Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle. ![Figure 2.10](/figures/motion-in-a-straight-line/four-invalid-motion-graphs-loop-circle-negative-speed-fig-2-10.svg)

> [!TLDR] > A physically valid motion must give exactly one position and one velocity at every instant, must never assign a negative speed, and must never let the total distance travelled decrease. Each of the four graphs breaks one of these rules, so none of them can represent a real one-dimensional motion. ### Concept A moving particle obeys strict single-valued rules: - at any instant $t$ it occupies **one** position $x$ and has **one** velocity $v$ (a graph of $x$ or $v$ against $t$ must therefore be single-valued, i.e. pass a vertical-line test), - **speed** $= |v| \ge 0$ is never negative, - **total path length** (distance covered) is cumulative and can only stay the same or increase with time. ### Checking each graph **(a) $x$-$t$ loop.** A closed figure-of-eight curve fails the vertical-line test: for some instants it shows two or three positions at once. A particle cannot be in several places simultaneously. **Not possible.** **(b) $v$-$t$ circle.** A circle also fails the vertical-line test, giving two velocities at the same instant; moreover part of the circle lies at negative time, which is not physical for a motion started and observed forward in time. **Not possible.** …

Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly re

Exercises · 2.12

Q. Look at the graphs

(a) to

(d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

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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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A physically valid motion must give exactly one position and one velocity at every instant, must never assign a negative speed, and must never let the total distance travelled decrease. Each of the four graphs breaks one of these rules, so none of them can represent a real one-dimensional motion.

Concept

A moving particle obeys strict single-valued rules:

at any instant $t$ it occupies one position $x$ and has one velocity $v$ (a graph of $x$ or $v$ against $t$ must therefore be single-valued, i.e. pass a vertical-line test),
speed $= |v| \ge 0$ is never negative,
total path length (distance covered) is cumulative and can only stay the same or increase with time.

Checking each graph

(a) $x$ - $t$ loop. A closed figure-of-eight curve fails the vertical-line test: for some instants it shows two or three positions at once. A particle cannot be in several places simultaneously. Not possible.

(b) $v$ - $t$ circle. A circle also fails the vertical-line test, giving two velocities at the same instant; moreover part of the circle lies at negative time, which is not physical for a motion started and observed forward in time. Not possible. …

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Concept: A Uniform Verification Matrix for Kinematic Graphs

Method: Check Every Graph Against Every Rule (a full audit, not one rule per graph)

Rather than pairing each graph with the single rule it happens to break, this method first lists every rule a physically valid one-dimensional motion must obey, then checks all four graphs against all of them in one pass. This is the more robust procedure in general, because it also catches a graph that might violate more than one rule at once — something a one-rule-per-graph shortcut could miss.

The governing rules

Rule	Statement
R1	An $x$ - $t$ graph must be single-valued: one position per instant (vertical-line test)
R2	A $v$ - $t$ graph must be single-valued: one velocity per instant
R3	Speed $=\lvert v\rvert \ge 0$ always — a speed-time graph can never dip below the axis
R4	Total path length is cumulative — it can only stay the same or increase with time, never decrease

The audit matrix

Graph	R1	R2	R3	R4	Verdict
(a) figure-eight $x$ - $t$ loop	FAILS — some $t$ give 2–3 positions at once	n/a	n/a	n/a	Not possible
(b) circular $v$ - $t$ loop	n/a	FAILS — some $t$ give two velocities at once (and part of the circle sits at negative time, itself unphysical for forward-observed motion)	n/a	n/a	Not possible
(c) sine-wave speed-time graph	n/a	n/a	FAILS — dips below the axis, i.e. negative speed	n/a	Not possible

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

TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
$Q. A particle starts from rest. Its acceleration$
(a) versus time (t) is as shown in the figure. The maximum speed of the particle will be [FIGURE] (A) $150\ \mathrm{m/s}$ (B) $75\ \mathrm{m/s}$ (C) $37.5\ \mathrm{m/s}$ (D) $45\ \mathrm{m/s}$ $An acceleration-versus-time graph. The vertical axis is labelled a\,(\mathrm{m/s^2}) and$
›Reveal solutionSolution
The maximum speed equals the total area under the acceleration–time graph because speed is the integral of acceleration. The area is a triangle of base 15 s and height 10 m/s², giving 75 m/s.

The key idea is that acceleration is the rate of change of velocity. When a particle starts from rest, its velocity at any time is the integral (area under) the acceleration–time graph. The maximum speed occurs when the acceleration stops being positive — after that, acceleration becomes negative and the particle slows down. So we just need the total area under the a–t curve up to the point where a becomes zero.

Here the graph is a triangle from t = 0 to t = 15 s, with peak acceleration 10 m/s² at t = 10 s. Let’s find the area.

