Chapter 2 Motion in a straight line

Chapter 3
Motion in a straight line

 Kinematics
Kinematics is the study of motion without going into its causes.
We shall treat the objects in motion as point objects. This approximation is valid so far as the size of
the object is much smaller than the distance it moves in a reasonable duration of time. In a good
number of situations in real-life, the size of objects can be neglected and they can be considered as
point-like objects without much error.
This chapter deals with motion along a straight line, i.e. rectilinear motion.
To describe motion we use two terms:
1. Reference point.
2. Position of object.

1. Reference point
In order to specify position, we need to use a reference point and a set of axes. It is convenient to
choose a rectangular coordinate system consisting of three mutually perpendicular axes, labelled X,
Y, and Z- axes. The point of intersection of these three axes is called origin (O) and serves as the
reference point. To measure time, we position a clock in this system. This coordinate system along
with a clock constitutes a frame of reference.
2. Position of object
How far object is situated from the point of reference is called position of object.
 Motion
Whenever there is a change in position of object with respect to reference point then the object is
said to be in motion.

 Distance or path length
Consider the motion of a car along a straight line. We choose the x-axis such that it coincides
with the path of the car’s motion and origin of the axis as the point from where the car started
moving, i.e. the car was at x = 0 at t = 0 (Fig. 3.1). Let P, Q and R represent the positions of the car
at different instants of time. Consider two cases of motion. In the first case, the car moves from O
to P. Then the distance moved by the car is OP = +360 m. This distance is called the path length
traversed by the car. In the second case, the car moves from O to P and then moves back from P
to Q. During this course of motion, the path length traversed is OP + PQ = + 360 m + (+120 m) =
+ 480 m. Path length is a scalar quantity — a quantity that has a magnitude only and no direction.

 Displacement
It is useful to define another quantity displacement as the change in position. Let x1 and x2 be the
positions of an object at time t1 and t2 . Then its displacement, denoted by ∆x, in time
∆t = (t 2 - t 1 ), is given by the difference between the final and initial positions
∆x = x2 − x1
If x2 > x1, ∆x is positive; and if x2 < x1 ∆x is negative.
Displacement has both magnitude and direction. Such quantities are represented by vectors. In one-
dimensional motion, there are only two directions (backward and forward, upward and downward)
in which an object can move, and these two directions can easily be specified by + and – signs. The
magnitude of displacement may or may not be equal to the path length traversed by an object.
For example: If an object starts from point O travels a distance l in +x direction and after that again
travels a distance l towards O. In this case the total distance travelled by object is l + l = 2l, but its
displacement is zero.

 Uniform and non uniform motion
If an object travels equal distance in equal interval of time then this type of motion is known as
uniform motion. Whereas if the object travels unequal distance in equal interval of time then this
type of motion in known as non uniform motion.

 Graphical methods to represent motion
There are two types of graphs which are used to represent the motion of any object:
1. Position – time graph (x – t graph): A graph plotted by taking distance or position of any object
at Y-axis and time on X-axis is known as position time graph.

Chapter 2 Motion in a straight line

 Velocity from distance time graph
If we say that an object is pursuing a uniform motion then its distance time graph shows a straight
line inclined at a particular angle to X-axis (time axis). The slope of such type of graph is used to
calculate the velocity of moving object.
Let the position time graph of an object pursuing uniform
motion is shown in diagram. Consider two points A and B on
the graph and draw perpendiculars to both time and distance
axis. Let s1 and s2 represents position of the object at time t1
and t2 then the slope of graph is given by:
2 1
2 1

 Graphical methods to represent motion
2. Velocity time graph: A graph plotted by taking velocity of any object at Y-axis and time on X-axis
is known as position time graph.

 Acceleration from velocity time graph
If we say that an object is pursuing motion whose velocity increases linearly with time then its
velocity time graph shows a straight line inclined at a particular angle to X-axis (time axis). The
slope of such type of graph is used to calculate the acceleration of moving object.
Let the velocity time graph of an object pursuing uniform
accelerated motion is shown in diagram. Consider two points
A and E on the graph and draw perpendiculars to both time
and velocity axis. Let u and v represents velocity of the object
at time t1 and t2 then the slope of graph is given by:
2 1

 Distance from velocity time graph
If we say that an object is pursuing motion whose velocity increases linearly with time then its
velocity time graph shows a straight line inclined at a particular angle to X-axis (time axis). The
distance travelled by object during a particular interval of time can also be calculated.
Let the velocity time graph of an object pursuing uniform
accelerated motion is shown in diagram. Consider two points
A and E on the graph and draw perpendiculars to both time
and velocity axis. Let u and v represents velocity of the object
at time t1 and t2 then the distance travelled during the time is
given by area under the curve.
This gives the distance travelled by object in time t2 – t1

