CHAPTER THREE
MOTION IN A STRAIGHT LINE
3.1 Introduction
3.2 Position, path length and
displacement
3.3 Average velocity and average
speed
3.4 Instantaneous velocity and
speed
3.5 Acceleration
3.6 Kinematic equations for
uniformly accelerated motion
3.7 Relative velocity
Summary
Points to ponder
Exercises
Additional exercises
Appendix 3.1
3.1 INTRODUCTION
Motion is common to everything in the universe. We walk,
run and ride a bicycle. Even when we are sleeping, air moves
into and out of our lungs and blood flows in arteries and
veins. We see leaves falling from trees and water flowing
down a dam. Automobiles and planes carry people from one
place to the other. The earth rotates once every twenty-four
hours and revolves round the sun once in a year. The sun
itself is in motion in the Milky Way, which is again moving
within its local group of galaxies.
Motion is change in position of an object with time. How
does the position change with time ? In this chapter, we shall
learn how to describe motion. For this, we develop the
concepts of velocity and acceleration. We shall confine
ourselves to the study of motion of objects along a straight
line, also known as rectilinear motion. For the case of
rectilinear motion with uniform acceleration, a set of simple
equations can be obtained. Finally, to understand the relative
nature of motion, we introduce the concept of relative velocity.
In our discussions, 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.
In Kinematics, we study ways to describe motion without
going into the causes of motion. What causes motion
described in this chapter and the next chapter forms the
subject matter of Chapter 5.
3.2 POSITION, PATH LENGTH AND DISPLACEMENT
Earlier you learnt that motion is change in position of an
object with time. In order to specify position, we need to use
a reference point and a set of axes. It is convenient to choose
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PHYSICS40
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 (see Chapter 4).
Displacement
It is useful to define another quantity
displacement as the change in position. Let
x
1
and x
2
be the positions of an object at time t
1
and t
2
. 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 = x
2
– x
1
(We use the Greek letter delta (∆) to denote a
change in a quantity.)
If x
2
> x
1
, ∆x is positive; and if x
2
< x
1
,
∆x is
negative.
Displacement has both magnitude and
direction. Such quantities are represented by
vectors. You will read about vectors in the next
chapter. Presently, we are dealing with motion
along a straight line (also called rectilinear
motion) only. 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. For example,
displacement of the car in moving from O to P is :
∆x = x
2
– x
1
= (+360 m) – 0 m = +360 m
The displacement has a magnitude of 360 m and
is directed in the positive x direction as indicated
by the + sign. Similarly, the displacement of the
car from P to Q is 240 m – 360 m = – 120 m. The
Fig. 3.1 x-axis, origin and positions of a car at different times.
a rectangular coordinate system consisting of
three mutually perpenducular 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. The coordinates (x, y. z) of an
object describe the position of the object with
respect to this coordinate system. To measure
time, we position a clock in this system. This
coordinate system along with a clock constitutes
a frame of reference.
If one or more coordinates of an object change
with time, we say that the object is in motion.
Otherwise, the object is said to be at rest with
respect to this frame of reference.
The choice of a set of axes in a frame of
reference depends upon the situation. For
example, for describing motion in one dimension,
we need only one axis. To describe motion in
two/three dimensions, we need a set of two/
three axes.
Description of an event depends on the frame
of reference chosen for the description. For
example, when you say that a car is moving on
a road, you are describing the car with respect
to a frame of reference attached to you or to the
ground. But with respect to a frame of reference
attached with a person sitting in the car, the
car is at rest.
To describe motion along a straight line, we
can choose an axis, say X-axis, so that it
coincides with the path of the object. We then
measure the position of the object with reference
to a conveniently chosen origin, say O, as shown
in Fig. 3.1. Positions to the right of O are taken
as positive and to the left of O, as negative.
Following this convention, the position
coordinates of point P and Q in Fig. 3.1 are +360
m and +240 m. Similarly, the position coordinate
of point R is –120 m.
Path length
Consider the motion of a car along a straight
line. We choose the x-axis such that it coincides
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MOTION IN A STRAIGHT LINE 41
negative sign indicates the direction of
displacement. Thus, it is not necessary to use
vector notation for discussing motion of objects
in one-dimension.
The magnitude of displacement may or may
not be equal to the path length traversed by
an object. For example, for motion of the car
from O to P, the path length is +360 m and the
displacement is +360 m. In this case, the
magnitude of displacement (360 m) is equal to
the path length (360 m). But consider the motion
of the car from O to P and back to Q. In this
case, the path length = (+360 m) + (+120 m) = +
480 m. However, the displacement = (+240 m) –
(0 m) = + 240 m. Thus, the magnitude of
displacement (240 m) is not equal to the path
length (480 m).
The magnitude of the displacement for a
course of motion may be zero but the
corresponding path length is not zero. For
example, if the car starts from O, goes to P and
then returns to O, the final position coincides
with the initial position and the displacement
is zero. However, the path length of this journey
is OP + PO = 360 m + 360 m = 720 m.
Motion of an object can be represented by a
position-time graph as you have already learnt
about it. Such a graph is a powerful tool to
represent and analyse different aspects of
motion of an object. For motion along a straight
line, say X-axis, only x-coordinate varies with
time and we have an x-t graph. Let us first
consider the simple case in which an object is
stationary, e.g. a car standing still at x = 40 m.
The position-time graph is a straight line parallel
to the time axis, as shown in Fig. 3.2(a).
If an object moving along the straight line
covers equal distances in equal intervals of
time, it is said to be in uniform motion along a
straight line. Fig. 3.2(b) shows the position-time
graph of such a motion.
Fig. 3.2 Position-time graph of (a) stationary object, and (b) an object in uniform motion.
Fig. 3.3 Position-time graph of a car.
t (s)
#
x
(m)
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PHYSICS42
Now, let us consider the motion of a car that
starts from rest at time t = 0 s from the origin O
and picks up speed till t = 10 s and thereafter
moves with uniform speed till t = 18 s. Then the
brakes are applied and the car stops at
t = 20 s and x = 296 m. The position-time graph
for this case is shown in Fig. 3.3. We shall refer
to this graph in our discussion in the following
sections.
3.3 AVERAGE VELOCITY AND AVERAGE
SPEED
When an object is in motion, its position
changes with time. But how fast is the position
changing with time and in what direction? To
describe this, we define the quantity 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 :
v
x x
t t
x
t
2 1
2 1
=
−
−
=
∆
∆
(3.1)
where x
2
and x
1
are the positions of the object
at time t
2
and t
1
, respectively. Here the bar over
the symbol for velocity is a standard notation
used to indicate an average quantity. The SI
unit for velocity is m/s or m s
–1
, although km h
–1
is used in many everyday applications.
