observer’s velocity. There are several cases which are given below.
1. When car A is in rest and car B is motion.
Arun Umrao
ave
ave a
av
v = 30
= 30m/s
m/s .
.

0.6. RELATIVE VELOCITY 53
vA = 0 vB
Figure 15:
In this case the relative velocity of car
vB with respect to car
vA is

vBA = velocity of car B - velocity of car A

vBA =
vB −
vA (62)
2. When car is in trawler.

vc=0

vt
Figure 16:
In this case the relative velocity of trawler w.r.t car is

vT C = velocity of trawler - velocity of car
Here velocities of car and trawler are same. Hence (velocity of trawler - velocity of
car = 0) ie

vT C = 0 (63)

vc

vt
Figure 17:
In ﬁgure (18), both trawler and car are in motion while car is in trawler’s frame of
reference. The resultant velocity of car with respect to the velocity of trawler is
vc
while the velocity of car with respect to the frame of reference at rest is
vT +
vc.

vc

vt
Figure 18:
Arun

Umrao
l
l
vT C
T C velocity of trawler velocity of c
velocity of trawler velocity of c
(

54
In ﬁgure (18), both trawler and car are in motion in opposite direction while car
is in trawler’s frame of reference. The resultant velocity of car with respect to the
velocity of trawler is −
vc while the velocity of car with respect to the frame of
reference at rest is
vT −
vc.
3. When both cars are moving in same direction.

vA
vB
Figure 19:
In this case the relative velocity of car B w.r.t car A is

vBA = velocity of car B - velocity of car A

vBA =
vB −
vA (64)
4. When both car are moving in opposite direction.

vA
vB
Figure 20:
If both cars are moving in opposite direction then assuming velocity direction of car
B as positive then velocity direction of car A is negative. In this case the relative
velocity of car B w.r.t car A is

vBA = velocity of car B - (- velocity of car A)

vBA =
vB − (−
vA)

vBA =
vB +
vA (65)

vA
vB
Figure 21:
Equation (65) is same for both cases when car A and car B are moving toward each
other or moving away from each other.
Arun
n
A
A
Arun Umrao
mra
Um
vA
https
h
h
https
tps mrao
mrao
m
mrao
ao
m

0.6.1 Equations of Motions for Relative Motion
Equations of linear motions are also used for relative motions. Take two cars moving with
initial velocities u1 and u2 respectively were accelerated for t seconds with accelerations
a1 and a2 respectively. After time t, their final velocities will be
v1 = u1 + a1t; v2 = u2 + a2t
Subtraction first equation from second equation, we have
v2 − v1 = u2 − u1 + (a2 − a1)t
Or
Δv = Δu + (Δa)t
Here Δv is final relative velocity, Δu is relative initial velocity and Δa is relative accel-
eration. This is first equation of relative motion.
Solved Problem 0.41 Two trains are running in parallel tracks in opposite direction with
speeds of 108km/h and 90km/h respectively. At a time t, engines of both trains are just
bruising to each other. After 9s, the trains crossed each other completely, Find the total
length of both trains.
Solution Both trains are running in opposite direction, hence their relative velocities
are 108 + 90 = 198km/h or 55m/s. If both trains crossed to each other in 9s then length
of both trains is l = vrel × t. It gives l = 55 × 9 i.e. 495m.
Solved Problem 0.42 Two trains of length 400m each are running in parallel tracks
towards each other with velocities of 144km/h and 72km/h respectively. Initially, engines
of both trains are just touching the two ends of the bridge having length 950m. Find the
time when both trains completely crossed to each.
Solution To cross the trains completely to each other, they should have to travel a
distance of 400 + 400 + 950 = 1750m. The relative speed of the trains are 144 + 72 =
216km/h or 60m/s. The time taken by trains to cross completely to each other is
t =
1750
60
It gives, 29.16s.
Solved Problem 0.43 Two vehicles were approaching to each other with velocities of
10m/s and 15m/s respectively. At time t, distance between them was x. If cars meet
each-other after 50s then find x.
Solution The cars are moving in opposite direction, hence their relative velocity is
10 + 15 = 25m/s. Initially the distance between the cars is xm. If cars meet after time
50s then distance x can be given as
x = vrel × t
It gives x = 25 × 50 = 1250m.
Arun
km/h
km/h
v Umrao
or 55
or m/s
/
×
× t
t It giv
It giv
rel g
p g g