 Speed
The distance travelled by an object in unit interval of time is known as speed. In other words we can
define speed as ratio of distance travelled by the object to time taken to cover that distance. Since
distance and time are scalar quantity therefore speed is also a scalar quantity.
S.I. unit of speed is m/s or m s-1 and its dimensional formula is [M0 L1 T-1]
In practical we always have a change in speed at every instant therefore we define a new quantity
which shows the average of all speeds, i.e. average speed.
Average speed is defined as the total path length travelled divided by the total time interval during
which the motion has taken place :

Average speed has obviously the same unit (m s–1) as that of speed. But it does not tell us in what
direction an object is moving. Thus, it is always positive.
 Velocity
The ratio of displacement of the object to total time taken to cover that path is known as velocity.
Hence velocity is rate of displacement of the object. Since displacement is a vector quantity and
time is a scaler quantity therefore velocity is a vector quantity. It is represented by ‘v’.
S.I. unit of velocity is m/s or m s-1 and its dimensional formula is [M0 L1 T-1]
As speed and velocity are almost same in nature, infect in one dimensional motion we consider
these speed and velocity to be same quantity only difference is presence of direction in velocity.

 Average velocity
Average velocity is defined as the change in position or displacement (∆x) divided by the time
intervals (∆t), in which the displacement occurs, i.e.,
2 1
2 1
where x2 and x1 are the positions of the object at time t2 and t1, respectively. Here the bar over the
symbol for velocity is a standard notation used to indicate an average quantity. In practical the
bigger unit for velocity is km/h or km h-1. If the motion of an object is along a straight line and in
the same direction, the magnitude of displacement is equal to the total path length. In that case, the
magnitude of average velocity is equal to the average speed. This is not always the case.

 INSTANTANEOUS VELOCITYAND SPEED
The velocity at an instant is defined as the limit of the average velocity as the time interval ∆t
becomes infinitesimally small. In other words:
∆t→
∆x
∆t
where the symbol
∆t→
stands for the operation of taking limit as ∆t→0 of the quantity on its right.
In the language of calculus, the quantity on the right hand side of Eq. is the differential coefficient
of x with respect to t and is denoted by . It is the rate of change of position with respect to time, at
that instant. Note that for uniform motion, velocity is the same as the average velocity at all instants.
Instantaneous speed or simply speed is the magnitude of velocity.

 Acceleration
Acceleration is defined as the rate of change of velocity of object. The average acceleration a over a
time interval is defined as the change of velocity divided by the time interval :
2 1
2 1
∆v
∆t
where v2 and v1 are the instantaneous velocities or simply velocities at time t2 and t1. It is the
average change of velocity per unit time. The SI unit of acceleration is m s–2 and its dimensional
formula is [M0 L1 T-2].
Instantaneous acceleration is defined in the same way as the instantaneous velocity :
∆t→
∆
∆t

 Kinematic equations for uniformly accelerated motion
 First equation of motion
Consider the velocity-time graph of an object that
moves under uniform acceleration. Let initial
velocity of the object is u (at point A) and then it
increases to v (at point B) in time t. The velocity
changes at a uniform rate ‘a’. The perpendicular
lines BC and BE are drawn from point B on the
time and the velocity axes respectively, so that the
initial velocity is represented by OA, the final
velocity is represented by BC and the time interval
t is represented by OC. BD = BC – CD, represents
the change in velocity in time interval t.

Let us draw AD parallel to OC. From the graph,
we observe that
BC = BD + DC = BD + OA
Substituting BC = v and OA = u,
we get, v = BD + u
or BD = v – u
From the velocity-time graph, the acceleration of
the object is given by:

 Second equation of motion

Let us draw AD parallel to OC. From the graph, we observe that
we get, v = BD + u
or BD = v – u
From the velocity-time graph, distance covered by object (x) in time t,
Distance covered = area under the curve
x = area of ∆ ADB + area of □ OADC
From first equation of motion we have, v = u + at,

This can also be written as,
0

 Third equation of motion

Let us draw AD parallel to OC. From the graph, we observe that
we get, v = BD + u
or BD = v – u
From the velocity-time graph, distance covered by object (x) in time t,
Distance covered = area under the curve
x = area of ⌂ OABC
x = sum of parallel sides × distance of parallel side

From 1st equation of motion we have,
Or
0

Hence the three equation of motion are written as,
0
0
0
The set of Equations weas obtained by assuming that at t = 0, the position of the particle, x is 0. We
can obtain a more general equation if we take the position coordinate at t = 0 as non-zero, say x0.
Then Equations are modified (replacing x by x – x0 ) to :
0
0 0
0 0

