Going down the ladder, read the slope. Going up the ladder, read the area.
x → v → a by slopes; a → v → x by areas. Every graph question in this chapter is one of
those two operations.
Graph questions look like a separate skill, but they are the same calculus as File 1
drawn instead of written. The derivative of a function is the slope of its graph, and the integral is
the area under it. Once you accept that, there is nothing left to memorise except which graph sits where.
animated The two directions. Move rightward by taking slopes (differentiating); move leftward by taking areas (integrating). The tangent sliding along the x–t parabola and the dot climbing the v–t line are the same information, shown twice.
Reading an x–t graph
Slope = velocity. Steep means fast; negative slope means moving towards decreasing x;
horizontal means at rest.
Curvature = acceleration. NCERT Fig. 2.2: the graph curves upward for positive acceleration,
downward for negative acceleration, and is a straight line for zero acceleration.
Intersection of two graphs = the two particles are at the same place at the same instant.
It does not mean equal velocity — that would need parallel lines.
The single biggest misconception
An x–t graph is not a picture of the path. A parabolic x–t graph describes motion
along a perfectly straight line with uniform acceleration. NCERT Exercise 2.13 exists purely to catch
this confusion, so expect a NEET version of it.
Reading a v–t graph
Slope gives acceleration; area gives displacement. The area rule is worth pausing on. NCERT proves it
for the easy case: constant velocity u for time T gives a rectangle of area uT, which is the
displacement. How can an area be a distance? Check the dimensions of the axes —
m s⁻¹ multiplied by s is m. The area of a graph inherits the product of the two axis units.
animated The four constant-acceleration cases from NCERT Fig. 2.3. Note (c): a negative velocity with a negative acceleration means the body is speeding up, moving backwards. And (d): the object reverses direction at t₁ while the acceleration never changes.
Sign of a does not tell you speeding up or slowing down
NCERT Points to Ponder 2 and 3. If a points the same way as v, the body speeds up; if it points the
opposite way, the body slows down. A body falling under gravity has negative acceleration (upward-positive
convention) and is speeding up; a body thrown upward has the same negative acceleration and is slowing down.
Graphs that cannot exist
Three tests kill an impossible graph instantly:
Is it double-valued in t? A particle has one position and one velocity per instant, so any
vertical line may cut the curve only once.
Does a magnitude go negative? Speed and total path length are magnitudes; they cannot dip
below zero.
Does a running total shrink? Total path length accumulates; it may stay flat but never
decrease.
animated NCERT Exercise 2.12: all four are impossible, and each fails a different test. Knowing which test each one fails is what turns this into a five-second question.
Why this topic is Priority 1
Graph items need almost no calculation, which makes them the fastest marks available and a direct lever
on blank rate: there is no arithmetic to get lost in, so an unattempted graph question is a mark simply
left on the table. NEET’s graph questions are also unusually close to the NCERT exercises —
Exercises 2.12, 2.13, 2.15, 2.16, 2.17 and 2.18 are the source material, and it is worth doing all six
in the book itself.
2 · Formula sheet
Slope of x–t
= velocity v
Tangent slope = instantaneous v. Chord slope = average v.
Curvature of x–t
= sign of acceleration
Curving up → a > 0. Curving down → a < 0. Straight → a = 0.
Slope of v–t
= acceleration a
Tangent slope at an instant.
Area under v–t
= displacement Δx
Signed: area below the axis is negative. NCERT Summary 5.
Area under a–t
= change in velocity Δv
Signed. Add v₀ to get v.
Uniform motion
x–t inclined line ; v–t line parallel to t-axis ; a = 0
Uniform acceleration
x–t parabola ; v–t inclined line ; a–t horizontal
Degree ladder 2, 1, 0.
Free fall from rest
a–t flat at −9.8 ; v–t line −9.8t ; y–t parabola −4.9t²
NCERT Fig. 2.7.