56
Solved Problem 0.44 A car of length l = 20m is approaching to a light pole of negligible
width. If speed of the car is 36km/h then find the time taken by the car to cross the pole.
Solution
l
Car is approaching to the pole of negligible width. Car has to travel a distance of its
length to cross the pole as shown in the figure given below.
l
Speed of the car is 36km/h or 10m/s. Now, the time taken by the car to cross the
pole is 20/10 = 2s.
Solved Problem 0.45 A plane is flying from city A to B, located at distance of 150km
with a speed of 250km/h. While plane was flying just above the city A, a laminar storm
hits it with speed of 20km/h from the direction of city B. Find the time taken by the
plane to reach city B.
Solution
Solved Problem 0.46 A train was running with speed of 126km/h towards a pole. If
length of train is 140m then find the time taken by the train to cross the pole.
Solution The velocity of the train is 126km/h or 35m/s. To cross the pole by train,
the train should travel by a distance equal to its length. So, time taken by train to cross
the pole is
t =
140
35
s
or, train will take 4s to cross the pole.
Solved Problem 0.47 Two trains A and B, each 400m long are running in parallel tracks
while B is ahead of A and they have same speeds. If train A takes 50s with an acceleration
of 1m/s2
to overtake train B. Find the original distance between them before train A
starts accelerating to overtake the train B.
Arun
n Umrao
U
i 36k /h 10 / N th ti t k b
36k /h 10 /

Solution Speed of both trains are equal initially. In time of 50s, distance covered by
train B is 50u while distance covered by the train A is
dB = 50u +
1
2
× 1 × 502
Now initial distance between two train was
d = dB − 400 − 400 − 50u
= 50u +
1
2
× 1 × 502
− 800 − 50u
= 450m
Initial distance between two trains was 450m.
Solved Problem 0.48 A car of length l1 = 20m is overtake to the standing car of length
l2 = 20m. If speed of overtaking car is 20m/s then ﬁnd the time in which car will overtake
to the standing car completed.
Solution According to the question, the moving car overtakes to the standing car
with a speed of 20m/s.
l1
v
l2
To overtake the car in rest by moving car, it should cover a distance of l1 + l2.
l1
v
l2
Now, the time taken by the car is
t =
l1 + l2
v
=
40
20
It gives, t = 2s, i.e. moving car shall overtake the car in rest in 2s.
Solved Problem 0.49 Two cars B and C are approaching to car A in opposite direction
with constant velocity 12m/s as shown in ﬁgure (22) and (23). Car B starts overtake
to the car A, running with speed of 10m/s without colliding with car C. Find the
acceleration of car B. Given, before acceleration of car B, distance between cars A and
B is 1500m and cars A and C is 1500m.
Solution
Arun
Arun
Arun Umrao
Umrao
Umrao
ao
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
ao
ao
ao
ao
ao
ao
ao
ao
ao
htt // it l / i /
co

58
A
v a
B
v
C
v
Figure 22: Positions of cars before overtaking in highway.
A
v a
B
v
C
v
Figure 23: Position of cars after overtaking in highway.
A
v a
B
v
C
v
A
B
C
d vt
d − vt
Solved Problem 0.50 Between two cities, A and B, taxis are ply in constant interval of
time T , with constant speed. A family car is driven with speed of 60km/h in the highway
joining the cities A and B, in the direction from city A to city B. The driver of the family
car observes that after each interval of 18 minutes, the taxi overtakes from his back side
while in each 6 minutes taxi passing the family car from the front side. Now, ﬁnd the
interval T .
Solution According the question, interval between two consecutive taxis ply between
cities A and B is T minutes. Velocity of taxis and family car are v and u respectively.
Arun
gure 2
gu
Umrao
3: Positio
3:

v u
v
For the ﬁrst case of family car and the taxis overtaking from the back side, at any
instant of time, t = 0, a taxi, x (say) is just overtaken to the family car f (say).
y
v
f
u
x
The the next taxi, y (say), which shall overtake the family car from backside is just
behind the T minutes. After 18 minutes taxi y will overtake the family car, f. In this
18 minutes, the family car shall also cover some distance in its direction. So the ﬁnal
positions shall be as
y Z f F Y
18v
18u
T v
The old and new positions of the family car and taxis are represented by ‘labels in
small letters’ and ‘labels in block letters’ respectively. Taking length of car negligible, we
have
T × v + 18 × u = 18 × v
It shall give
T × v = 18 × v − 18 × u
Now, in case of family car and taxis passing the car from front side.
f
u
x y
v
Arun
Arun Umrao
Umrao
tps://s
ps:
ps:
tp ://s
Z google
oog
oog
g gle
f m/view
/vie
/vie
m/ ew
F /arunu
aru
aru
/a unu
Y
Y
Y

