JEE Mains Physics · Motion in a Straight Line

Motion Graphs: Slopes and Areas

The slope of a position–time graph is the velocity and the slope of a velocity–time graph is the acceleration; the area under a velocity–time graph is the displacement, and the area under an acceleration–time graph is the change in velocity.

All Motion in a Straight Line notes JEE Mains Physics strategy

Why this matters

Twenty PYQs, nineteen of them multiple choice, three from 2026, and nineteen carry a graph. Six take areas under a velocity–time graph, ten match a graph's shape to the motion, and four read a graph of velocity against position. Two questions decide every one of them: is the answer a slope or an area, and where does the velocity change sign?

Concept 1 of 3: Areas under velocity–time and acceleration–time graphs

Area under a velocity–time graph is velocity × time, which is a displacement. Area below the time axis is negative: the body is moving backwards. So the displacement is the signed sum of the areas, while the distance adds them all as positive. Split the graph wherever the line crosses the axis, find each piece, and decide which total the question wants.

Definition

Displacement $=$ signed area under v–t. Distance $=$ sum of the sizes of the areas.
Average velocity $= \dfrac{\text{displacement}}{\text{total time}}$ ; average speed $= \dfrac{\text{distance}}{\text{total time}}$ .
Slope of v–t $=$ acceleration.
Area under a–t $=$ change in velocity, $v_2 - v_1$ .
Pieces are triangles $\tfrac{1}{2}bh$ , rectangles bh and trapezia $\tfrac{1}{2}(a + b)h$ .
A line crossing zero: find where with the slope before splitting the area.

Areas

\Delta x = \int v\,dt \qquad \Delta v = \int a\,dt

Worked example

A body's velocity rises uniformly from 0 to

12\ \text{m/s}

in 4 s, stays at

12\ \text{m/s}

until

t = 10\ \text{s}

, then falls uniformly to

-6\ \text{m/s}

t = 16\ \text{s}

. Find the displacement, the distance and the average velocity over the 16 s.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 2 · Q4Moderate

Example 1 · Motion in a Straight Line · Motion Graphs: Slopes and Areas

The velocity

(v)

versus time

(t)

plot of a particle is shown in the figure, for a time interval of 40s. The total distance travelled by the particle and the average velocity during this period are, respectively

\_\_\_\_

Area gives displacement, not distance

The statement "area under a v–t graph is the distance" is false as soon as the graph dips below the axis. Distance needs every area counted as positive.

Average velocity comes from areas on a v–t graph

Average velocity is total displacement over total time, so on a v–t graph it comes from the areas, not from a slope. When the areas above and below the axis are equal, the average velocity is zero even though the body has moved.

Drill 5 more on areas under velocity–time and acceleration–time graphs

Concept 2 of 3: Matching graph shapes to the motion

Each graph is the slope of the one before it: v–t is the slope of x–t, and a–t is the slope of v–t. So a straight x–t line means constant velocity, a curving one means acceleration, and its steepness at any instant is the velocity then. Read the shapes in the table across a row, and use the list to rule out graphs that no real motion can produce.

Definition

Slope of x–t at an instant $=$ instantaneous velocity; chord between two times $=$ average velocity.
x–t: concave up (opening upward) means $a > 0$ ; concave down means $a < 0$ .
Slope of a momentum–time graph $=$ force: steepest part, largest force.
Falling with drag $F = -kv$ : $v = \dfrac{mg}{k}\left(1 - e^{-kt/m}\right)$ , rising from 0 and levelling at $mg/k$ .
Impossible graphs: two values at one time, time running backwards, total distance decreasing, or speed below zero.

Motion	Position–time	Velocity–time	Acceleration–time
At rest	Horizontal line	On the time axis (v = 0)	On the time axis (a = 0)
Constant velocity	Straight sloping line	Horizontal line	On the time axis (a = 0)
Speeding up from rest, constant a	Parabola opening upward, $x \propto t^{2}$	Straight line through the origin	Horizontal line above the axis
Slowing to rest, constant a	Curve bending over, flat where it stops	Straight line falling to zero	Horizontal line below the axis
Thrown up and caught (up positive)	Downward-opening parabola	Straight line from +u to −u, zero at the top	Horizontal line at −g The velocity line crosses zero at the top, but the acceleration line never moves.
Falling from rest with drag kv (down positive)	Curve that straightens into a line of slope mg/k	Rises from 0 and levels off at mg/k	Starts at g and falls towards zero
Constant velocity, then reversed at equal speed	Rising line, then falling line	Positive constant, then negative constant	Zero, with a spike at the reversal

Each column is the slope of the column to its left.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q86Moderate

Example 2 · Motion in a Straight Line · Motion Graphs: Slopes and Areas

The displacement

x

versus time graph is shown below. (A) The average velocity during 0 to 3 s is

10\text{ }m/s

(B) The average velocity during 3 to 5 s is

0\text{ }m/s

t = 2\text{ }s

5\text{ }m/s

(D) The average velocity during 5 to 7 s and instantaneous velocity at

t = 6.5\text{ }s

are equal (E) The average velocity from

t = 0

t = 9\text{ }s

is zero Choose the correct answer from the options given below:

Average velocity is a chord, not a tangent

Between two times, join the two points on the x–t graph and take that slope. The tangent at one instant gives the instantaneous velocity instead.