Like displacement, average velocity is also a
vector quantity. But as explained earlier, for
motion in a straight line, the directional aspect
of the vector can be taken care of by + and –
signs and we do not have to use the vector
notation for velocity in this chapter.
Fig. 3.4 The average velocity is the slope of line P
1
P
2
.
Consider the motion of the car in Fig. 3.3. The
portion of the x-t graph between t = 0 s and t = 8
s is blown up and shown in Fig. 3.4. As seen
from the plot, the average velocity of the car
between time t = 5 s and t = 7 s is :
(
((
(
)
))
)
(
((
( )
))
)
1–
12
12
sm 8.7
s 57
m 010427
=
==
=
−
−−
−
=
==
=
−
−−
−
=
==
=
.
.
tt
xx
v
Geometrically, this is the slope of the straight
line P
1
P
2
connecting the initial position
1
P
to
the final position P
2
as
shown in Fig. 3.4.
The average velocity can be positive or negative
depending upon the sign of the displacement. It
is zero if the displacement is zero. Fig. 3.5 shows
the x-t graphs for an object, moving with positive
velocity (Fig. 3.5a), moving with negative velocity
(Fig. 3.5b) and at rest (Fig. 3.5c).
Average velocity as defined above involves
only the displacement of the object. We have seen
earlier that the magnitude of displacement may
be different from the actual path length. To
describe the rate of motion over the actual path,
we introduce another quantity called 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
Total path length
Total time
interval
=
(3.2)
Average speed has obviously the same unit
(m s
–1
) as that of velocity. But it does not tell us
in what direction an object is moving. Thus, it
is always positive (in contrast to the average
velocity which can be positive or negative). 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
Fig. 3.5 Position-time graph for an object (a) moving
with positive velocity, (b) moving with
negative velocity, and (c) at rest.
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MOTION IN A STRAIGHT LINE 43
t
is equal to the average speed. This is not always
the case, as you will see in the following example.
Example 3.1 A car is moving along a
straight line, say OP in Fig. 3.1. It moves
from O to P in 18 s and returns from P to Q
in 6.0 s. What are the average velocity
and average speed of the car in going (a)
from O to P ? and (b) from O to P and back
to Q ?
Answer (a)
Average velocity
Displacement
Time inter
val
=
1
+ 360 m
20 m s
18 s
v
−
= = +
Average speed
Path length
Time inter
val
=
1
360 m
= 20 m s
18 s
−
=
Thus, in this case the average speed is equal to
the magnitude of the average velocity.
(b) In this case,
( )
240 m
18 6.0 s
Displacement
Average velocity =
Time interval
+
=
+
-1
=+10 m s
OP + PQ
Path length
Average speed = =
Time interval t
∆
(
)
-1
360+120 m
= = 20 m s
24 s
Thus, in this case the average speed is not equal
to the magnitude of the average velocity. This
happens because the motion here involves
change in direction so that the path length is
greater than the magnitude of displacement.
This shows that speed is, in general, greater
than the magnitude of the velocity. t
If the car in Example 3.1 moves from O to P
and comes back to O in the same time interval,
average speed is 20 m/s but the average velocity
is zero !
3.4 INSTANTANEOUS VELOCITY AND SPEED
The average velocity tells us how fast an object
has been moving over a given time interval but
does not tell us how fast it moves at different
instants of time during that interval. For this,
we define instantaneous velocity or simply
velocity v at an instant t.
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,
v lim
x
t
=
t
0
∆
∆
∆
→
(3.3a)
=
d
d
x
t
(3.3b)
where the symbol
lim
t
0
∆
→
stands for the operation
of taking limit as ∆t
g
0 of the quantity on its
right. In the language of calculus, the quantity
on the right hand side of Eq. (3.3a) is the
differential coefficient of x with respect to t and
is denoted by
d
d
x
t
(see Appendix 3.1). It is the
rate of change of position with respect to time,
at that instant.
We can use Eq. (3.3a) for obtaining the value
of velocity at an instant either graphically or
numerically. Suppose that we want to obtain
graphically the value of velocity at time t = 4 s
(point P) for the motion of the car represented
in Fig. 3.3. The figure has been redrawn in
Fig.3.6 choosing different scales to facilitate the
Fig. 3.6 Determining velocity from position-time
graph. Velocity at t = 4 s is the slope of the
tangent to the graph at that instant.
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PHYSICS44
calculation. Let us take ∆t = 2 s centred at
t = 4 s. Then, by the definition of the average
velocity, the slope of line P
1
P
2
( Fig. 3.6) gives
the value of average velocity over the interval
3 s to 5 s. Now, we decrease the value of ∆t from
2 s to 1 s. Then line P
1
P
2
becomes Q
1
Q
2
and its
slope gives the value of the average velocity over
the interval 3.5 s to 4.5 s. In the limit ∆t → 0,
the line P
1
P
2
becomes tangent to the position-
time curve at the point P and the velocity at t =
4 s is given by the slope of the tangent at that
point. It is difficult to show this process
graphically. But if we use numerical method
to obtain the value of the velocity, the
meaning of the limiting process becomes
clear. For the graph shown in
Fig. 3.6, x = 0.08 t
3
. Table 3.1 gives the value of
∆x/∆t calculated for ∆t equal to 2.0 s, 1.0 s, 0.5
s, 0.1 s and 0.01 s centred at t = 4.0 s. The
second and third columns give the value of t
1
=
t
t
2
−
∆
and
t t
t
2
2
= +
∆
and the fourth and
the fifth columns give the corresponding values
of x, i.e. x (t
1
) = 0.08
t
1
3
and x (t
2
) = 0.08
t
2
3
. The
sixth column lists the difference ∆x = x (t
2
) – x
(t
1
) and the last column gives the ratio of ∆x and
∆t, i.e. the average velocity corresponding to the
value of ∆t listed in the first column.
We see from Table 3.1 that as we decrease
the value of ∆t from 2.0 s to 0.010 s, the value of
the average velocity approaches the limiting
value 3.84 m s
–1
which is the value of velocity at
t = 4.0 s, i.e. the value of
d
d
x
t
at t = 4.0 s. In this
manner, we can calculate velocity at each
instant for motion of the car shown in Fig. 3.3.
For this case, the variation of velocity with time
is found to be as shown in Fig. 3.7.
Fig. 3.7 Velocity–time graph corresponding to motion
shown in Fig. 3.3.
The graphical method for the determination
of the instantaneous velocity is always not a
convenient method. For this, we must carefully
plot the position–time graph and calculate the
value of average velocity as ∆t becomes smaller
and smaller. It is easier to calculate the value
of velocity at different instants if we have data
of positions at different instants or exact
expression for the position as a function of time.
