The variation of a quantity $A$ with a quantity $B$ describes the motion of a particle along a straight line. When $A$ (vertical axis) is plotted against $B$ (horizontal axis) the graph is a straight line of positive slope that does not pass through the origin — it meets the vertical ($A$) axis at a positive intercept and then rises linearly. Choose the correct statement(s). (Note: more than one option may be correct.) (a) Quantity B may represent time. (b) Quantity A is velocity if the motion is uniform. (c) Quantity A is displacement if the motion is uniform. (d) Quantity A is velocity if the motion is uniformly accelerated.

The variation of a quantity A with a quantity B describes the motion of a particle along a straight line. When

NCERT Exemplar · Q7

A

(a) Quantity B may represent time.

(b) Quantity A is velocity if the motion is uniform.

(d) Quantity A is velocity if the motion is uniformly accelerated.

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Concept understanding — Uniformly Accelerated Motion

Uniformly Accelerated Motion

Imagine you're sitting in a train that starts moving from a station. At first, it crawls — then it picks up speed smoothly, second by second. If you watch the speedometer, you might see it climb by the same amount every second: 0 to 10 km/h, then 10 to 20, then 20 to 30. That steady, predictable increase is the heart of uniformly accelerated motion.

The Intuition

When something moves with uniform acceleration, its velocity changes by the same amount in every equal interval of time. The change is constant — not faster one second and slower the next.

Think of a ball rolling down a gentle, straight ramp. It starts from rest. In the first second, it gains some speed. In the next second, it gains exactly the same amount of speed again. The acceleration — the rate of change of velocity — is fixed.

Note

"Uniform" here means "constant" or "unchanging." It does not mean the speed is constant. In fact, the speed is changing — but the rate at which it changes is constant.

The Precise Statement

Uniformly accelerated motion is motion in a straight line where the acceleration $a$ is constant in both magnitude and direction.

Mathematically, if $v$ is velocity at time $t$ , and $u$ is the initial velocity (at $t = 0$ ), then:

a = \frac{v - u}{t} = \text{constant}

This single idea leads to the three famous equations of motion (for constant acceleration):

v = u + at

Here:

$u$ = initial velocity (at $t = 0$ )
$v$ = velocity at time $t$
$a$ = constant acceleration
$s$ = displacement in time $t$

What It Looks Like in Real Life

Situation	Acceleration	Why it's (approximately) uniform
A car accelerating on a highway	~2–3 m/s²	Engine provides roughly constant force
A ball dropped from a height	9.8 m/s² downward	Gravity is nearly constant near Earth's surface
A train starting from a station	~0.5 m/s²	Controlled by the driver to be smooth

Watch out

Not all motion is uniformly accelerated. A car stopping suddenly has deceleration that changes — it's not uniform. A roller coaster has acceleration that varies wildly. Uniform acceleration is an ideal model that works beautifully for many real situations (like free fall) but not all.

The Key Insight

The word "uniform" refers to the acceleration, not the velocity. If acceleration is constant, then:

Velocity changes linearly with time (a straight line on a $v$ - $t$ graph)
Displacement changes quadratically with time (a parabola on an $s$ - $t$ graph)

This is why the equations above are so powerful: they let you predict position and velocity at any instant, as long as acceleration stays constant.

Important

For uniformly accelerated motion, the $v$ - $t$ graph is always a straight line. The slope of that line equals the acceleration. If the graph is curved, acceleration is not uniform. …

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A straight line with a positive intercept represents any relation $A = mB + c$ with $c\neq 0$ . If $B$ is time, this is exactly displacement for uniform motion ( $x=x_0+vt$ ) and velocity for uniformly accelerated motion ( $v=u+at$ ). A constant velocity (uniform motion) would give a horizontal line instead. Hence the correct choices are (a), (c) and (d).

What the graph says

The plotted line obeys

A = mB + c, \qquad m>0,\; c>0,

so $A$ grows linearly with $B$ and is already positive when $B=0$ .

