The position–time ($x$ versus $t$) graph of a particle moving in a straight line has the following shape, reading left to right. It begins at point A at $t=0$ with a horizontal tangent (the curve is momentarily flat) at a small positive $x$. It rises to a rounded maximum at point B, then falls to a short flat portion at point C where the tangent is again horizontal (a momentary point of inflection). From C it descends to a local minimum (trough) at point D, whose height still lies above the starting level A. Finally it rises steeply to a high maximum at point E on the right. Choose the correct statement(s). (Note: more than one option may be correct.) (a) The particle was released from rest at $t = 0$. (b) At B, the acceleration $a > 0$. (c) At C, the velocity and the acceleration vanish. (d) Average velocity for the motion between A and D is positive. (e) The speed at D exceeds that at E.

The position–time (x versus t) graph of a particle moving in a straight line has the following shape, reading

NCERT Exemplar · Q8

x

(a) The particle was released from rest at

t = 0

(b) At B, the acceleration

a > 0

(d) Average velocity for the motion between A and D is positive.

(e) The speed at D exceeds that at E.

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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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On an $x$ – $t$ graph, velocity $=$ slope and acceleration $=$ rate of change of slope (curvature). Applying this: A has zero slope (released from rest), B is a maximum (so $a<0$ , not $>0$ ), C is a flat inflection ( $v=0$ and $a=0$ ), and D sits above A so the displacement — and hence the average velocity — from A to D is positive. The correct statements are (a), (c) and (d).

Concept

For $x(t)$ : instantaneous velocity $v=\dfrac{dx}{dt}$ is the tangent slope, and acceleration $a=\dfrac{d^2x}{dt^2}$ tells how the slope changes (concave up $\Rightarrow a>0$ , concave down $\Rightarrow a<0$ ).

Testing each statement

(a) Released from rest at $t=0$ . At A the tangent is horizontal, so $v=0$ at the start — the particle begins from rest. Correct.
(b) At B, $a>0$ . B is a maximum of $x$ : the slope changes from positive to negative, so the curve is concave down and $a<0$ there. Incorrect.
(c) At C, velocity and acceleration vanish. C is a flat point of inflection: the tangent is horizontal ( $v=0$ ) and the curvature changes sign through zero ( $a=0$ ). Correct. …

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Concept: The First and Second Derivative Test, Applied Point by Point

Method: A Formal Curve-Sketching Classification Table (name and apply the standard calculus test at each labelled point, rather than describing each point narratively)

Calculus provides a standard, named procedure for classifying any point on a smooth curve — the first derivative test ( $x'=dx/dt$ , giving the slope/velocity) and the second derivative test ( $x''=d^2x/dt^2$ , giving the concavity/acceleration). This method builds one table applying both tests formally to every labelled point, instead of describing each point's behaviour in separate prose.

The classification table

Point	Feature on the $x$ - $t$ curve	$x'=v$ (1st derivative test)	$x''=a$ (2nd derivative test)
A ( $t=0$ )	horizontal tangent (curve momentarily flat)	$v=0$	curve concave down here (about to fall into a max shape) $\Rightarrow$ not evaluated further — position is what starts the motion
B	rounded maximum	$v=0$ (horizontal tangent at any smooth max)	local max $\Rightarrow$ curve concave down $\Rightarrow a<0$
C	flat inflection (horizontal tangent, momentary pause in curvature)	$v=0$	inflection point where curvature changes sign $\Rightarrow a=0$ exactly here
D	local minimum (trough, still above A's level)	$v=0$	local min $\Rightarrow$ curve concave up $\Rightarrow a>0$ (not asked directly, but follows the same test)
E	steep rise to a high maximum	(not a stationary point within the given data — approaching a max)	—

Applying the test to each statement

(a) Released from rest at $t=0$ . By the first derivative test, $A$ has a horizontal tangent, so $v(0)=0$ — the particle starts from rest. Correct.
(b) At B, $a>0$ . The second derivative test at a local maximum always gives concave-down, i.e. $a<0$ there — the opposite of the claim. Incorrect.
(c) At C, velocity and acceleration vanish. C is explicitly a flat inflection: first derivative test gives $v=0$ (horizontal tangent) and, since it's the point where concavity switches from one sign to the other, the second derivative passes through exactly $a=0$ there. Correct. …

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Concept

Testing each statement

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