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 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 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. 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.
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| Fig 1: x-axis, origin and positions of a car at different times. |
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 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. 2(a).
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| Fig. 2: Position-time graph of (a) stationary object, and (b) an object in uniform motion. |
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. 2(b) shows the position-time
graph of such a motion.
 |
| Fig 3: Position-time graph of a car |
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. We shall refer to this graph in our discussion in the following sections.
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