Intersection of two x–t graphs
same position at the same instant
NOT the same velocity. Same velocity means parallel.
v–t crossing the time axis
direction reverses; displacement is extreme there
Impossible graphs
double-valued in t ; negative speed ; decreasing path length
The three tests from Exercise 2.12.
Sharp kinks
not physical — v and a change continuously
Textbook kinks are idealisations.
3 · Exceptions and traps
Trap 1 — graph mistaken for trajectory
An x–t curve shows how position changes with time, not the shape of the route. In this whole
chapter the route is a straight line, always.
Trap 2 — intersection read as equal velocity
Crossing means same position. Parallel means same velocity. These are different geometric facts.
Trap 3 — unsigned area
Area below the time axis is negative displacement but positive distance. If a v–t line crosses
the axis, the two answers will differ — check which one the question wants.
Trap 4 — value confused with slope
At the peak of a v–t curve the velocity is largest and the acceleration is zero.
At the peak of an x–t curve the position is largest and the velocity is zero.
Trap 5 — forgetting the intercept
A v–t line that starts above the origin encloses a trapezium, not a triangle. Missing the
intercept loses the v₀t term.
Exception — kinks are idealisations
Several NCERT figures show sharp corners. The text is explicit that in any realistic situation the
functions are differentiable everywhere and the graphs are smooth, because velocity and acceleration
cannot change abruptly at an instant.
4 · Numbers and shapes to remember
Situation
x–t
v–t
a–t
At rest
horizontal line
on the axis
on the axis
Uniform velocity
inclined straight line
line parallel to t-axis
on the axis
Uniform acceleration
parabola
inclined straight line
line parallel to t-axis
Free fall from rest
y = −4.9t²
v = −9.8t
flat at −9.8
Bouncing ball (speed–time)
—
sawtooth, peaks decaying
flat at g between bounces
5 · Scientists NEET names
Galileo Galilei (1564–1642)
The same name as the previous two files, and for a reason that belongs here: it was Galileo who
established that the right graph to look at is velocity against time, not velocity against
distance. His inclined-plane experiments showed dv/dt is constant in free fall
while dv/dx is not.
6 · Twenty worked questions
Graph items in NEET stay very close to NCERT Exercises 2.12,
2.13, 2.15, 2.16, 2.17 and 2.18, so those exercise numbers are cited in the stems where relevant.
Sketch each graph on rough paper before choosing — drawing it yourself is faster than staring at it.
Q1NEET-style
The position–time graph of a particle is the straight line shown. Its velocity is
(a) 2 m s⁻¹
(b) 5 m s⁻¹
(c) 7.5 m s⁻¹
(d) 20 m s⁻¹
Given
x–t straight line through (2 s, 10 m) and (6 s, 30 m)
Asked
velocity
Concept to use
On an x–t graph the slope is the velocity. A straight line means constant velocity, so the slope is the same everywhere.
Formula or rule
v = Δx/Δt
Baby steps
Δx = 30 − 10 = 20 m.
Δt = 6 − 2 = 4 s.
v = 20/4 = 5 m s⁻¹.
Answer
5 m s⁻¹
Shortcut
Use the two marked points, never the origin, unless the line actually passes through it. Here x/t at the second point would give 5 by luck; at the first it would give 5 too, but only because the line happens to pass through the origin when extended.
Trap
Reading the y-value at one point (30 m) as the velocity. A position is not a velocity — only the slope is.
Q2NEET-style
A body’s velocity–time graph is shown. The distance it travels in 15 s is
(a) 150 m
(b) 180 m
(c) 300 m
(d) 210 m
Given
v–t trapezium: 0→20 m s⁻¹ in 4 s, constant 20 m s⁻¹ to 10 s, 20→0 by 15 s
Asked
distance travelled
Concept to use
The area under a v–t curve is the displacement. Since v never goes negative here, distance and displacement are equal.