Then, we calculate ∆x/∆t from the data for
decreasing the value of ∆t and find the limiting
value as we have done in Table 3.1 or use
differential calculus for the given expression and
calculate
d
d
x
t
at different instants as done in
the following example.
Table 3.1 Limiting value of
∆
∆
x
t
at t = 4 s
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MOTION IN A STRAIGHT LINE 45
t
Example 3.2 The position of an object
moving along x-axis is given by x = a + bt
2
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 between t = 2.0 s and t = 4.0 s ?
Answer In notation of differential calculus, the
velocity is
(
)
v
x
t
t
a bt 2b t =
2
= = + =
d
d
d
d
5.0 t m s
-1
At t = 0 s, v = 0 m s
–1
and at t = 2.0 s,
v = 10 m s
-1
.
(
)
(
)
4.0 2.0
4.0 2.0
x x
Average velocity
−
=
−
16 – – 4
6.0
2.0
a b a b
b
+
= = ×
-1
6.0 2.5 =15 m s
= ×
t
From Fig. 3.7, we note that during the period
t =10 s to 18 s the velocity is constant. Between
period t =18 s to t = 20 s, it is uniformly
decreasing and during the period t = 0 s to t
= 10 s, it is increasing. 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. For example, a velocity of
+ 24.0 m s
–1
and a velocity of – 24.0 m s
–1
— both
have an associated speed of 24.0 m s
-1
. It should
be noted that though average speed over a finite
interval of time is greater or equal to the
magnitude of the average velocity,
instantaneous speed at an instant is equal to
the magnitude of the instantaneous velocity at
that instant. Why so ?
3.5 ACCELERATION
The velocity of an object, in general, changes
during its course of motion. How to describe this
change? Should it be described as the rate of
change in velocity with distance or with time ?
This was a problem even in Galileo’s time. It was
first thought that this change could be described
by the rate of change of velocity with distance.
But, through his studies of motion of freely falling
objects and motion of objects on an inclined
plane, Galileo concluded that the rate of change
of velocity with time is a constant of motion for
all objects in free fall. On the other hand, the
change in velocity with distance is not constant
– it decreases with the increasing distance of fall.
This led to the concept of acceleration as the rate
of change of velocity with time.
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 v
v
a
t t t
∆
= =
∆
(3.4)
where v
2
and v
1
are the instantaneous velocities
or simply velocities at time t
2
and t
1
. It is the
average change of velocity per unit time. The SI
unit of acceleration is m s
–2
.
On a plot of velocity versus time, the average
acceleration is the slope of the straight line
connecting the points corresponding to (v
2
, t
2
)
and (v
1
, t
1
). The average acceleration
for velocity-time graph shown in Fig. 3.7 for
different time intervals 0 s - 10 s, 10 s – 18 s,
and 18 s – 20 s are :
0 s - 10 s
(
)
( )
–1
–2
24 – 0 m s
2.4 m s
10 – 0 s
a
= =
10 s - 18 s
(
)
( )
–1
–2
24 – 24 m s
0 m s
18 – 10 s
a
= =
18 s - 20 s
(
)
( )
–1
–2
0 – 24 m s
– 12 m s
20 – 18 s
a
= =
Fig. 3.8 Acceleration as a function of time for motion
represented in Fig. 3.3.
Instantaneous acceleration is defined in the same
way as the instantaneous velocity :
d
d
t 0
v v
a lim
t t
∆ →
∆
= =
∆
(3.5)
The acceleration at an instant is the slope of
the tangent to the v–t curve at that instant. For
the v–t curve shown in Fig. 3.7, we can obtain
acceleration at every instant of time. The
resulting a – t curve is shown in Fig. 3.8. We see
a (m s
–2
)
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PHYSICS46
that the acceleration is nonuniform over the
period 0 s to 10 s. It is zero between 10 s and
18 s and is constant with value –12 m s
–2
between 18 s and 20 s. When the acceleration
is uniform, obviously, it equals the average
acceleration over that period.
Since velocity is a quantity having both
magnitude and direction, a change in velocity
may involve either or both of these factors.
Acceleration, therefore, may result from a
change in speed (magnitude), a change in
direction or changes in both. Like velocity,
acceleration can also be positive, negative or
zero. Position-time graphs for motion with
positive, negative and zero acceleration are
shown in Figs. 3.9 (a), (b) and (c), respectively.
Note that the graph curves upward for positive
acceleration; downward for negative
acceleration and it is a straight line for zero
acceleration. As an exercise, identify in Fig. 3.3,
the regions of the curve that correspond to these
three cases.
Although acceleration can vary with time,
our study in this chapter will be restricted to
motion with constant acceleration. In this case,
the average acceleration equals the constant
value of acceleration during the interval. If the
velocity of an object is v
o
at t = 0 and v at time t,
we have
or
0
0
0
v v
a , v v a t
t
−
= = +
−
(3.6)
Fig. 3.9 Position-time graph for motion with
(a) positive acceleration; (b) negative
acceleration, and (c) zero acceleration.
Let us see how velocity-time graph looks like
for some simple cases. Fig. 3.10 shows velocity-
time graph for motion with constant acceleration
for the following cases :
(a) An object is moving in a positive direction
with a positive acceleration, for example
the motion of the car in Fig. 3.3 between
t = 0 s and t = 10 s.
(b) An object is moving in positive direction
with a negative acceleration, for example,
motion of the car in Fig 3.3 between
t = 18 s and 20 s.
(c) An object is moving in negative direction
with a negative acceleration, for example
the motion of a car moving from O in Fig.
3.1 in negative x-direction with
increasing speed.
(d) An object is moving in positive direction
till time t
1
,
and then turns back with the
same negative acceleration, for example
the motion of a car from point O to point
Q in Fig. 3.1 till time t
1
with decreasing
speed and turning back and moving with
the same negative acceleration.
An interesting feature of a velocity-time graph
for any moving object is that the area under the
curve represents the displacement over a
given time interval. A general proof of this
Fig. 3.10 Velocity–time graph for motions with
constant acceleration. (a) Motion in positive
direction with positive acceleration,
(b) Motion in positive direction with
negative acceleration, (c) Motion in negative
direction with negative acceleration,
(d) Motion of an object with negative
acceleration that changes direction at time
t
1
. Between times 0 to t
1
, its moves in
positive x - direction and between t
1
and
t
2
it moves in the opposite direction.
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MOTION IN A STRAIGHT LINE 47
statement requires use of calculus. We can,
however, see that it is true for the simple case of
an object moving with constant velocity u. Its
velocity-time graph is as shown in Fig. 3.11.
Fig. 3.11 Area under v–t curve equals displacement
of the object over a given time interval.
The v-t curve is
a straight line parallel to the
time axis and the area under it between t = 0
and t = T
is the area of the rectangle of height u
and base T. Therefore, area = u × T = uT which
is the displacement in this time interval. How
come in this case an area is equal to a distance?