60
The taxi x just passes the family car f at any instant of time t = 0. Now, after
6 minutes the family car shall move forward direction and in the same time the next
approaching taxi y (say) shall pass the family car as shown below:
f F
X x Y y Z
6u 6v
T v
The old and new positions of the family car and taxis are represented by ‘labels in
small letters’ and ‘labels in block letters’ respectively. Now, from the above figure,
T × v = 6 × u + 6 × v
Now, solving the two equations of taxi and family car overtaking, we have
18 × v − 18 × u = 6 × u + 6 × v
It shall give relation, v = 2u. Now, the ply time is
T × 2u = 6 × u + 6 × 2u
It gives T = 9 minutes.
Solved Problem 0.51 An airplane flies between two cities at distance D. Speed of plane is
vp while wind blows directly from first city to second city with speed of va. How long does
it take for the airplane to make a round trip between the two cities? Also find average
speed and velocity.
Solution
a. Velocity of the airplane and air are vp and va respectively. When airplane travels
from first city to second, its relative velocity is vp + va. Now time take to cover distance
D is
t1 =
D
vp + va
Similarly when airplane heads to first city from second city, its relative velocity becomes
vp − va and time taken by it is
t2 =
D
vp − va
Total time in a round trip is T = t1 + t2 and
T =
D
vp + va
+
D
vp − va
= D
2vp
v2
p − v2
a
Arun Umrao
T
T
utes.
utes.

0.7. MOTION IN RIVER 61
b. Average speed of the plane is
vpavg =
2D
T
=
v2
p − v2
a
vp
As net displacement in a round trip of the airplane is zero, hence average velocity of the
airplane is “zero”.
Solved Problem 0.52 A roller is made by joining together two cones at their vertices O.
It is kept on two rails AB and CD which are placed asymmetrically (see figure), with its
axis perpendicular to CD and its center O at the center of line joining AB and CD (see
figure). It is given a light push so that it starts rolling with its center O moving parallel
to CD in the direction shown. In which direction the roller tend.
Solution
A
B
C
D
O
A
B
C
D
O
One rail is sloped rightward as shown in the figure. When the roller is rolled over the
rail, instantaneous circle of rolling over slanted rail decreases continuously and ultimately
potential energy. This is why the roller tends to turn left.
Solved Problem 0.53 Two trains of length 140m each, are running in parallel track with
speeds of 50km/h and 45km/h respectively. Train with speed 45km/h is ahead to other.
Initially, distance between the trains is 250m. Find the time in seconds after which guard
of trailing train will say hello to the guard of leading train.
Solution
0.7 Motion in River
Transportation in a river is take place by two method. One is along its width, where
goods and products are cross over to the river by means of boat and ferries. Second along
its length, where goods and products are transported from one place to other by means
of river.
Arun Umrao
g
g
D
D
D
D e
e B
B

62
0.7.1 First Type
Assume that a boat carry goods and products from one bank to other bank. Its speed is
constant and it is equal to vb.
vr
d
N
db
vb
dr
R
θ
α
O
P Q
Figure 24: Possible positions of the boat while it is sailed in a river.
There are two cases by which a boat can cross the river. One when it sails normal to
the river stream and reach opposite point Q, second it sails towards point P and crosses
river normally. These two cases are discusses bellow:
When Boat Sails Towards P and Crosses the River Normally In this case boat is sailed
at an angle α with the normal and boat crosses river in its shortest distance i.e. normally.
The velocity of river is vr and velocity of boat is vb, hence applying the trigonometric
relation for triangle OPN-
sin α =
vr
vb
vN
vb
vr
α
O
P N
(a)
d
db
dr
α
O
P N
(b)
Figure 25: Figure (a) shows the velocity of boat and river in vector form. Figure (b)
shows the distance covered by boat along the ﬂow of river current and its motion.
In ﬁgure (24) the complete displacement of boat and river is shown. Actually these
distance are shown for the time in which boat crosses the river. Now the component of
boat velocity along the normal direction of river stream is
vN = vb cos α
The time taken by boat to cross the river normally is
t =
d
vN
Arun Umrao
v
vb
b