Formula or rule
area = ½(sum of parallel sides) × height, or split into pieces
Baby steps
Triangle 1: ½(4)(20) = 40 m.
Rectangle: (10 − 4)(20) = 120 m.
Triangle 2: ½(5)(20) = 50 m.
Total = 40 + 120 + 50 = 210 m.
Answer
210 m
Shortcut
Trapezium formula in one step: ½(15 + 6)(20) = 210 m, where 15 s is the whole base and 6 s is the flat top.
Trap
Multiplying peak velocity by total time (20 × 15 = 300 m). That assumes the body moved at 20 m s⁻¹ the whole time, which it did not during the ramps.
Q3NEET-style
The velocity–time graph of a particle is shown. Over the interval 0 to 8 s, its displacement and the distance it travels are respectively
(a) 0 m and 0 m
(b) 80 m and 80 m
(c) 0 m and 80 m
(d) 40 m and 80 m
Given
v–t straight line from +20 m s⁻¹ at t = 0 to −20 m s⁻¹ at t = 8 s, crossing zero at t = 4 s
Asked
displacement and distance over 0–8 s
Concept to use
Area below the time axis counts as negative displacement but as positive distance. This one distinction is the whole question.
Formula or rule
displacement = signed area ; distance = Σ|area|
Baby steps
Area above axis (0–4 s): ½(4)(20) = +40 m.
Area below axis (4–8 s): ½(4)(20) = −40 m.
Displacement = 40 − 40 = 0.
Distance = 40 + 40 = 80 m.
Answer
displacement 0 m, distance 80 m
Shortcut
Whenever a v–t line crosses the time axis, expect the two answers to differ. Find the crossing time first, then treat each side separately.
Trap
Reporting distance as zero because the areas cancel. The particle went out 40 m and came back 40 m — it certainly travelled. Distance can never be zero for a body that moved.
Q4NEET-style
The acceleration–time graph of a particle starting from rest is shown. Its velocity at t = 7 s is
(a) 4 m s⁻¹
(b) 8 m s⁻¹
(c) 12 m s⁻¹
(d) 20 m s⁻¹
Given
a = +4 m s⁻² for 0–3 s, then a = −2 m s⁻² for 3–7 s; v = 0 at t = 0
Asked
velocity at t = 7 s
Concept to use
The area under an a–t graph is the change in velocity, with sign. Add the initial velocity at the end.
Formula or rule
Δv = ∫ a dt = signed area
Baby steps
First block: (+4)(3) = +12 m s⁻¹.
Second block: (−2)(4) = −8 m s⁻¹.
Net Δv = +4 m s⁻¹.
v = 0 + 4 = 4 m s⁻¹.
Answer
4 m s⁻¹
Shortcut
The a–t graph is the least useful of the three — but the rule is symmetric: area of a–t gives Δv, area of v–t gives Δx.
Trap
Adding the magnitudes (12 + 8 = 20) and ignoring the sign of the second block. Option (d) is that error.
Q5NEET-style
Two particles P and Q move along a straight line; their position–time graphs are shown. At the point M,
(a) they have the same velocity
(b) P overtakes Q for the second time
(c) both are momentarily at rest
(d) they collide or are at the same position
Given
two intersecting straight x–t lines, meeting at M
Asked
the physical meaning of the intersection
Concept to use
An intersection on an x–t graph means equal position at the same time. Equal velocity would mean equal slope, i.e. parallel lines — not crossing lines.
Formula or rule
same (t, x) point ⇒ same place at the same instant
Baby steps
At M both graphs give the same x for the same t.
So the two particles are at the same location at that instant.
Their slopes are clearly different (one positive, one negative), so their velocities differ.
Straight lines mean constant velocities, so neither is at rest.
Answer
They are at the same position at that instant
Shortcut
Crossing = same place. Parallel = same velocity. Two rules, and they never overlap.