Velocity zero does not make the acceleration zero

Where the v–t line crosses the axis the body is at rest for an instant, but the slope there, the acceleration, is unchanged.

Drill 9 more on matching graph shapes to the motion

Concept 3 of 3: Velocity–position graphs: a = v dv/dx

When velocity is plotted against position, its slope dv/dx is not the acceleration. Multiply by the velocity: a = v dv/dx. A straight falling v–x line therefore gives an acceleration that changes with x. If v² is plotted against x instead, a straight line means constant acceleration, and its slope is 2a.

Definition

$a = v\dfrac{dv}{dx} = \dfrac{1}{2}\dfrac{d(v^{2})}{dx}$ .
$v = v_0 - kx$ (straight, falling): $a = k^{2}x - kv_0$ , a rising a–x line with a negative intercept, zero where v = 0.
$v = v_0 + kx$ (straight, rising): $a = kv_0 + k^{2}x$ , a rising line with a positive intercept.
$v^{2} = u^{2} + 2ax$ is a straight $v^{2}$ –x line: slope $2a$ , intercept $u^{2}$ .
v constant over a stretch: a = 0 there.

Acceleration from velocity and position

a = v\frac{dv}{dx} = \frac{1}{2}\frac{d(v^{2})}{dx}

Worked example

A particle's velocity falls in a straight line from

12\ \text{m/s}

at x = 0 to zero at x = 6 m. Find its acceleration at x = 0, x = 3 m and x = 6 m, and describe the a–x graph.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 23 · Q24Moderate

Example 3 · Motion in a Straight Line · Motion Graphs: Slopes and Areas

A particle is moving with constant acceleration '

a

'. Following graph shows

v^{2}

versus

x

(displacement) plot. The acceleration of the particle is

\_\_\_\_

m/s^{2}

The slope of v–x is not the acceleration

dv/dx has units of 1/s, not m/s². Multiply by v first. A straight v–x line does not mean constant acceleration.

Halving the slope of v² against x

v² = u² + 2ax, so the slope is 2a. Reading the slope as a doubles the answer.

Drill 3 more on velocity–position graphs: a = v dv/dx

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Areas under velocity–time and acceleration–time graphs
Areas
$\Delta x = \int v\,dt \qquad \Delta v = \int a\,dt$
Velocity–position graphs: a = v dv/dx
Acceleration from velocity and position
$a = v\frac{dv}{dx} = \frac{1}{2}\frac{d(v^{2})}{dx}$

Reference tables (1)

Matching graph shapes to the motion7 rows

Motion	Position–time	Velocity–time	Acceleration–time
At rest	Horizontal line	On the time axis (v = 0)	On the time axis (a = 0)
Constant velocity	Straight sloping line	Horizontal line	On the time axis (a = 0)
Speeding up from rest, constant a	Parabola opening upward, $x \propto t^{2}$	Straight line through the origin	Horizontal line above the axis
Slowing to rest, constant a	Curve bending over, flat where it stops	Straight line falling to zero	Horizontal line below the axis
Thrown up and caught (up positive)	Downward-opening parabola	Straight line from +u to −u, zero at the top	Horizontal line at −g The velocity line crosses zero at the top, but the acceleration line never moves.
Falling from rest with drag kv (down positive)	Curve that straightens into a line of slope mg/k	Rises from 0 and levels off at mg/k	Starts at g and falls towards zero
Constant velocity, then reversed at equal speed	Rising line, then falling line	Positive constant, then negative constant	Zero, with a spike at the reversal

Each column is the slope of the column to its left.

Watch out for (6)

Area gives displacement, not distance
→ Areas under velocity–time and acceleration–time graphs
Average velocity comes from areas on a v–t graph
→ Areas under velocity–time and acceleration–time graphs
Average velocity is a chord, not a tangent
→ Matching graph shapes to the motion
Velocity zero does not make the acceleration zero
→ Matching graph shapes to the motion
The slope of v–x is not the acceleration
→ Velocity–position graphs: a = v dv/dx
Halving the slope of v² against x
→ Velocity–position graphs: a = v dv/dx

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