Think! Note the dimensions of quantities on
the two coordinate axes, and you will arrive at
the answer.
Note that the x-t, v-t, and a-t graphs shown
in several figures in this chapter have sharp
kinks at some points implying that the
functions are not differentiable at these
points. In any realistic situation, the
functions will be differentiable at all points
and the graphs will be smooth.
What this means physically is that
acceleration and velocity cannot change
values abruptly at an instant. Changes are
always continuous.
3.6 KINEMATIC EQUATIONS FOR
UNIFORMLY ACCELERATED MOTION
For uniformly accelerated motion, we can derive
some simple equations that relate displacement
(x), time taken (t), initial velocity (v
0
), final
velocity (v) and acceleration (a). Equation (3.6)
already obtained gives a relation between final
and initial velocities v and v
0
of an object moving
with uniform acceleration
a :
v = v
0
+ at (3.6)
This relation is graphically represented in Fig. 3.12.
The area under this curve is :
Area between instants 0 and t = Area of triangle
ABC + Area of rectangle OACD
( )
–
0 0
1
v v t + v t
2
=
Fig. 3.12 Area under v-t curve for an object with
uniform acceleration.
As explained in the previous section, the area
under v-t curve represents the displacement.
Therefore, the displacement x of the object is :
( )
1
–
2
0 0
x v v t + v t
=
(3.7)
But
v
v
a t
0
−
=
Therefore,
2
0
1
2
x a t + v t
=
or,
2
0
1
2
x v t at
= +
(3.8)
Equation (3.7) can also be written as
0
2
v + v
x t v t
= =
(3.9a)
where,
0
2
v v
v
+
=
(constant acceleration only)
(3.9b)
Equations (3.9a) and (3.9b) mean that the object
has undergone displacement x with an average
velocity equal to the arithmetic average of the
initial and final velocities.
From Eq. (3.6), t = (v – v
0
)/a. Substituting this in
Eq. (3.9a), we get
x v t
v v v v
a
v v
a
= =
+
−
=
−
0 0
2
0
2
2 2
2 2
0
2
v v ax
= +
(3.10)
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PHYSICS48
t
This equation can also be obtained by
substituting the value of t from Eq. (3.6) into
Eq. (3.8). Thus, we have obtained three
important equations :
0
v v at
= +
2
0
1
2
x v t at
= +
2 2
0
2
v v ax
= +
(3.11a)
connecting five quantities v
0
, v, a, t and x. These
are kinematic equations of rectilinear motion
for constant acceleration.
The set of Eq. (3.11a) were 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 x
0
. Then Eqs. (3.11a) are
modified (replacing x by x – x
0
) to :
0
v v at
= +
2
0 0
1
2
x x v t at
= + +
(3.11b)
2 2
0 0
2 ( )
v v a x x
= + −
(3.11c)
Example 3.3 Obtain equations of motion
for constant acceleration using method of
calculus.
Answer By definition
d
d
v
a
t
=
dv = a dt
Integrating both sides
d dv a t
v
v t
0
0
∫ ∫
=
=
∫
a t
t
d
0
(a is constant)
0
–
v v at
=
0
v v at
= +
Further,
d
d
x
v
t
=
dx = v dt
Integrating both sides
dx
x
x
0
∫
=
∫
v t
t
d
0
= +
( )
∫
v at t
t
0
0
d
2
0 0
1
–
2
x x v t a t
= +
x =
2
0 0
1
2
x v t a t
+ +
We can write
d d d d
d d d d
v v x v
a v
t x t x
= = =
or, v dv = a dx
Integrating both sides,
v v a x
v
v
x
x
d d
0 0
∫ ∫
=
( )
2 2
0
0
–
–
2
v v
a x x
=
(
)
2 2
0 0
2 –
v v a x x
= +
The advantage of this method is that it can be
used for motion with non-uniform acceleration
also.
Now, we shall use these equations to some
important cases. t
Example 3.4 A ball is thrown vertically
upwards with a velocity of 20 m s
–1
from
the top of a multistorey building. The
height of the point from where the ball is
thrown is 25.0 m from the ground. (a) How
high will the ball rise ? and (b) how long
will it be before the ball hits the ground?
Take g = 10 m s
–2
.
Answer (a) Let us take the y-axis in the
vertically upward direction with zero at the
ground, as shown in Fig. 3.13.
Now v
o
= + 20 m s
–1
,
a = – g = –10 m s
–2
,
v = 0 m s
–1
If the ball rises to height y from the point of
launch, then using the equation
(
)
0
2 2
0
v v 2 a y – y
= +
we get
0 = (20)
2
+ 2(–10)(y – y
0
)
Solving, we get, (y – y
0
) = 20 m.
(b) We can solve this part of the problem in two
ways. Note carefully the methods used.
t
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MOTION IN A STRAIGHT LINE 49
t
Fig. 3.13
FIRST METHOD : In the first method, we split
the path in two parts : the upward motion (A to
B) and the downward motion (B to C) and
calculate the corresponding time taken t
1
and
t
2
. Since the velocity at B is zero, we have :
v = v
o
+ at
0 = 20 – 10t
1
Or, t
1
= 2 s
This is the time in going from A to B. From B, or
the point of the maximum height, the ball falls
freely under the acceleration due to gravity. The
ball is moving in negative y direction. We use
equation
2
0 0
1
2
y y v t at
= + +
We have, y
0
= 45 m, y = 0, v
0
= 0, a = – g = –10 m s
–2
0 = 45 + (½) (–10) t
2
2
Solving, we get t
2
= 3 s
Therefore, the total time taken by the ball before
it hits the ground = t
1
+
t
2
=
2
s
+ 3 s = 5 s.
SECOND METHOD : The total time taken can
also be calculated by noting the coordinates of
initial and final positions of the ball with respect
to the origin chosen and using equation
2
0 0
1
2
y y v t at
= + +
Now y
0
=
25 m y = 0 m
v
o
= 20 m s
-1
, a = –10m s
–2
, t = ?
0 = 25 +20 t + (½) (-10) t
2
Or, 5t
2
– 20t – 25 = 0
Solving this quadratic equation for t, we get
t = 5s
Note that the second method is better since we
do not have to worry about the path of the motion
as the motion is under constant acceleration.
t
Example 3.5 Free-fall : Discuss the
motion of an object under free fall. Neglect
air resistance.
Answer An object released near the surface of
the Earth is accelerated downward under the
influence of the force of gravity. The magnitude
of acceleration due to gravity is represented by
g. If air resistance is neglected, the object is
said to be in free fall. If the height through
which the object falls is small compared to the
earth’s radius, g can be taken to be constant,
equal to 9.8 m s
–2
. Free fall is thus a case of
motion with uniform acceleration.