Trap
Reading an intersection as ‘same speed’. That is the single most common graph-reading error in NEET, and it also appears in relative-velocity questions.
Q6NEET-style
The x–t graph of a particle is a parabola opening upward (curving upward) with a horizontal tangent at t = 0. This tells us that the particle has
(a) constant positive acceleration
(b) zero acceleration
(c) constant negative acceleration
(d) increasing acceleration
Given
x–t curve bends upward; slope zero at t = 0
Asked
the nature of the acceleration
Concept to use
Curvature of the x–t graph carries the sign of the acceleration. NCERT Fig. 2.2: curving upward = positive a, curving downward = negative a, straight line = zero a.
Formula or rule
a = d²x/dt² = curvature
Baby steps
A horizontal tangent at t = 0 means the particle starts from rest.
The slope then grows steadily, so v increases uniformly.
A uniformly increasing v means constant positive a.
A parabola is exactly what x = ½at² plots as.
Answer
constant positive acceleration
Shortcut
Under constant acceleration, x–t is a parabola and v–t is a straight line. NCERT states this pair in Summary point 4. Reading which one is which decides many questions.
Trap
Calling the parabola ‘increasing acceleration’ because the curve gets steeper. The steepening is the velocity increasing; the acceleration is the rate of steepening, which is constant.
Q7NEET-style
For an object in uniform motion (constant velocity), the correct pair of graph shapes is
(a) x–t a straight line through the origin; v–t a straight line inclined to the t-axis
(b) x–t a parabola; v–t a straight line inclined to the t-axis
(c) x–t a straight line parallel to the t-axis; v–t a parabola
(d) x–t a straight line inclined to the t-axis; v–t a straight line parallel to the t-axis
Given
uniform motion
Asked
shapes of the x–t and v–t graphs
Concept to use
Uniform motion means constant v, hence zero a. Constant slope on x–t; constant height on v–t.
Formula or rule
NCERT Summary 4: for uniform motion a = 0, x–t is a straight line inclined to the time axis, v–t is a straight line parallel to the time axis
Baby steps
Constant v ⇒ constant slope of x–t ⇒ a straight, inclined line.
Constant v also means the v–t graph is a horizontal line at that value.
‘Parallel to the time axis’ is the phrase NCERT uses. Option (d).
Answer
x–t inclined straight line; v–t parallel to the t-axis
Shortcut
Build a two-row table in your memory: uniform motion → line / horizontal line; uniform acceleration → parabola / inclined line.
Trap
Requiring the x–t line to pass through the origin. It need not — the particle may start at any x₀. Only the slope matters for velocity.
Q8NEET-style
Which of the following graphs cannot possibly represent one-dimensional motion of a particle? (i) an x–t curve that loops back on itself; (ii) a v–t graph in the shape of a circle; (iii) a speed–time graph dipping below the time axis; (iv) a total-path-length–time graph that decreases in places.
(a) only (i) and (ii)
(b) only (iii) and (iv)
(c) only (i), (ii) and (iii)
(d) all four
Given
four candidate graphs (NCERT Exercise 2.12)
Asked
which are physically impossible
Concept to use
In one dimension a particle has exactly one position and one velocity at each instant, speed is a magnitude, and path length can only accumulate.
Formula or rule
single-valued functions of time; |v| ≥ 0; path length is non-decreasing
Baby steps
(i) A loop gives two positions at the same t — impossible.
(ii) A circle gives two velocities at the same t — impossible.
(iii) Speed is a magnitude, so it can never be negative — impossible.
(iv) Total path length is a running total of distance covered; it can stay flat but never decrease — impossible. So all four.
Answer
all four
Shortcut
Three quick tests on any graph: does a vertical line cut it twice? does a magnitude go negative? does a running total shrink? Any ‘yes’ kills the graph.
Trap
Accepting (iv) because a particle can turn around. Turning back reduces the displacement, not the path length. Path length counts every metre walked, in both directions.