Testing each statement

(a) $B$ may represent time. Nothing forbids the horizontal axis from being time; a linear $A$ – $t$ relation is perfectly physical. Correct.
(b) $A$ is velocity if the motion is uniform. In uniform motion the velocity is constant, so a velocity–time graph is a horizontal line ( $m=0$ ), not the sloped line shown. Incorrect. …

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Concept: A Complete 2 $\times$ 2 Matrix of Kinematic Graphs

Method: Test the Full Matrix (position/velocity $\times$ uniform/accelerated motion), Not Just the Listed Options

Rather than testing statements (a)–(d) one at a time as they're given, this method first builds the complete set of possibilities — every combination of {which quantity is plotted} $\times$ {what kind of motion it is} — and only then reads off which of those combinations match a straight line with a positive, non-zero intercept. This surfaces an extra insight the given options don't explicitly ask about, but which explains why the matrix comes out the way it does.

The 2 $\times$ 2 matrix (taking $B=$ time, since option (a) allows it)

	Uniform motion (constant $v$ )	Uniformly accelerated motion (constant $a$ )
$A=$ position $x$	$x = x_0 + vt$ — linear, slope $=v$ , intercept $=x_0$	$x = x_0+ut+\tfrac12at^2$ — quadratic, not a straight line at all
$A=$ velocity $v$	$v = \text{constant}$ — a horizontal line, slope $=0$	$v = u+at$ — linear, slope $=a$ , intercept $=u$

Reading the matrix against the given graph (straight line, positive slope, positive intercept)

Position, uniform motion ( $x=x_0+vt$ ): a straight line with slope $v>0$ (possible) and intercept $x_0$ (can be positive). Matches the graph → statement (c) is correct.
Position, accelerated motion ( $x=x_0+ut+\tfrac12at^2$ ): this cell of the matrix is not even linear — it's a parabola. The graph shown is a straight line, so this combination is automatically ruled out on shape grounds alone (not one of the listed options, but the matrix makes clear why "position under acceleration" was never a candidate to begin with).
Velocity, uniform motion ( $v=\text{const}$ ): this cell is a horizontal line (slope exactly $0$ ). The given graph has a positive slope, so this does not match. Statement (b) is incorrect. …

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

AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$n^{th}$
›Reveal solutionSolution
Using the distance-in- $n^{th}$ -second formula for two different seconds and subtracting eliminates both $u$ and $n$ , directly giving $a=25/3\approx8.33\ \mathrm{ms^{-2}}$ — option (A).

Concept and Intuition

The distance covered in the $n^{th}$ second of uniformly accelerated motion is $s_n=u+\dfrac{a}{2}(2n-1)$ . Because this expression is linear in $n$ , subtracting the value at $n$ from the value at $n+3$ removes both the unknown initial velocity $u$ and the unknown $n$ itself, leaving a direct equation for $a$ .

Step-by-Step Solution

Distance in $n^{th}$ second: $u+\dfrac{a}{2}(2n-1)=100$ .

Distance in $(n+3)^{th}$ second: $u+\dfrac{a}{2}\big(2(n+3)-1\big)=125$ .

Subtract (2) − (1): $\dfrac{a}{2}\Big[(2n+5)-(2n-1)\Big]=25$ .

Simplify the bracket: $(2n+5)-(2n-1)=6$ . …
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AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$t_1$
›Reveal solutionSolution
Because the particle speeds up under constant acceleration, it covers the second half of the distance faster than the first half, so $t_1 > t_2$ .

Concept and Intuition

For motion starting from rest under constant acceleration, distance covered grows with the square of elapsed time ( $d = \tfrac12 a t^2$ ), meaning the particle is moving slowly at first and progressively faster later. Since the average speed during the first half of the journey is lower than during the second half, covering the same distance (half the total) takes longer in the first half than in the second.

Step-by-Step Solution

Let total distance be $D$ , and let the particle start from rest with constant acceleration $a$ .

Time to cover the first half, $D/2$ : from $\tfrac12 a t_1^2 = D/2$ , get $t_1 = \sqrt{D/a}$ .

Time to cover the full distance $D$ : from $\tfrac12 a T^2 = D$ , get $T = \sqrt{2D/a}$ . …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$1:1$
›Reveal solutionSolution
Using the standard "distance in nth second" formula for free fall from rest, the 2nd-to-5th second ratio is $1:3$ .