 RELATIVE VELOCITY
Consider yourself travelling in a train and being overtaken by another train moving in the same direction
as you are. While that train must be travelling faster than you to be able to pass you, it does seem slower
to you than it would be to someone standing on the ground and watching both the trains. In case both the
trains have the same velocity with respect to the ground, then to you the other train would seem to be
not moving at all. To understand this we introduce a concept of relative velocity:
“The relative velocity of an object A with respect to another object B is the velocity that object A would
appear to have to an observer situated on object B moving along with it.”
Consider two objects A and B moving uniformly with average velocities vA and vB in one dimension,
say along x-axis. If xA(0) and xB(0) are positions of objects A and B, respectively at time t = 0, their
positions xA(t) and xB(t) at time t are given by:
xA (t ) = xA (0) + vA t
xB (t) = xB (0) + vB t

Then, the displacement from object A to object B is given by:
xBA(t) = xB (t) – xA (t)
= [ xB (0) – xA (0) ] + (vB – vA) t.
It tells us that as seen from object A, object B has a velocity vB – vA because the displacement from A
to B changes steadily by the amount vB – vA in each unit of time. We say that the velocity of object B
relative to object A is vB – vA :
vBA = vB – vA
Similarly, velocity of object A relative to object B is:
vAB = vA – vB
This shows:
vBA = – vAB

If vB = vA, vB – vA = 0. Then, from Eq. (3.13), xB (t) – xA (t) = xB (0) – xA (0). Therefore, the two
objects stay at a constant distance (xB (0) – xA (0)) apart, and their position–time graphs are straight
lines parallel to each other as shown in Fig. 3.16. The relative velocity vAB or vBA is zero in this case

If vA > vB, vB – vA is negative. One graph is steeper than the other and they meet at a common point

Suppose vA and vB are of opposite signs. In this case, the magnitude of vBA or vAB ( = 30 m s–1) is
greater than the magnitude of velocity of A or that of B. If the objects under consideration are two
trains, then for a person sitting on either of the two, the other train seems to go very fast.

 Derivatives and differentiation
Differentiation allows us to find rates of change. For example, it allows us to find the rate of change of
velocity with respect to time (which is acceleration). It also allows us to find the rate of change of x
with respect to y, which on a graph of y against x is the gradient of the curve. Suppose we have a
quantity y whose value depends upon a single variable x, and is expressed by an equation defining y
as some specific function of x. This is represented as:
y = f (x)
Consider the point P on the curve y = f (x)
whose coordinates are (x, y) and another point
Q where coordinates are (x + ∆x, y + ∆y). The
slope of the line joining P and Q is given by:
∆x→
∆y
∆x
∆y
∆x

Since y = f (x) and y + ∆y = f (x + ∆x), we can write the definition of the derivative as:
∆x→
∆y
∆x
∆x
∆x
 Basic formulae for differentiation

Que. The position of any object is a function of time and it is given by
Find the function for velocity of object and the velocity of object at time t = 5s.
Sol. Given,
,
We know that velocity of any object is given by
Therefore the velocity of object at time t=5s is,
v = 75+ 20 – 1 = 94m /s

Que. The velocity of any object is a function of time and it is given by
Where j =1 m–1.Find the function for acceleration of object and the acceleration of object at time t=3s.
Sol. Given,
We know that acceleration of any object is given by
Therefore the velocity of object at time t = 5s is,
a = 30 + 1 = 31 m /s2

 Integration
Differential Calculus is centred on the concept of the derivative. The original motivation for the
derivative was the problem of defining tangent lines to the graphs of functions and calculating the
slope of such lines. Integral Calculus is motivated by the problem of defining and calculating the area
of the region bounded by the graph of the functions. The development of integral calculus arises out
of the efforts of solving the problems of the following types:
(a) the problem of finding a function whenever its derivative is given,
(b) The problem of finding the area bounded by the graph of a function under certain conditions.
These two problems lead to the two forms of the integrals, e.g., indefinite and definite integrals,
which together constitute the Integral Calculus.
Integration is the inverse process of differentiation. Instead of differentiating a function, we are given
the derivative of a function and asked to find its primitive, i.e., the original function. Such a process is
called integration or anti differentiation.

Note that for any real number C, treated as constant function, its derivative is zero and hence, we can
write:
Thus, anti derivatives (or integrals) of the above cited functions are not unique. Actually, there exist
infinitely many anti derivatives of each of these functions which can be obtained by choosing C
arbitrarily from the set of real numbers. For this reason C is customarily referred to as arbitrary
constant. In fact, C is the parameter by varying which one gets different anti derivatives (or integrals)
of the given function.
Remember this constant needs to be added only when we have integrated a function indefinitely. For
definite integration this constant is not added.

 Implementation of Integration in physics
We know that acceleration of any body at a very particular instant is defined as the derivative of its
velocity, therefore we can obtain the velocity function of any object in motion by integrating the
acceleration function with respect to time. i.e.,
Similarly in order to obtain the distance travelled by object using a velocity time graph we can have:
Remember this is just an generalised expression to calculate area under the curve accurately we use
definite integral.