We assume that the motion is in y-direction,
more correctly in –y-direction because we
choose upward direction as positive. Since the
acceleration due to gravity is always downward,
it is in the negative direction and we have
a = – g = – 9.8 m s
–2
The object is released from rest at y = 0. Therefore,
v
0
= 0 and the equations of motion become:
v = 0 – g t = –9.8 t m s
–1
y = 0 – ½ g t
2
= –4.9 t
2
m
v
2
= 0 – 2 g y = –19.6 y m
2
s
–2
These equations give the velocity and the
distance travelled as a function of time and also
the variation of velocity with distance. The
variation of acceleration, velocity, and distance,
with time have been plotted in Fig. 3.14(a), (b)
and (c).
(a)
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PHYSICS50
t
(b)
(c)
Fig. 3.14 Motion of an object under free fall.
(a) Variation of acceleration with time.
(b) Variation of velocity with time.
(c) Variation of distance with time t
Example 3.6 Galileo’s law of odd
numbers : “The distances traversed, during
equal intervals of time, by a body falling
from rest, stand to one another in the same
ratio as the odd numbers beginning with
unity [namely, 1: 3: 5: 7…...].” Prove it.
Answer Let us divide the time interval of
motion of an object under free fall into many
equal intervals
τ
and find out the distances
traversed during successive intervals of
time. Since initial velocity is zero, we have
Using this equation, we can calculate the
position of the object after different time
intervals, 0, τ, 2τ, 3τ… which are given in
second column of Table 3.2. If we take
(–1/
2) gτ
2
as y
0
— the position coordinate after
first time interval τ, then third column gives
the positions in the unit of y
o
. The fourth
column gives the distances traversed in
successive τs. We find that the distances are
in the simple ratio 1: 3: 5: 7: 9: 11… as shown
in the last column. This law was established
by Galileo Galilei (1564-1642) who was the first
to make quantitative studies of free fall. t
Example 3.7 Stopping distance of
vehicles : When brakes are applied to a
moving vehicle, the distance it travels before
stopping is called stopping distance. It is
an important factor for road safety and
depends on the initial velocity (v
0
) and the
braking capacity, or deceleration, –a that
is caused by the braking. Derive an
expression for stopping distance of a vehicle
in terms of v
o
and
a.
Answer Let the distance travelled by the vehicle
before it stops be d
s
. Then, using equation of
motion v
2
= v
o
2
+ 2 ax, and noting that v = 0, we
have the stopping distance
d
v
a
s
=
–
0
2
2
Thus, the stopping distance is proportional to
the square of the initial velocity. Doubling the
t
Table 3.2
y gt= −
1
2
2
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MOTION IN A STRAIGHT LINE 51
t
initial velocity increases the stopping distance
by a factor of 4 (for the same deceleration).
For the car of a particular make, the braking
distance was found to be 10 m, 20 m, 34 m and
50 m corresponding to velocities of 11, 15, 20
and 25 m/s which are nearly consistent with
the above formula.
Stopping distance is an important factor
considered in setting speed limits, for example,
in school zones. t
Example 3.8 Reaction time : When a
situation demands our immediate
action, it takes some time before we
really respond. Reaction time is the
time a person takes to observe, think
and act. For example, if a person is
driving and suddenly a boy appears on
the road, then the time elapsed before
he slams the brakes of the car is the
reaction time. Reaction time depends
on complexity of the situation and on
an individual.
You can measure your reaction time
by a simple experiment. Take a ruler
and ask your friend to drop it vertically
through the gap between your thumb
and forefinger (Fig. 3.15). After you
catch it, find the distance d travelled
by the ruler. In a particular case, d was
found to be 21.0 cm. Estimate reaction
time.
Or,
Given d = 21.0 cm and g = 9.8 m s
–2
the reaction
time is
t
3.7 RELATIVE VELOCITY
You must be familiar with the experience of
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 such observations, we now
introduce the concept of relative velocity.
Consider two objects A and B moving
uniformly with average velocities v
A
and v
B
in
one dimension, say along x-axis. (Unless
otherwise specified, the velocities mentioned in
this chapter are measured with reference to the
ground). If x
A
(0) and x
B
(0) are positions of objects
A and B, respectively at time t = 0, their positions
x
A
(t) and x
B
(t) at time t are given by:
x
A
(t ) = x
A
(0) + v
A
t (3.12a)
x
B
(t) = x
B
(0) + v
B
t (3.12b)
Then, the displacement from object A to object
B is given by
x
BA
(t) = x
B
(t) – x
A
(t)
= [ x
B
(0) – x
A
(0) ] + (v
B
– v
A
) t. (3.13)
Equation (3.13) is easily interpreted. It tells us
that as seen from object A, object B has a
velocity v
B
– v
A
because the displacement from
A to B changes steadily by the amount v
B
– v
A
in
each unit of time. We say that the velocity of
object B relative to object A is v
B
– v
A
:
v
BA
= v
B
– v
A
(3.14a)
Similarly, velocity of object A relative to object B
is:
v
AB
= v
A
– v
B
(3.14b)
Fig. 3.15 Measuring the reaction time.
Answer The ruler drops under free fall.
Therefore, v
o
= 0, and a = –g = –9.8 m s
–2
. The
distance travelled d and the reaction time t
r
are
related by
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PHYSICS52
This shows: v
BA
= – v
AB
(3.14c)
Now we consider some special cases :
(a) If v
B
= v
A
, v
B
–
v
A
= 0. Then, from Eq. (3.13), x
B
(t) – x
A
(t) = x
B
(0) – x
A
(0). Therefore, the two
objects stay at a constant distance (x
B
(0) – x
A
(0)) apart, and their position–time graphs are
straight lines parallel to each other as shown
in Fig. 3.16. The relative velocity v
AB
or v
BA
is
zero in this case.
(b) If v
A
> v
B
, v
B
– v
A
is negative. One graph is
steeper than the other and they meet at a
common point. For example, suppose v
A
= 20 m s
-1
and x
A
(0) = 10 m; and v
B
= 10 m s
-1
, x
B
(0) = 40
m; then the time at which they meet is t = 3 s
(Fig. 3.17). At this instant they are both at a
position x
A
(t) = x
B
(t) = 70 m. Thus, object A
overtakes object B at this time. In this case,v
BA
= 10 m s
–1
– 20 m s
–1
= – 10 m s
–1
=– v
AB
.
(c) Suppose v
A
and v
B
are of opposite signs. For
example, if in the above example object A is
moving with 20 m s
–1
starting at x
A
(0) = 10 m
and object B is moving with – 10 m s
–1
starting
at x
B
(0) = 40 m, the two objects meet at t = 1 s
(Fig. 3.18). The velocity of B relative to A,
v
BA
= [–10 – (20)] m s
–1
= –30 m s
–1
= – v
AB
. In this
case, the magnitude of v
BA
or v
AB
( = 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.