Concept and Intuition

For uniformly accelerated motion starting from rest, the distance covered in the $n$ -th second (not up to the $n$ -th second) is $s_n=u+\frac{g}{2}(2n-1)=\frac{g}{2}(2n-1)$ when $u=0$ . This grows linearly in $n$ (odd multiples of $g/2$ ), which is why later seconds cover much more distance than earlier ones even though the ratio is smaller than $n^2$ (that would be the ratio of total distances fallen, not per-second distances).

Step-by-Step Solution

Displacement in the $n$ -th second (free fall from rest): $s_n=\dfrac{g}{2}(2n-1)=g\left(n-\dfrac12\right)$ .

For $n=2$ : $s_2=g(2-0.5)=1.5g$ .

For $n=5$ : $s_5=g(5-0.5)=4.5g$ . …
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AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
$n^{th}$
›Reveal solutionSolution
Using $v=an$ to find $a$ , then the "displacement in the $k$ th second" formula for both $n$ and $n-1$ and adding, gives $\frac{2v(n-1)}{n}$ .

Concept and Intuition

For uniformly accelerated motion starting from rest, the displacement covered during the $k$ -th second (not up to it) is a standard kinematic quantity: $s_k = u+\frac{a}{2}(2k-1)$ , which for $u=0$ simplifies to $\frac{a}{2}(2k-1)$ . The problem gives velocity at time $n$ (not displacement), which is what lets us solve for $a$ first.

Step-by-Step Solution

From rest, $v=u+at=0+an=an \Rightarrow a=\dfrac{v}{n}$ .

Displacement during the $n$ -th second: $s_n=\dfrac a2(2n-1)$ .

Displacement during the $(n-1)$ -th second: $s_{n-1}=\dfrac a2\big(2(n-1)-1\big)=\dfrac a2(2n-3)$ .

Total: $s_n+s_{n-1}=\dfrac a2\big[(2n-1)+(2n-3)\big]=\dfrac a2(4n-4)=2a(n-1)$ . …
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AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
$\mathrm{ms^{-1}}$
›Reveal solutionSolution
This tests deriving acceleration from a velocity-displacement relation using $a = V\frac{dV}{dx}$ . The relation gives a constant acceleration of $4\ \mathrm{ms^{-2}}$ .

Concept and Intuition

When velocity is given as a function of position, $V(x)$ , the acceleration is most conveniently found via $a = V\dfrac{dV}{dx} = \dfrac{1}{2}\dfrac{d(V^2)}{dx}$ , avoiding the need to first find $x(t)$ .

Step-by-Step Solution

Given $2V = \sqrt{37+32x}$ , square both sides: $4V^2 = 37 + 32x$ .

So $V^2 = \dfrac{37+32x}{4} = 9.25 + 8x$ .

Differentiate with respect to $x$ : $\dfrac{d(V^2)}{dx} = 8$ .

Acceleration: $a = \dfrac{1}{2}\dfrac{d(V^2)}{dx} = \dfrac{1}{2}(8) = 4\ \mathrm{ms^{-2}}$ . …
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AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
$\dfrac{5}{4}$
›Reveal solutionSolution
Directly apply the standard "distance in the $n$ th second" kinematics formula for uniformly accelerated motion from rest.

Concept and Intuition

The distance covered during the $n$ th second (not up to it) is a well-known derived formula from the equations of motion: $s_n=u+\dfrac a2(2n-1)$ , obtained by subtracting the distance up to $(n-1)$ seconds from the distance up to $n$ seconds.

Step-by-Step Solution

Given $u=0$ (starts from rest), $a=\dfrac54$ m/s², and we want the distance in the 3rd second ( $n=3$ ).

Apply $s_n=u+\dfrac{a}{2}(2n-1)=0+\dfrac{5/4}{2}(2\cdot3-1)=\dfrac{5}{8}(5)$ .

$s_3=\dfrac{25}{8}$ m.

Common Mistakes …
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AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
$X$
›Reveal solutionSolution
Since distance from rest scales as $t^2$ , the distance covered in each successive equal time interval grows — here the second interval covers $3X$ , three times the first.