Que. The acceleration of a car moving in straight line is given by:
find the velocity of car at t = 2 s. if initial velocity of car was 5 m /s.
Ans. Given that,
The velocity of car can be given by:
Using

To find the value of C we use initial conditions, i.e. at t = 0, v = 5 m /s.
C = 5
Hence the velocity function of car is given by:
Now at t = 2 s we have:
/s

Que. Derive first equation of motion by method of calculus.
Ans. By definition of acceleration we know that,
Integrating both the sides we have
Let us consider that acceleration is uniform throughout the motion then,
0
0
0 0

Que. Derive second equation of motion by method of calculus.
Ans. By definition of velocity we have:
Integrating both the sides we have,
0 0
0 0
x = x0 + v0 (t – 0 ) +at (t – 0 )
0 0
2
Using first equation of motion
v = v0 +at

Que. Derive third equation of motion by method of calculus.
Ans. By definition of acceleration we have:
Integrating both the sides, we have,

0 0
0 0
0 0
The advantage of this method is that it can be used for motion with non-uniform acceleration also.

 Summary and formulas that we have learned so far.
 An object is said to be in motion if its position changes with time. The position of the object can be
specified with reference to a conveniently chosen origin. For motion in a straight line, position to the
right of the origin is taken as positive and to the left as negative.
 Path length is defined as the total length of the path traversed by an object. Displacement is the change
in position : ∆x = x2 – x1. Path length is greater or equal to the magnitude of the displacement
between the same points.
 An object is said to be in uniform motion in a straight line if its displacement is equal in equal
intervals of time. Otherwise, the motion is said to be non-uniform.
 Average velocity is the displacement divided by the time interval in which the displacement occurs :
∆x
∆t
On an x-t graph, the average velocity over a time interval is the slope of the line connecting the initial
and final positions corresponding to that interval.

 Instantaneous velocity or simply velocity is defined as the limit of the average velocity as the time
interval ∆t becomes infinitesimally small :
∆t→
∆x
∆t
 Average acceleration is the change in velocity divided by the time interval during which the change
occurs :
∆v
∆t
 Instantaneous acceleration is defined as the limit of the average acceleration as the time interval ∆t
goes to zero:
∆t→
∆v
∆t

 For objects in uniformly accelerated rectilinear motion, the five quantities, displacement x, time taken
t, initial velocity v0, final velocity v and acceleration a are related by a set of simple equations called
kinematic equations of motion :
0
0 0
2
0 0
if the position of the object at time t = 0 is 0. If the particle starts at x = x0 , x in above equations is
replaced by (x – x0).
 Velocity of object seen from other object in motion is known as relative velocity.

Que. A jet airplane travelling at the speed of 500 km h–1 ejects its products of combustion at the speed of
1500 km h–1 relative to the jet plane. What is the speed of the latter with respect to an observer on
the ground ?

Que. Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of
72 km h–1 in the same direction, with A ahead of B. The driver of B decides to overtake A and
accelerates by 1 m s–2. If after 50 s, the guard of B just brushes past the driver of A, what was the
original distance between them ?

Que. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h–1.
Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1.
What is the
(a) magnitude of average velocity
(b) average speed of the man over the interval of time
(i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ?

Que. The position of an object moving along x-axis is given by x = a + bt2
where a = 8.5 m, b = 2.5 m s–2 and t is measured in seconds.
What is its velocity at t = 0 s and t = 2.0 s . What is the average velocity b/w t = 2.0 s and t = 4.0 s ?

Que. The position of an object moving along x-axis is given by x = a + bt2
where a = 8.5 m, b = 2.5 m s–2 and t is measured in seconds.
What is its velocity at t = 0 s and t = 2.0 s . What is the average velocity b/w t = 2.0 s and t = 4.0 s ?
Ans. In notation of differential calculus, the velocity is

Que. Read each statement below carefully and state with reasons and examples, if it is true or false ;
A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant.
(b) with zero speed may have non-zero velocity,
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration must be speeding up.

Que. A car start from rest and acquire a velocity of 54 km/h in 2 sec. Find
(i) the acceleration
(ii) distance travelled by car assume motion of car is uniform.

Que. A car start from rest and acquire a velocity of 54 km/h in 2 sec. Find
(i) the acceleration
(ii) distance travelled by car assume motion of car is uniform
Ans. Given that, u = 0 m /s
/
×
/s
t = 2 s
(i) By the definition of acceleration we have:
a = 7.5 m /s2

(ii) By third equation of motion we have,
x = 15 m
Hence distance travelled during this time = 15 m.

Chapter 2 Motion in a straight line

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