Note that Eq. (3.14) are valid even if v
A
and v
B
represent instantaneous velocities.
Example 3.9 Two parallel rail tracks run
north-south. Train A moves north with a
speed of 54 km h
–1
, and train B moves south
with a speed of 90 km h
–1
. What is the
(a) velocity of B with respect to A ?,
(b) velocity of ground with respect to B ?,
and
(c) velocity of a monkey running on the
roof of the train A against its motion
(with a velocity of 18 km h
–1
with
respect to the train A) as observed by
a man standing on the ground ?
Answer Choose the positive direction of x-axis
to be from south to north. Then,
Fig. 3.16 Position-time graphs of two objects with
equal velocities.
Fig. 3.17 Position-time graphs of two objects with
unequal velocities, showing the time of
meeting.
Fig. 3.18 Position-time graphs of two objects with
velocities in opposite directions, showing
the time of meeting.
t(s)
t
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MOTION IN A STRAIGHT LINE 53
v
A
= + 54 km h
–1
= 15 m s
–1
v
B
= – 90 km h
–1
= – 25 m s
–1
Relative velocity of B with respect to A = v
B
– v
A
=
– 40 m s
–1
, i.e. the train B appears to A to move
with a speed of 40 m s
–1
from north to south.
Relative velocity of ground with respect to
B = 0 – v
B
= 25 m s
–1
.
In (c), let the velocity of the monkey with respect
to ground be v
M
.
Relative velocity of the monkey
with respect to A,
v
MA
= v
M
– v
A
= –18 km h
–1
=–5 ms
–1
. Therefore,
v
M
= (15 – 5) m s
–1
= 10 m s
–1
.
t
SUMMARY
1. 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.
2. Path length is defined as the total length of the path traversed by an object.
3. Displacement is the change in position : ∆x = x
2
– x
1
.
Path length is greater or equal to
the magnitude of the displacement between the same points.
4. 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.
5. Average velocity is the displacement divided by the time interval in which the
displacement occurs :
v
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.
6. Average Speed is the ratio of total path length traversed and the corresponding time
interval.
The average speed of an object is greater or equal to the magnitude of the average
velocity over a given time interval.
7. Instantaneous velocity or simply velocity is defined as the limit of the average velocity as
the time interval ∆t becomes infinitesimally small :
d
d
t 0 t 0
x x
v lim v lim
t t
∆ → ∆ →
∆
= = =
∆
The velocity at a particular instant is equal to the slope of the tangent drawn on
position-time graph at that instant.
8. Average acceleration is the change in velocity divided by the time interval during which
the change occurs :
v
a
t
∆
=
∆
9. Instantaneous acceleration is defined as the limit of the average acceleration as the time
interval ∆t goes to zero :
d
d
t 0 t 0
v v
a lim a lim
t t
∆ → ∆ →
∆
= = =
∆
The acceleration of an object at a particular time is the slope of the velocity-time
graph at that instant of time. For uniform motion, acceleration is zero and the x-t
graph is a straight line inclined to the time axis and the v-t graph is a straight line
2020-21
PHYSICS54
parallel to the time axis. For motion with uniform acceleration, x-t graph is a parabola
while the v-t graph is a straight line inclined to the time axis.
10. The area under the velocity-time curve between times t
1
and t
2
is equal to the displacement
of the object during that interval of time.
11. For objects in uniformly accelerated rectilinear motion, the five quantities, displacement
x, time taken t, initial velocity v
0
, final velocity v and acceleration a are related by a set
of simple equations called kinematic equations of motion :
v = v
0
+ at
x v t
1
2
at
0
2
= +
v v 2ax
2
0
2
= +
if the position of the object at time t = 0 is 0. If the particle starts at x = x
0
, x in above
equations is replaced by (x – x
0
).
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MOTION IN A STRAIGHT LINE 55
POINTS TO PONDER
1. The path length traversed by an object between two points is, in general, not the same
as the magnitude of displacement. The displacement depends only on the end points;
the path length (as the name implies) depends on the actual path. In one dimension,
the two quantities are equal only if the object does not change its direction during the
course of motion. In all other cases, the path length is greater than the magnitude of
displacement.
2. In view of point 1 above, the average speed of an object is greater than or equal to the
magnitude of the average velocity over a given time interval. The two are equal only if
the path length is equal to the magnitude of displacement.
3. The origin and the positive direction of an axis are a matter of choice. You should first
specify this choice before you assign signs to quantities like displacement, velocity
and acceleration.
4. If a particle is speeding up, acceleration is in the direction of velocity; if its speed is
decreasing, acceleration is in the direction opposite to that of the velocity. This
statement is independent of the choice of the origin and the axis.
5. The sign of acceleration does not tell us whether the particle’s speed is increasing or
decreasing. The sign of acceleration (as mentioned in point 3) depends on the choice
of the positive direction of the axis. For example, if the vertically upward direction is
chosen to be the positive direction of the axis, the acceleration due to gravity is
negative. If a particle is falling under gravity, this acceleration, though negative,
results in increase in speed. For a particle thrown upward, the same negative
acceleration (of gravity) results in decrease in speed.
6. The zero velocity of a particle at any instant does not necessarily imply zero acceleration
at that instant. A particle may be momentarily at rest and yet have non-zero
acceleration. For example, a particle thrown up has zero velocity at its uppermost
point but the acceleration at that instant continues to be the acceleration due to
gravity.
7. In the kinematic equations of motion [Eq. (3.11)], the various quantities are algebraic,
i.e. they may be positive or negative. The equations are applicable in all situations
(for one dimensional motion with constant acceleration) provided the values of different
quantities are substituted in the equations with proper signs.
8. The definitions of instantaneous velocity and acceleration (Eqs. (3.3) and (3.5)) are
exact and are always correct while the kinematic equations (Eq. (3.11)) are true only
for motion in which the magnitude and the direction of acceleration are constant
during the course of motion.
EXERCISES
3.1 In which of the following examples of motion, can the body be considered
approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
3.2 The position-time (x-t) graphs for two children A and B returning from their school
O to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct
entries in the brackets below ;
(a) (A/B) lives closer to the school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) (A/B) walks faster than (B/A)
(d) A and B reach home at the (same/different) time
(e) (A/B) overtakes (B/A) on the road (once/twice).
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PHYSICS56
3.3 A woman starts from her home at 9.00 am, walks with a speed of 5 km h
–1
on a
straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and
returns home by an auto with a speed of 25 km h
–1
. Choose suitable scales and
plot the x-t graph of her motion.
3.4 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward,
followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1
m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically
and otherwise how long the drunkard takes to fall in a pit 13 m away from the
start.
