Figure 2.11 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 $t 0$? If not, suggest a suitable physical context for this graph. ![Figure 2.11](/figures/motion-in-a-straight-line/x-t-graph-flat-then-parabolic-fig-2-11.svg)

Figure 2.11 shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph th

Exercises · 2.13

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Concept understanding — One Dimensional Motion

One Dimensional Motion

Imagine you're standing on a long, perfectly straight railway track. A train moves along it — it can only go forward or backward. It cannot turn left, right, up, or down. That's the core idea: motion confined to a single straight line.

The Intuition

In the real world, a ball thrown across a room moves in three dimensions — it goes forward, sideways, and up-down. But many problems in physics are simpler. We deliberately restrict motion to one dimension (1D) to understand the fundamental laws without the clutter of angles and curves.

Think of:

A car moving on a straight highway (no turns).
A lift going up or down a shaft.
A ball dropped straight down from a height.
A puck sliding on a frictionless straight track.

In each case, the object's position can be described by just one number — its distance from a fixed point (the origin) along that line.

The Precise Statement

x

This means:

The path is a straight line.
The direction is either positive (say, to the right or upward) or negative (left or downward).
All vector quantities (displacement, velocity, acceleration) have only two possible directions — forward or backward.

The Three Key Quantities

To describe 1D motion precisely, we use three quantities:

Position ( $x$ or $y$ ) — where the object is relative to the origin.

Example: $x = +5\ \text{m}$ means 5 metres to the right of the origin.
Displacement ( $\Delta x$ ) — change in position:

\Delta x = x_{\text{final}} - x_{\text{initial}}

This is a vector — it has a sign. If you move from $x=2\ \text{m}$ to $x=7\ \text{m}$ , $\Delta x = +5\ \text{m}$ . If you move back to $x=3\ \text{m}$ , $\Delta x = -4\ \text{m}$ .

Velocity ( $v$ ) — rate of change of position:

v = \frac{\Delta x}{\Delta t}

Average velocity has a sign. Instantaneous velocity is the slope of the position-time graph.

Acceleration ( $a$ ) — rate of change of velocity:

a = \frac{\Delta v}{\Delta t}

Again, a signed quantity. Positive acceleration doesn't always mean speeding up — it means velocity is becoming more positive (or less negative).

The Equations of Motion (Constant Acceleration)

For the special (and very common) case of constant acceleration, we have three equations that connect these quantities. They are the equations of motion for 1D:

v = u + at

Where:

$u$ = initial velocity
$v$ = final velocity
$a$ = constant acceleration
$t$ = time
$s$ = displacement

Watch out

These equations only work when acceleration is constant. If acceleration changes, you cannot use them directly — you'd need calculus or graphical methods.

A Simple Example …

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The statement confuses an $x$ - $t$ graph with a spatial trajectory. This is one-dimensional motion, so the particle always moves along one straight line; the graph only tells us how its coordinate $x$ varies with time. Being flat ( $x=0$ ) for $t<0$ means it is at rest at the origin, and the parabola for $t>0$ means $x\propto t^2$ , i.e. uniformly accelerated motion from rest.

Concept

An $x$ - $t$ graph plots the single coordinate $x$ against time; its shape is not the geometric path travelled through space. In one dimension the path is always a straight line. The graph's slope $\dfrac{dx}{dt}$ is the velocity and its curvature reflects the acceleration.

Why the statement is wrong

For $t < 0$ : $x = 0$ (constant). A horizontal $x$ - $t$ line means zero velocity, i.e. the particle is at rest at the origin. It is not 'moving in a straight line'; it is not moving at all.
For $t > 0$ : $x$ rises as an upward-opening parabola. Comparing with the kinematic equation for constant acceleration from rest,

x = \tfrac{1}{2}\,a\,t^2,

the parabolic $x$ - $t$ curve simply signifies uniform acceleration, with velocity $v = at$ increasing linearly. The particle is still moving along the same straight line in space, only speeding up.

A suitable physical context …

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Concept: A Graph's Shape Lives in the Plane; the Trajectory Lives on the Line

Method: Domain–Range Mapping Argument (the graph is 2-D; the physical path is 1-D, by definition — no case-by-case sign reasoning needed)

Instead of walking through the flat segment and the rising segment separately and describing what each means physically, this method makes one general observation about any $x$ - $t$ graph of one-dimensional motion, which settles the question immediately and for every possible graph shape, not just this particular parabola-after-a-flat-line one.

Steps

Treat position as a function of time, $x = f(t)$ . The graph of $f$ is the set of points $\{(t, f(t))\}$ — this lives in the two-dimensional $t$ - $x$ plane, and its shape can be anything a function's graph can be: straight, curved, parabolic, wavy.
The physical trajectory, by contrast, is the RANGE of $f$ — the set $\{x = f(t) : t \text{ in the domain}\}$ . This is a subset of the real number line (the $x$ -axis alone), because "one-dimensional motion" means, by definition, that the particle's only degree of freedom is its single coordinate $x$ .
A subset of a line is always confined to that line, regardless of how curvy the graph of $f$ against $t$ looks. A parabola-shaped $x$ - $t$ graph does not mean a parabola-shaped trajectory — the trajectory can only ever be a stretch of the single $x$ -axis, because there is no second spatial dimension for it to curve into.
Apply this to the given graph. For $t<0$ : $x=0$ (flat line) — range is the single point $\{0\}$ , i.e. the particle sits at the origin, at rest (zero velocity, since the graph's slope is zero). For $t>0$ : $x$ increases along an upward-curving path that steepens — comparing to $x = \tfrac12 a t^2$ (displacement from rest under constant acceleration $a$ ), this is the signature of uniformly accelerated motion starting from rest, not a hint about spatial curvature. The RANGE for $t>0$ is still just an increasing stretch of the same $x$ -axis. …

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Q.Figure 2.11 shows the xxx-ttt 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<0t < 0t<0 and on a parabolic path for t>0t > 0t>0? If not, suggest a suitable physical context for this graph.

Concept

Why the statement is wrong

A suitable physical context …

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Q.Figure 2.11 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.

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