Concept and Intuition

For motion starting from rest under constant acceleration, $s = \tfrac12 a t^2$ . Because displacement depends on the square of elapsed time (not linearly), equal time intervals do NOT cover equal distances — later intervals cover progressively more distance. This is the basis of the classic '1:3:5:7…' ratio for distances in successive equal time intervals starting from rest.

Step-by-Step Solution

Distance in the first $t$ seconds: $s_1 = \tfrac12 a t^2 = X$ .

Total distance in the first $2t$ seconds: $s_2 = \tfrac12 a (2t)^2 = 4 \cdot \tfrac12 a t^2 = 4X$ .

Distance covered specifically in the next $t$ seconds (i.e., from $t$ to $2t$ ) is $s_2 - s_1 = 4X - X = 3X$ . …
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AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
$\dfrac{(v+u)}{2}$
›Reveal solutionSolution
Using $v^2=u^2+2as$ for each point along the train, the middle compartment's speed comes out as the RMS-average of $u$ and $v$ : $\sqrt{(u^2+v^2)/2}$ .

Concept and Intuition

Every point along the train passes the same fixed tree, but at different points during the train's (uniformly accelerated) motion, having traveled different distances from some common reference. Since $v^2=u^2+2as$ is linear in the distance $s$ travelled, and the middle compartment is exactly halfway (in distance) between the engine and the last compartment, its $v^2$ value is the average of the engine's and the last compartment's $v^2$ values (not the average of their speeds directly).

Step-by-Step Solution

Let $L$ be the length of the train (distance from engine to last compartment). The engine crosses the tree with speed $u$ ; travelling this whole extra length $L$ further, the last compartment crosses with speed $v$ . By $v^2=u^2+2as$ : $v^2-u^2=2aL$ .

The middle compartment crosses the tree after the same point in the train has travelled only $L/2$ past where the engine started at speed $u$ . Let its speed be $w$ : $w^2-u^2=2a\left(\dfrac L2\right)=aL$ .

From step 1, $aL=\dfrac{v^2-u^2}{2}$ .

Substitute into step 2: $w^2=u^2+\dfrac{v^2-u^2}{2}=\dfrac{2u^2+v^2-u^2}{2}=\dfrac{u^2+v^2}{2}$ . …
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AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
$Q. The velocity$
(v) of a particle starting from rest increases linearly with time (t) as v = 4t, where v is in $\text{ms}^{-1}$ and t is in second. The distance covered by the particle in the first 4 seconds is (A) 16 m (B) 32 m (C) 8 m (D) 64 m
›Reveal solutionSolution
Integrate the velocity (or use $s=\frac12at^2$ for uniform acceleration) over the given time to get the distance.

Concept and Intuition

$v=4t$ means the particle has constant acceleration $a=4\ \text{ms}^{-2}$ , starting from rest. Distance is the area under the $v$ – $t$ graph, i.e. $\int v\,dt$ .

Step-by-Step Solution

$v(t)=4t$ , so acceleration $a=4\ \text{ms}^{-2}$ , and $u=0$ (starts from rest).

Distance $s=\displaystyle\int_0^4 v\,dt=\int_0^44t\,dt=\left[2t^2\right]_0^4=2(16)-0=32$ m. …
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AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
$2:3$
›Reveal solutionSolution
Combine $v=\sqrt{2gh}$ with $p=mv$ and plug in the given mass and height ratios.

Concept and Intuition

For free fall from rest, the speed on reaching the ground depends only on the height fallen: $v = \sqrt{2gh}$ . Momentum then combines both mass and this speed, so the ratio of momenta needs both the mass ratio and the square-root of the height ratio.

Step-by-Step Solution

Let $m_1:m_2 = 2:3$ and $h_1:h_2 = 9:16$ .

Speeds on landing: $v_1 = \sqrt{2gh_1}$ , $v_2=\sqrt{2gh_2}$ , so $\dfrac{v_1}{v_2} = \sqrt{\dfrac{h_1}{h_2}} = \sqrt{\dfrac{9}{16}} = \dfrac{3}{4}$ . …
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What the graph says

Testing each statement

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