Q9NEET-style
A ball is dropped from a height onto a hard floor and bounces repeatedly, losing a fixed fraction of its speed at each collision. Its speed–time graph is
(a) a set of straight-line segments with equal peaks
(b) a smooth curve decreasing to zero
(c) a sawtooth of straight-line segments whose peaks decrease after each bounce
(d) a horizontal line
Given
ball dropped, bounces, loses a fixed fraction of speed at each impact (NCERT Exercise 2.8 setup)
Asked
shape of the speed–time graph
Concept to use
Between bounces the only acceleration is g, so speed changes linearly with time. At each impact the speed drops abruptly to a fraction of its value.
Formula or rule
speed = |v|, and between impacts |dv/dt| = g
Baby steps
Free fall from rest: speed rises linearly from 0 to the impact speed.
At impact the speed jumps down to (fraction) × impact speed.
Rising after the bounce, speed falls linearly to zero at the top, then rises again.
Each successive peak is smaller, giving a decaying sawtooth of straight segments.
Answer
a sawtooth of straight segments with decreasing peaks
Shortcut
‘Loses one tenth of its speed’ means each peak is 0.9 of the previous one — a geometric sequence of peaks joined by straight lines of slope g.
Trap
Drawing curves instead of straight lines. Under constant g the speed–time graph between bounces is straight; only the height–time graph is curved.
Q10NEET-style
On a position–time graph, the instantaneous velocity of a particle at a given instant is
(a) the value of x at that instant divided by t
(b) the slope of the tangent drawn to the graph at that instant
(c) the slope of the chord joining the end points of the graph
(d) the area under the graph up to that instant
Given
an arbitrary x–t graph
Asked
how to read instantaneous velocity off it
Concept to use
Instantaneous velocity is the limit of the average velocity as the interval shrinks to zero — and geometrically, a chord becomes a tangent in that limit.
Formula or rule
v = dx/dt = slope of the tangent
Baby steps
Average velocity over an interval = slope of the chord.
As Δt → 0 the chord rotates into the tangent at that point.
So instantaneous velocity = slope of the tangent. NCERT Summary point 2 says exactly this.
Answer
the slope of the tangent drawn at that instant
Shortcut
Chord → average. Tangent → instantaneous. Area belongs to v–t and a–t graphs, never to x–t.
Trap
Choosing ‘area under the graph’ by reflex. The area under an x–t graph has dimensions of metre-second and no physical meaning here.
Q11NEET-style
An x–t graph has a sharp kink (a corner) at one point. NCERT notes that in any realistic situation such a kink cannot occur, because
(a) velocity and acceleration cannot change abruptly at an instant; changes are always continuous
(b) the graph must always be a straight line
(c) position cannot be negative
(d) the particle would have to be at two places at once
Given
an x–t graph with a sharp corner
Asked
why real graphs are smooth
Concept to use
A kink means the slope jumps from one value to another with no time in between — an instantaneous change of velocity, which would require infinite acceleration.
Formula or rule
NCERT: ‘acceleration and velocity cannot change values abruptly at an instant. Changes are always continuous’
Baby steps
At a kink the left-hand and right-hand slopes differ, so the function is not differentiable there.
A discontinuous jump in velocity implies an infinite acceleration for zero time.
Real interactions take a finite time, so real graphs are smooth.
The kinks in textbook figures are idealisations for simplicity.
Answer
Velocity and acceleration cannot change abruptly; changes are always continuous
Shortcut
This exact wording appears in the box just before section 2.4 in NCERT. It is prime material for an assertion-reason item.
Trap
Assuming the kinked graph is simply wrong. It is not wrong — it is an approximation, and NCERT is careful to say the real functions would be differentiable everywhere.