3.5 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 ?
3.6 A car moving along a straight highway with speed of 126 km h
–1
is brought to a
stop within a distance of 200 m. What is the retardation of the car (assumed
uniform), and how long does it take for the car to stop ?
3.7 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 ?
3.8 On a two-lane road, car A is travelling with a speed of 36 km h
–1
. Two cars B and
C approach car A in opposite directions with a speed of 54 km h
–1
each. At a
certain instant, when the distance AB is equal to AC, both being 1 km, B decides
to overtake A before C does. What minimum acceleration of car B is required to
avoid an accident ?
3.9 Two towns A and B are connected by a regular bus service with a bus leaving in
either direction every T minutes. A man cycling with a speed of 20 km h
–1
in the
direction A to B notices that a bus goes past him every 18 min in the direction of
his motion, and every 6 min in the opposite direction. What is the period T of the
bus service and with what speed (assumed constant) do the buses ply on the
road?
3.10 A player throws a ball upwards with an initial speed of 29.4 m s
–1
.
(a) What is the direction of acceleration during the upward motion of the ball ?
(b) What are the velocity and acceleration of the ball at the highest point of its
motion ?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its
highest point, vertically downward direction to be the positive direction of
x-axis, and give the signs of position, velocity and acceleration of the ball
during its upward, and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the
player’s hands ? (Take g = 9.8 m s
–2
and neglect air resistance).
Fig. 3.19
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MOTION IN A STRAIGHT LINE 57
3.11 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.
3.12 A ball is dropped from a height of 90 m on a floor. At each collision with the floor,
the ball loses one tenth of its speed. Plot the speed-time graph of its motion
between t = 0 to 12 s.
3.13 Explain clearly, with examples, the distinction between :
(a) magnitude of displacement (sometimes called distance) over an interval of time,
and the total length of path covered by a particle over the same interval;
(b) magnitude of average velocity over an interval of time, and the average speed
over the same interval. [Average speed of a particle over an interval of time is
defined as the total path length divided by the time interval]. Show in both (a)
and (b) that the second quantity is either greater than or equal to the first.
When is the equality sign true ? [For simplicity, consider one-dimensional
motion only].
3.14 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, and
(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 ? [Note: You will appreciate from this exercise why it
is better to define average speed as total path length divided by time, and not
as magnitude of average velocity. You would not like to tell the tired man on
his return home that his average speed was zero !]
3.15 In Exercises 3.13 and 3.14, we have carefully distinguished between average speed
and magnitude of average velocity. No such distinction is necessary when we
consider instantaneous speed and magnitude of velocity. The instantaneous speed
is always equal to the magnitude of instantaneous velocity. Why ?
3.16 Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of
these cannot possibly represent one-dimensional motion of a particle.
Fig. 3.20
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PHYSICS58
Fig. 3.24
Fig. 3.21
3.17 Figure 3.21 shows the x-t plot of one-dimensional
motion of a particle. Is it correct to say from the
graph that the particle moves in a straight line for
t < 0 and on a parabolic path for t >0 ? If not, suggest
a suitable physical context for this graph.
3.18 A police van moving on a highway with a speed of
30 km h
–1
fires a bullet at a thief’s car speeding away
in the same direction with a speed of 192 km h
–1
. If
the muzzle speed of the bullet is 150 m s
–1
, with
what speed does the bullet hit the thief’s car ? (Note:
Obtain that speed which is relevant for damaging
the thief’s car).
3.19 Suggest a suitable physical situation for each of the
following graphs (Fig 3.22):
Fig. 3.22
3.20 Figure 3.23 gives the x-t plot of a particle executing one-dimensional simple
harmonic motion. (You will learn about this motion in more detail in Chapter14).
Give the signs of position, velocity and acceleration variables of the particle at
t = 0.3 s, 1.2 s, – 1.2 s.
Fig. 3.23
3.21 Figure 3.24 gives the x-t plot of a
particle in one-dimensional motion.
Three different equal intervals of time
are shown. In which interval is the
average speed greatest, and in which
is it the least ? Give the sign of average
velocity for each interval.
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MOTION IN A STRAIGHT LINE 59
3.22 Figure 3.25 gives a speed-time graph of
a particle in motion along a constant
direction. Three equal intervals of time
are shown. In which interval is the
average acceleration greatest in
magnitude ? In which interval is the
average speed greatest ? Choosing the
positive direction as the constant
direction of motion, give the signs of v
and a in the three intervals. What are
the accelerations at the points A, B, C
and D ?
Additional Exercises
3.23 A three-wheeler starts from rest, accelerates uniformly with 1 m s
–2
on a straight
road for 10 s, and then moves with uniform velocity. Plot the distance covered by
the vehicle during the n
th
second (n = 1,2,3….) versus n. What do you expect this
plot to be during accelerated motion : a straight line or a parabola ?
3.24 A boy standing on a stationary lift (open from above) throws a ball upwards with
the maximum initial speed he can, equal to 49 m s
–1
. How much time does the ball
take to return to his hands? If the lift starts moving up with a uniform speed of
5 m s
-1
and the boy again throws the ball up with the maximum speed he can, how
long does the ball take to return to his hands ?
3.25 On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed
9 km h
–1
(with respect to the belt) between his father and mother located 50 m apart
on the moving belt. The belt moves with a speed of 4 km h
–1
. For an observer on a
stationary platform outside, what is the
(a) speed of the child running in the direction of motion of the belt ?.
(b) speed of the child running opposite to the direction of motion of the belt ?
(c) time taken by the child in (a) and (b) ?
Which of the answers alter if motion is viewed by one of the parents ?
Fig. 3.26
3.26 Two stones are thrown up simultaneously from the edge of a cliff 200 m high with
initial speeds of 15 m s
–1
and 30 m s
–1
. Verify that the graph shown in Fig. 3.27
correctly represents the time variation of the relative position of the second stone
with respect to the first. Neglect air resistance and assume that the stones do not
rebound after hitting the ground. Take g = 10 m s
–2
. Give the equations for the
linear and curved parts of the plot.
Fig. 3.25
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PHYSICS60
Fig. 3.27
3.27 The speed-time graph of a particle moving along a fixed direction is shown in
Fig. 3.28. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s,
(b) t = 2 s to 6 s.
Fig. 3.28
What is the average speed of the particle over the intervals in (a) and (b) ?