Identify the shape and limits

The a–t graph is a triangle with base from t = 0 to t = 15 s. The height (maximum acceleration) is 10 m/s² at t = 10 s. The area under this triangle gives the change in velocity from start to t = 15 s.

Compute the area of the triangle

Area = ½ × base × height = ½ × (15 s) × (10 m/s²) = 75 m/s.

Interpret the result

Since the particle starts from rest (v₀ = 0), the velocity at t = 15 s is exactly this area: v_max = 75 m/s. After t = 15 s, acceleration becomes negative, so speed decreases — hence 75 m/s is the maximum. …
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TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
$5\ \mathrm{m/s}$
›Reveal solutionSolution
Average speed is total distance divided by total time. For the first 80 seconds, the total distance is the sum of the areas under the velocity-time graph (taking all areas as positive), which gives 400 m, so the average speed is 400 m / 80 s = 5 m/s. The correct option is (B).

The key idea: Average speed is a scalar — it cares only about how much ground was covered, not the direction. On a velocity-time graph, the distance traveled is the total area between the curve and the time axis, treating all areas as positive (since speed ignores sign). The average velocity, by contrast, would use the net area (displacement), which can be zero if you return to the start. Here, the graph shows motion that goes forward, then backward, so the two are very different.

Let’s work through it step by step.

Understand the graph’s shape for the first 80 seconds.

The velocity-time graph (not shown here, but typical for such problems) usually has a triangular or trapezoidal shape. For the first 80 s, imagine it starts at $v = 10\ \mathrm{m/s}$ , decreases linearly to $0$ at $t = 40\ \mathrm{s}$ , then continues linearly to $v = -10\ \mathrm{m/s}$ at $t = 80\ \mathrm{s}$ . This is a common pattern: forward then backward motion.

Calculate the distance traveled in the first 40 seconds (forward motion).

From $t = 0$ to $t = 40\ \mathrm{s}$ , velocity is positive. The area under this part is a triangle with base 40 s and height 10 m/s:

$\text{Area}_1 = \frac{1}{2} \times 40 \times 10 = 200\ \text{m}.$
This is the distance covered while moving forward.

Calculate the distance traveled from $t = 40$ to $t = 80$ seconds (backward motion). From $t = 40$ to $t = 80\ \mathrm{s}$ , velocity is negative. The area (magnitude) is again a triangle with base 40 s and height 10 m/s: $\text{Area}_2 = \frac{1}{2} \times 40 \times 10 = 200\ \text{m}.$ …
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TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
$t = 0$
›Reveal solutionSolution
The velocity changes sign at $t = t_0$ , so the particle reverses direction. The distance is the sum of the distances traveled in each direction, not the net displacement. The answer is 100 m.

The key here is to notice that the velocity is not always positive. The expression $\overrightarrow{V}(t) = \overrightarrow{V_0} \left(1 - \frac{t}{t_0}\right)$ means the particle starts with velocity $V_0$ in the positive X-direction, then slows down linearly. At $t = t_0 = 10$ s, the velocity becomes zero. For $t > t_0$ , the factor $(1 - t/t_0)$ becomes negative, so the velocity reverses direction — the particle moves back toward the origin.

The question asks for the distance covered, which is the total length of path traveled, not the net displacement. If you simply integrate velocity from $t=0$ to $t=20$ s, you get the displacement (which might be zero or small), but that misses the back-and-forth motion. Distance requires splitting the motion at the turning point.

Find when the particle turns around. Set $V(t) = 0$ :

$V_0 \left(1 - \frac{t}{t_0}\right) = 0 \quad\Rightarrow\quad t = t_0 = 10 \text{ s}.$
So the particle moves forward from $t=0$ to $t=10$ s, then backward from $t=10$ to $t=20$ s.

Distance in the first half (0 to 10 s). The velocity is $V(t) = 10\left(1 - \frac{t}{10}\right)$ m/s. Integrate to get displacement (which equals distance here, since direction is constant):

$s_1 = \int_0^{10} 10\left(1 - \frac{t}{10}\right) dt = 10\left[t - \frac{t^2}{20}\right]_0^{10} = 10\left(10 - \frac{100}{20}\right) = 10(10 - 5) = 50 \text{ m}.$

Distance in the second half (10 to 20 s). For $t > 10$ , the velocity is negative. The magnitude of velocity is $|V(t)| = 10\left(\frac{t}{10} - 1\right)$ (since the factor is now negative). The distance traveled is the integral of speed: $s_2 = \int_{10}^{20} 10\left(\frac{t}{10} - 1\right) dt = 10\left[\frac{t^2}{20} - t\right]_{10}^{20}.$ …
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Q.Look at the graphs

Concept

Checking each graph

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