Q12NEET-style
An x–t graph has a negative slope over some interval. During that interval the particle is
(a) moving in the direction of decreasing x
(b) at rest
(c) decelerating
(d) moving with negative acceleration
Given
x–t graph with negative slope
Asked
what the particle is doing
Concept to use
The sign of the slope of x–t is the sign of the velocity, which in one dimension is the direction of motion. Acceleration is a separate matter — it comes from the curvature.
Formula or rule
v = dx/dt < 0 ⇒ motion towards decreasing x
Baby steps
Negative slope means x is decreasing with time.
So the particle moves in the negative x direction.
Nothing here tells us about the acceleration — that would need the curvature.
A straight negative-slope line would in fact have zero acceleration.
Answer
moving in the direction of decreasing x
Shortcut
Slope sign → direction. Curvature sign → acceleration. Keep the two readings separate and most graph questions collapse.
Trap
Equating ‘negative velocity’ with ‘deceleration’. NCERT Points to Ponder 3: the sign of acceleration does not by itself tell you whether the particle is speeding up or slowing down.
Q13NEET-style
A particle’s v–t graph is a straight line with a positive intercept on the v-axis and a negative slope, crossing the time axis at t = t₁. The particle
(a) is at rest at t = t₁ and remains at rest afterwards
(b) reverses its direction of motion at t = t₁
(c) has zero acceleration at t = t₁
(d) reaches its minimum displacement at t = t₁
Given
v–t straight line, positive v₀, constant negative a, crossing zero at t₁
Asked
what happens at t₁
Concept to use
v changing sign means the direction of motion reverses. The acceleration stays constant and non-zero throughout — that is what a straight, sloping v–t line means.
Formula or rule
sign change of v ⇒ reversal ; slope of v–t = a
Baby steps
Before t₁: v > 0, moving forward and slowing.
At t₁: v = 0, momentarily at rest.
After t₁: v < 0, moving backward and speeding up.
The slope (and hence a) is unchanged the whole time, so it is not zero at t₁.
Answer
reverses its direction of motion at t₁
Shortcut
This is NCERT Fig. 2.3(d): motion in the positive direction until t₁, then back the other way with the same negative acceleration. Displacement is maximum at t₁, not minimum.
Trap
Concluding the particle stops for good. It is at rest for an instant only, exactly like a ball at the top of a throw.
Q14NEET-style
An x–t plot shows a particle in one-dimensional motion over three equal time intervals. In interval 1 the curve rises gently, in interval 2 it rises steeply to a peak, and in interval 3 it falls very steeply. The interval with the greatest average speed and the sign of the average velocity in interval 3 are
(a) interval 2; positive
(b) interval 1; negative
(c) interval 3; negative
(d) interval 3; positive
Given
x–t plot, three equal intervals of the shape described (NCERT Exercise 2.17)
Asked
interval of greatest average speed, and the sign of average velocity in interval 3
Concept to use
Average speed over equal time intervals is set by the steepness of the chord, regardless of sign. Average velocity takes its sign from whether x rose or fell.
Formula or rule
average speed ∝ |slope of chord| ; sign of v̄ = sign of Δx
Baby steps
The three intervals are equal in duration, so comparing chords compares average speeds directly.
Interval 3 has the steepest chord, so its average speed is the greatest.
In interval 3 the particle moves from the peak downwards, so Δx < 0.
Hence the average velocity there is negative. (Interval 2 is the least steep of 2 and 3, and NCERT’s answer for the least average speed is interval 2.)
Answer
interval 3, with a negative average velocity
Shortcut
Equal time intervals → just eyeball which chord is steepest. You never need to compute anything for this question type.
Trap
Ranking by the height reached rather than by the change in height. The peak sits in interval 2, but the greatest change happens in interval 3.
