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

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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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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 27 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following is an example of non-uniform motion?$
(a) A car travelling at a constant speed on a straight road
(b) A car accelerating from rest
(c) A car maintaining a steady speed around a circular track
(d) A car coming to a stop at a traffic light
›Reveal solutionSolution
Non-uniform motion is motion in which speed changes with time (unequal distances in equal time intervals). A car speeding up from rest is the standard textbook example.

Uniform motion = constant speed, equal distances covered in equal time intervals (option a and, in the basic sense taught in this chapter, option c -- uniform circular motion at constant speed -- are both treated as "uniform" speed cases here).

Non-uniform motion = speed is NOT constant -- the object covers unequal distances in equal time intervals.

(a) Constant speed on a straight road -- uniform motion.

(b) Accelerating from rest -- speed rises continuously from 0, a clean, unambiguous case of non-uniform motion. This is the example most commonly used in this chapter to introduce accelerated (non-uniform) motion. …
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CBSE 2026Set ANNUAL1 mark
$Q. What does the area under velocity-time graph represents?$
›Reveal solutionSolution

Area under a $v$ - $t$ graph = displacement, because $v\,dt$ (a thin vertical strip's area) is exactly the small displacement $dx$ .

Velocity is defined as $v=\dfrac{dx}{dt}$ , so a small displacement over a tiny time interval $dt$ is $dx=v\,dt$ . On a velocity-time graph, $v\,dt$ is precisely the area of a thin vertical strip of width $dt$ and height $v$ . Summing (integrating) all such strips from time $t_1$ to $t_2$ gives:

$\Delta x=\int_{t_1}^{t_2} v\,dt = \text{area under the } v\text{-}t \text{ graph between } t_1 \text{ and } t_2$

…

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CBSE 2026Set sz1 markMCQ
$Q. The area under velocity-time graph represents:$
(a) acceleration
(b) force
(c) displacement
(d) work
›Reveal solutionSolution

The area under a v-t graph gives displacement, since displacement = integral of velocity dt.

Displacement s = integral of v dt over the time interval. Geometrically, this integral is exactly the area bounded by the velocity curve and the time axis (with sign, since …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The area enclosed by velocity-time graph with time axis represents -$
(a) velocity
(b) displacement
(c) acceleration
(d) work
›Reveal solutionSolution

The area under a velocity-time graph, between the curve and the time axis, equals the displacement of the body.

For a small time interval dt, if the velocity is v, the small displacement covered is dx = v × dt — exactly the area of a thin vertical strip of the v-t graph of height v and width dt. Adding up (integrating) all such strips between two times t1 and t2 gives the total displacement:

x = ∫ v dt (from t1 to t2) …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. For which of the following position-time (x-t) graphs acceleration is zero ?$
(a) concave-upward (U-shaped) x-t curve
(b) dome-shaped (concave-downward) x-t curve
(c) straight line rising with time
(d) oscillating wave-like x-t curve
›Reveal solutionSolution
Only a straight-line x-t graph gives constant velocity and zero acceleration. Answer (C).

In a position-time graph:

the slope (dx/dt) gives velocity,

if the slope changes with time, the velocity changes, meaning there is acceleration.

A curved x-t graph (concave up, dome-shaped, or oscillating) has a continuously changing slope, so the velocity changes and acceleration is non-zero. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. The area under the velocity-time (v-t) graph of a moving body represents$
(a) velocity of the body
(b) acceleration of the body
(c) kinetic energy of the body
(d) displacement of the body
›Reveal solutionSolution

Area under a v-t graph = displacement. Answer (D).

On a velocity-time graph, a small strip of width dt and height v has area v dt. Summing (integrating) all such strips gives the total displacement:

displacement = integral of v dt = area under the v-t curve.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following is an example of non-uniform motion?$
(a) A car travelling at a constant speed on a straight road
(b) A car accelerating from rest
(c) A car maintaining a steady speed around a circular track
(d) A car coming to a stop at a traffic light
›Reveal solutionSolution

Non-uniform motion means the body covers unequal distances in equal time intervals, i.e. its speed changes with time -- exactly what happens as a car accelerates from rest.

Uniform motion: a body covers equal displacements in equal intervals of time, moving at constant speed along a straight path -- option (a) is uniform motion.

Non-uniform (accelerated) motion: the speed changes with time. A car accelerating from rest goes from 0 speed to increasing speed over time -- unequal distances are covered in successive equal time intervals -- so this is the clearest, textbook example of non-uniform motion.

…

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CBSE 2025Set ANNUAL1 mark
$Q. Answer in one word or one sentence: What does the slope of velocity-time graph represent?$
›Reveal solutionSolution

The slope of a v-t graph gives the instantaneous acceleration.

Acceleration is defined as the rate of change of velocity with time, a = dv/dt. On a velocity-time (v-t) graph, the slope at any point is exactly dv/dt at that instant. So the slope of the v-t graph directly represents the instantaneous acceleration of the particle — a steeper slope means larger acceleration, a negative slope means deceleration (or acceleration op …

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CBSE 2025Set ANNUAL1 mark
$Q. State True or False: A particle in one-dimensional motion with zero speed may have non-zero velocity.$
›Reveal solutionSolution

The statement is False: zero speed always means zero velocity.

Speed is defined as the magnitude of the velocity vector: speed = |velocity|. If the speed of a particle is zero, then the magnitude of its velocity vector is zero, and a vector with zero magnitude is the zero vector itself — it has no direction and represents no motion. Therefore a …

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CBSE 2025Set sz1 markMCQ
$Q. When the distance travelled by a body is proportional to the time taken, what happens to its speed? (A) Becomes zero (B) Increases (C) Remains the same (D) Decreases$
›Reveal solutionSolution

Distance proportional to time (s = kt) implies a constant velocity, so the speed remains the same.

If the distance travelled s is directly proportional to time t, we can write $s = kt$ where k is a constant.

Speed is the rate of change of distance: $v = \dfrac{ds}{dt} = k$ , which is a constant independent of time. …

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CBSE 2025Set ANN1 markMCQ
$Q. A graph of position-time of an object is given. Choose the correct answer related to this graph.$
(a) velocity of object increases
(b) object is stationary
(c) object has constant velocity
›Reveal solutionSolution

A straight-line position-time graph has a constant slope, so the object moves with constant (uniform) velocity.

On a position-time (x-t) graph, the instantaneous velocity at any instant equals the slope of the graph at that point: v = dx/dt.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The area between velocity-time graph and time axis is equal to (A) acceleration (B) distance covered (C) change in velocity (D) change in acceleration$
›Reveal solutionSolution

Area under a $v$ - $t$ graph gives distance/displacement covered.

Displacement is defined as $x=\int v\,dt$ . Geometrically, this integral is exactly the area enclosed between the velocity curve and the time axis over the interval considered. It is not acceleration (that's the slop …

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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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