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.

Chhattisgarh CgbseMCQ· 1mImportance★★★★★est

63% · 32/51 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ6 — At a metro station, a girl walks up a stationary escalator in time t₁.…→Next questionQ8 — The position–time (x versus t) graph of a particle moving in a straigh…

What the graph says

Testing each statement

🎓Unlock everything free for 14 days

Unlock everything free for 14 days