Q15NEET-style
A speed–time graph of a particle moving along a constant direction has smooth maxima at points B and D and a smooth minimum at C, with A on a flat portion. The acceleration at the points A, B, C and D is
(a) positive at all four
(b) positive at A and C, negative at B and D
(c) negative at all four
(d) zero at all four
Given
speed–time graph, A on a flat part, B and D maxima, C a minimum (NCERT Exercise 2.18)
Asked
acceleration at A, B, C, D
Concept to use
Acceleration is the slope of the speed–time graph. At a smooth maximum or minimum the tangent is horizontal, so the slope is zero.
Formula or rule
a = dv/dt = slope of the v–t curve
Baby steps
At a peak (B, D) the curve momentarily stops rising and starts falling: slope = 0.
At a trough (C) it stops falling and starts rising: slope = 0.
At A the graph is flat: slope = 0.
So the acceleration is zero at all four points.
Answer
zero at all four points
Shortcut
Turning points on any graph mean zero slope. On a v–t graph that means zero acceleration; on an x–t graph it would mean zero velocity.
Trap
Confusing the value of v with the slope of v. At B the speed is at its largest, which tempts you to say the acceleration is large too. It is exactly zero.
Q16NEET-style
Figure 2.13 in NCERT gives the x–t plot of a particle in simple harmonic motion, x = A cos(2πt/T) with T = 4 s. At t = 0.3 s, the signs of position, velocity and acceleration are respectively
(a) +, −, −
(b) +, +, +
(c) −, −, +
(d) −, +, +
Given
SHM x–t plot with period 4 s, evaluated at t = 0.3 s
Asked
signs of x, v and a
Concept to use
For SHM, a = −ω²x, so acceleration always has the opposite sign to position. Velocity comes from the slope.
Formula or rule
x = A cosωt ; v = −Aωsinωt ; a = −ω²x
Baby steps
At t = 0.3 s (well before the quarter-period at 1 s), the particle is still on the positive side: x > 0.
It is moving from the crest back towards zero, so the graph is falling: v < 0.
Since a = −ω²x and x is positive, a < 0.
Signs: +, −, −.
Answer
+, −, −
Shortcut
In SHM, x and a are always opposite in sign. Fix x from the graph, flip it for a, and read v from whether the curve is rising or falling. Two of the three answers come free.
Trap
Reading the velocity from the position rather than from the slope. Being at a positive x says nothing about which way the particle is heading.
Q17NEET-style
A velocity–time graph is a straight line that does not pass through the origin, cutting the v-axis at 6 m s⁻¹ and passing through the point (4 s, 18 m s⁻¹). The acceleration and the displacement in the first 4 s are
(a) 3 m s⁻², 72 m
(b) 4.5 m s⁻², 48 m
(c) 3 m s⁻², 48 m
(d) 3 m s⁻², 24 m
Given
v–t line: intercept 6 m s⁻¹, passing through (4 s, 18 m s⁻¹)
Asked
acceleration and displacement over 0–4 s
Concept to use
Slope gives acceleration; area gives displacement. The area is a trapezium because the line does not start at zero.
Formula or rule
a = Δv/Δt ; x = ½(v₀ + v)t
Baby steps
a = (18 − 6)/4 = 3 m s⁻².
Area = trapezium = ½(6 + 18)(4).
= ½(24)(4) = 48 m.
Cross-check: x = 6(4) + ½(3)(16) = 24 + 24 = 48 m. ✓
Answer
3 m s⁻² and 48 m
Shortcut
The trapezium area ½(u+v)tis the equation x = v̄t in geometric form. Reading the graph and using the formula are the same act.
Trap
Treating the whole area as a triangle (giving 36 m) by forgetting the 6 m s⁻¹ head start. Always check whether the line starts at the origin.
Q18NEET-style
For an object in free fall from rest, the correct set of graph shapes for acceleration, velocity and distance against time is
(a) all three straight lines
(b) a horizontal line, a straight inclined line, and a parabola
(c) a parabola, a straight inclined line, and a horizontal line
(d) a horizontal line, a parabola, and a cubic
Given
free fall from rest (NCERT Fig. 2.7)
Asked
shapes of the a–t, v–t and y–t graphs
Concept to use
Each graph is the accumulated area of the one before it, so each is one degree higher in t: constant, then linear, then quadratic.