0.7. MOTION IN RIVER 63
Solved Problem 0.54 A swimmer wants to cross river in shortest distance. If width and
velocity of river are 80m and 4m/s respectively then find the time in which (s)he can
cross the river. The stroke capacity of the swimmer is 5m/s.
Solution Swimmer wants to cross the river normally, then he should swim the river
making an angle with shortest distance against the flow of river. As given, swimmer’s
stroke is 5m/s, river velocity is 4m/s that gives the velocity vector of swimmer along the
shorted distance is 3m/s. Now the required time, taken by swimmer for swim the river
normally is
t =
80
3
≈ 26.66s
It is required result.
When Boat Sail Normal to the River Stream : In this case boat is sailed normal to
the river stream and boat crosses river in the direction of vector
OQ =
R. The velocity of
river is vr and velocity of boat is vb, hence applying the trigonometric relation for triangle
OQN
vb
vR
vr
θ
O
Q
N
(a)
d
db
dR
dr
θ
O
Q
N
(b)
Figure 26: Figure (a) shows the velocity of boat and river in vector form. Figure (b)
shows the distance covered by boat along the flow of river current and its motion.
tan θ =
vr
vb
The width of river is d but boat’s motion is along the
R and it covers total distance dR.
Now the distance dR is given by
d = dR cos θ
Or
dR =
d
cos θ
This the actual distance covered by the boat in crossing the river.
When Boat Crosses the River in Least Time : When boat has to cross the river in least
time, it must cross the width of river with maximum speed. Let the boat sails against
the river drift at an angle α.
r
Ar mrao
Um
w/arunumrao
sites.google
https://s
vR e.com/view/ar
w/ar
a
w/a

64
d
vN
vb
vr
α
O
P N
The time taken by it to cross the river is t, so
t =
d
vN
=
d
vb cos α
From this relation, t is minimum when denominator is maximum. As vb is constant, so
cos α must be maximum.
cos α = 1 ⇒ α = 0
It means, the boat should sail normal to the river drift.
Solved Problem 0.55 A boad has speed of 5 m/s in still water. It crosses a river of width
500 meter in shortest distance in 125 seconds. Find the velocity of the river.
Solution The boat crosses river normally in 125 seconds. It means, boat must sail
against the river drift as shown in the following ﬁgure.
d
vN
vb
vr
α
O
P N
The normal velocity of the boat is
vN = vb cos α =
d
t
It gives
5 cosα =
500
125
⇒ cos α =
4
5
Using the properties of triangle,
sin α =
3
5
=
vr
vb
So, the velocity of the river is vr = 3m/s.
Arun
rift as
ift as
Umrao
shown in
shown in

0.8. MOTION IN LIFT 65
0.7.2 Second Type
Assume A and B are two places and a boat start its journey from point A and reaches
to point B. The speed of boat is vbf along the direction of river stream and is vbr in the
opposite direction of river stream. Velocity of river is constant vr.
A B
d
vbf
vbr
vr
Figure 27: Boat sails in a river toward and opposite to the ﬂow of river current.
When boat makes a trip from A to B, its resultant speed is
v
= vbf + vr
When it makes a trip from B to A, then its resultant speed is
v
= vbr − vr
The time taken by boat to make a complete round trip is
t = tf + tr
t =
d
v
+
d
v
Here tf and tr are the time taken by boat during A to B and B to A trip respectively.
Now,
t =
d
vbf + vr
+
d
vbr − vr
(66)
If velocity of boat remains constant throuout the trip and it is vb then equation (66)
becomes
t =
d
vb + vr
+
d
vb − vr
t =
2vbd
v2
bf − v2
r
(67)
0.8 Motion in Lift
Lift is used to lift objects in vertical positions. Lift moves in vertical plane with constant
acceleration. A person using lift feel apparent increase or decrease of one’s weight. There
are three cases in which apparent weight can be measured.
Arun Umrao
f

66
0.8.1 Lift is In Rest
mg
R
Figure 28:
Suppose an object is placed in lift in rest position. The weight of object applying a force
mg on ﬂoor of lift. Lift also applying equal and opposite reaction force on the particle.
As the particle is in rest, hence net force is zero. Now
R − mg = 0
R = mg (68)
0.8.2 Lift Moving Upward Direction
mg
R
a
Figure 29:
Suppose an object is placed in lift that is moving in upward direction. The weight of
object is applying a force mg on the ﬂoor of lift. As lift is moving vertically upward
direction hence reaction force applied on the particle is greater than the weight of the
particle. particle is moving vertically upward with constant acceleration, hence net force
on the particle is in motion vertically upward direction. Now net force in the particle is
F = R − mg
Arun Umrao
oving Upward Direction
oving Upward Direction

Principles of Linear Motion of Objects - Physics - Explained by Arun Umrao

Principles of Linear Motion of Objects - Physics - Explained by Arun Umrao

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Principles of Linear Motion of Objects - Physics - Explained by Arun Umrao

Similar to Principles of Linear Motion of Objects - Physics - Explained by Arun Umrao (20)

More from ssuserd6b1fd

More from ssuserd6b1fd (20)

Recently uploaded

Recently uploaded (20)

Principles of Linear Motion of Objects - Physics - Explained by Arun Umrao