3.28 The velocity-time graph of a particle in one-dimensional motion is shown in
Fig. 3.29 :
Fig. 3.29
Which of the following formulae are correct for describing the motion of the particle
over the time-interval t
1
to t
2
:
(a) x(t
2
) = x(t
1
) + v (t
1
) (t
2
– t
1
) +(½) a (t
2
– t
1
)
2
(b) v(t
2
) = v(t
1
) + a (t
2
– t
1
)
(c) v
average
= (x(t
2
) – x(t
1
))/(t
2
– t
1
)
(d) a
average
= (v(t
2
) – v(t
1
))/(t
2
– t
1
)
(e) x(t
2
) = x(t
1
) + v
average
(t
2
– t
1
) + (½) a
average
(t
2
– t
1
)
2
(f) x(t
2
) – x(t
1
) = area under the v-t curve bounded by the t-axis and the dotted line
shown.
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MOTION IN A STRAIGHT LINE 61
APPENDIX 3.1 : ELEMENTS OF CALCULUS
Differential Calculus
Using the concept of ‘differential coefficient’ or ‘derivative’, we can easily define velocity and
acceleration. Though you will learn in detail in mathematics about derivatives, we shall introduce
this concept in brief in this Appendix so as to facilitate its use in describing physical quantities
involved in motion.
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) (1)
This relationship can be visualised by drawing a graph of function y = f (x) regarding y and x as
Cartesian coordinates, as shown in Fig. 3.30 (a).
(a) (b)
Fig. 3.30
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
yyy
x
y
tan
∆
−∆+
=
∆
∆
=
θ
(2)
Suppose now that the point Q moves along the curve towards P. In this process, ∆y and ∆x
decrease and approach zero; though their ratio
∆
∆
y
x
will not necessarily vanish. What happens
to the line PQ as ∆y→ 0, ∆x→ 0. You can see that this line becomes a tangent to the curve at
point P as shown in Fig. 3.30(b). This means that tan
θ
approaches the slope of the tangent at
P, denoted by m:
x 0 0
( )
lim lim
x
y y y y
m
x x
∆ → ∆ →
∆ + ∆ −
= =
∆ ∆
(3)
The limit of the ratio ∆y/∆x as ∆x approaches zero is called the derivative of y with respect to x
and is written as dy/dx. It represents the slope of the tangent line to the curve y = f (x) at the
point (x, y).
Since y = f (x) and y + ∆y = f (x + ∆x), we can write the definition of the derivative as:
∆
∆+
=
∆
∆
==
→∆→∆
x
xfxxf
x
y
)(
x
)(–)(
limlim
dx
d
dx
dy
00x
xf
Given below are some elementary formulae for derivatives of functions. In these u (x) and v (x)
represent arbitrary functions of x, and a and b denote constant quantities that are independent
of x. Derivatives of some common functions are also listed .
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PHYSICS62
x
u
a
x
u
d
d
d
) (a d
=
;
d d d
= .
d d d
u u x
t x t
x
u
v
x
v
u
x
uv)
d
d
d
d
d
d(
+=
;
(
)
2
d /
1 d d
–
d d d
u v
u v
u
x x x
v
=
xv
xu
v
u
dd
dd
d
d
=
( )
xx
x
cossin
d
d
=
;
(
)
d
cos – sin
d
x x
x
=
2
d
(tan ) sec
d
x x
x
=
;
2
d
(cot ) – cos
d
x ec x
x
=
x xx
x
sec tan ) (sec
d
d
=
;
2
d
(cosec ) – cot co sec
d
x x x
x
=
x
u
u nu
x
nn
d
d
)(
d
d
1–
=
;
d 1
ln
d
( u)
u u
=
d
(e ) e
d
u u
u
=
In terms of derivatives, instantaneous velocity and acceleration are defined as
t
x
t
x
v
t
d
d
lim
0
=
∆
∆
=
→∆
2
2
0
d d
lim
d d
t
v v x
a
t t t
∆ →
∆
= = =
∆
Integral Calculus
You are familiar with the notion of area. The formulae for areas of simple geometrical figures are
also known to you. For example, the area of a rectangle is length times breadth and that of a
triangle is half of the product of base and height. But how to deal with the problem of determination
of area of an irregular figure? The mathematical notion of integral is necessary in connection with
such problems.
Let us take a concrete example. Suppose a variable force f (x) acts on a particle in its motion
along x - axis from x = a to x = b. The problem is to determine the work done (W) by the force on the
particle during the motion. This problem is discussed in detail in Chapter 6.
Figure 3.31 shows the variation of F(x) with x. If the force were constant, work would be simply
the area F (b-a) as shown in Fig. 3.31(i). But in the general case, force is varying .
Fig. 3.31
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MOTION IN A STRAIGHT LINE 63
To calculate the area under this curve [Fig. 3.31 (ii)], let us employ the following trick. Divide the
interval on x-axis from a to b into a large number (N) of small intervals: x
0
(=a) to x
1
, x
1
to x
2
; x
2
to x
3
,
................................ x
N-1
to x
N
(=b). The area under the curve is thus divided into N strips. Each strip
is approximately a rectangle, since the variation of F(x) over a strip is negligible. The area of the i
th
strip shown [Fig. 3.31(ii)] is then approximately
xxFxxxFA
iiiii
∆==∆ )()–)((
1–
where ∆x is the width of the strip which we have taken to be the same for all the strips. You may
wonder whether we should put F(x
i-1
) or the mean of F(x
i
) and F(x
i-1
) in the above expression. If we
take N to be very very large (N
→∞
), it does not really matter, since then the strip will be so thin that
the difference between F(x
i
) and F(x
i-1
) is vanishingly small. The total area under the curve then is:
∑∑
==
∆=∆=
N
i
i
N
i
i
xxFAA
11
)(
The limit of this sum as N
→∞
is known as the integral of F(x) over x from a to b. It is given a special
symbol as shown below:
∫
=
b
a
dxxFA )(
The integral sign
∫
looks like an elongated S, reminding us that it basically is the limit of the sum
of an infinite number of terms.
A most significant mathematical fact is that integration is, in a sense, an inverse of differentiation.
Suppose we have a function g (x) whose derivative is f (x), i.e.
dx
xdg
xf
)(
)( =
The function g (x) is known as the indefinite integral of f (x) and is denoted as:
∫
= dxxfxg )()(
An integral with lower and upper limits is known as a definite integral. It is a number. Indefinite
integral has no limits; it is a function.
A fundamental theorem of mathematics states that
)(–)()()( agbg xgdx xf
b
a
b
a
≡=
∫
As an example, suppose f (x) = x
2
and we wish to determine the value of the definite integral from
x =1 to x = 2. The function g (x) whose derivative is x
2
is x
3
/3. Therefore,
3
7
3
1
–
3
8
2
1
2
1
2
===
∫
3
x
dx x
3
Clearly, to evaluate definite integrals, we need to know the corresponding indefinite integrals. Some
common indefinite integrals are
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PHYSICS64
( ) ln ( )
1
x
x x xd 0= >
∫
This introduction to differential and integral calculus is not rigorous and is intended to convey to
you the basic notions of calculus.
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