Formula or rule
a = −g ; v = −gt ; y = −½gt²
Baby steps
a is constant at −9.8 m s⁻²: a horizontal line.
v = −9.8t: a straight line through the origin, sloping down.
y = −4.9t²: a parabola.
So: horizontal, inclined straight line, parabola.
Answer
a horizontal line, a straight inclined line, and a parabola
Shortcut
Degree ladder: 0, 1, 2. Whenever acceleration is constant, the three graphs are always constant → linear → parabolic, in that order.
Trap
Putting the parabola on the velocity graph. The parabola always belongs to the position graph under constant acceleration.
Q19NEET-style
On a velocity–time graph, the area between the curve and the time axis from t₁ to t₂ is equal to
(a) the distance travelled between t₁ and t₂, always
(b) the average velocity over the interval
(c) the displacement between t₁ and t₂
(d) the change in acceleration over the interval
Given
any v–t curve
Asked
physical meaning of the area under it
Concept to use
NCERT Summary point 5 states it exactly: the area under the velocity–time curve between two times equals the displacement during that interval.
Formula or rule
Δx = ∫ v dt = signed area
Baby steps
Split the interval into thin strips of width dt; each contributes v dt, a tiny displacement.
Summing them gives the total displacement.
Area below the axis is negative displacement, so the sum is signed.
It equals the distance only when v never changes sign.
Answer
the displacement between t₁ and t₂
Shortcut
NCERT shows why with the simplest case: constant velocity u for time T gives a rectangle of area uT — which is exactly the displacement. Check the axes’ dimensions (m s⁻¹ × s = m) and the ‘area equals a distance’ puzzle dissolves.
Trap
Choosing ‘distance, always’. The word always is the giveaway — absolute-word options in graph questions are usually the planted trap.
Q20NEET-style
A particle moves so that its x–t graph is a straight line for t < 0 and a parabola for t > 0, with the two joining smoothly at the origin. A suitable physical context is
(a) a car moving on a circular track
(b) a body moving at constant velocity that is given a constant force from t = 0 onwards, such as a ball rolling on a smooth floor that then rolls down an incline
(c) a ball thrown up, which is impossible to represent this way
(d) a particle at rest throughout
Given
x–t straight for t < 0, parabolic for t > 0 (NCERT Exercise 2.13)
Asked
a suitable physical situation
Concept to use
Straight x–t = constant velocity, zero acceleration. Parabolic x–t = constant non-zero acceleration. So something starts accelerating the body at t = 0.
Formula or rule
curvature of x–t ⇒ acceleration
Baby steps
For t < 0: constant slope ⇒ uniform velocity, no net force.
For t > 0: parabola ⇒ constant acceleration, so a constant force begins acting.
A body sliding at constant speed on a smooth horizontal floor that then reaches an incline fits exactly.
Note the graph does not mean the path is straight then curved — the motion is one-dimensional throughout.
Answer
a body in uniform motion that starts experiencing a constant force at t = 0
Shortcut
NCERT’s own warning on this exercise: an x–t graph is not a picture of the path. A parabolic x–t graph describes straight-line motion with uniform acceleration.
Trap
Answering that the particle ‘moves in a straight line then on a parabolic path’. Confusing the graph with the trajectory is exactly what Exercise 2.13 is designed to catch.
Answer key
bQ1
dQ2
cQ3
aQ4
dQ5
aQ6
dQ7
dQ8
cQ9
bQ10
aQ11
aQ12
bQ13
cQ14
dQ15
aQ16
cQ17
bQ18
cQ19
bQ20
NEET 2027 preparation — built for Aamirah Fathima. Source: NCERT Physics Class XI,
Chapter 2 (Reprint 2026–27). Single self-contained file — print directly from any browser.