Graphs — NEET Physics MCQs with Solutions
Free NEET Physics Graphs MCQs with step-by-step solutions (36 questions). Part of Motion in 1 Dimension. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Graphs · easy · theory
The slope of a position–time ($x$–$t$) graph gives
A. Force
B. Velocity ✓ Correct
C. Distance
D. Acceleration
Solution: Slope $= \frac{dx}{dt} = v$.
Q2 — Graphs · easy · theory
The slope of a velocity–time ($v$–$t$) graph gives
A. Velocity
B. Acceleration ✓ Correct
C. Position
D. Displacement
Solution: Slope $= \frac{dv}{dt} = a$.
Q3 — Graphs · easy · theory
The area under a velocity–time graph (with sign) gives
A. Acceleration
B. Force
C. Displacement ✓ Correct
D. Average speed
Solution: $\int v\,dt = $ displacement; the total unsigned area gives distance.
Q4 — Graphs · medium · theory
The area under an acceleration–time graph gives
A. Position
B. Distance
C. Change in velocity ✓ Correct
D. Displacement
Solution: $\int a\,dt = \Delta v$.
Q5 — Graphs · easy · theory
A straight-line position–time graph (with non-zero slope) indicates
A. Increasing acceleration
B. Uniform acceleration
C. Rest
D. Uniform velocity ✓ Correct
Solution: Constant slope = constant velocity.
Q6 — Graphs · easy · theory
A position–time graph parallel to the time axis means the body is
A. Falling freely
B. At rest ✓ Correct
C. Moving uniformly
D. Accelerating
Solution: Position never changes — zero slope, zero velocity.
Q7 — Graphs · medium · theory
A parabolic position–time graph ($x \propto t^2$) indicates
A. Uniform velocity
B. Rest
C. Decreasing speed always
D. Uniform acceleration ✓ Correct
Solution: $x = \frac{1}{2}at^2$ — the signature of constant acceleration from rest.
Q8 — Graphs · medium · theory
A horizontal line on a velocity–time graph means
A. Zero acceleration (uniform velocity) ✓ Correct
B. Infinite acceleration
C. Uniform acceleration
D. The body is at rest
Solution: Velocity unchanging in time — its slope (acceleration) is zero.
Q9 — Graphs · easy · theory
A straight line through the origin on a velocity–time graph represents
A. Uniform velocity
B. Uniform acceleration starting from rest ✓ Correct
C. Rest
D. Non-uniform acceleration
Solution: $v = at$: starts at $v=0$ and grows linearly — constant $a$.
Q10 — Graphs · medium · theory
A position–time graph can never be exactly perpendicular to the time axis because that would mean
A. Zero velocity
B. Infinite velocity ✓ Correct
C. Negative time
D. Zero position
Solution: A vertical jump in $x$ at one instant implies change of position in zero time — infinite speed.
Q11 — Graphs · medium · theory
The position–time graphs of two cars intersect at time $t_0$. At $t_0$ the cars have the same
A. Velocity
B. Position (they meet) ✓ Correct
C. Acceleration
D. Speed
Solution: Same $x$ at the same $t$ = same place at that moment; their slopes (velocities) may differ.
Q12 — Graphs · easy · theory
A negative slope on a position–time graph indicates
A. Negative acceleration
B. Deceleration
C. Rest
D. Motion in the negative direction ✓ Correct
Solution: Slope is velocity; negative slope = negative velocity.
Q13 — Graphs · easy · theory
The part of a velocity–time graph lying BELOW the time axis represents
A. Impossible motion
B. Rest
C. Motion in the opposite (negative) direction ✓ Correct
D. Deceleration only
Solution: $v < 0$ — the body moves the other way; that area counts as negative displacement.
Q14 — Graphs · easy · theory
The velocity–time graph of a body falling freely from rest is
A. A horizontal line
B. An exponential curve
C. A parabola
D. A straight line through the origin ✓ Correct
Solution: $v = gt$ — linear with slope $g \approx 10$ m/s².
Q15 — Graphs · medium · theory
For a ball thrown vertically upward, the velocity–time graph (taking up as positive) is
A. A horizontal line
B. A straight line of negative slope crossing the time axis at the top ✓ Correct
C. A parabola
D. A V-shaped line
Solution: $v = u - gt$: constant slope $-g$; $v$ passes through zero at the highest point and turns negative.
Q16 — Graphs · medium · theory
For a ball thrown up and returning, the SPEED–time graph looks like
A. A horizontal line
B. A circle
C. A V shape (decreasing to zero, then increasing) ✓ Correct
D. A single straight line
Solution: Speed $= |v|$: it falls to zero at the top and rises again — the negative part flips up, making a V.
Q17 — Graphs · easy · theory
The total DISTANCE from a velocity–time graph equals
A. The signed area
B. The maximum velocity
C. The total area, counting parts below the axis as positive ✓ Correct
D. The slope
Solution: Distance accumulates regardless of direction, so all areas are added as positive.
Q18 — Graphs · medium · theory
If the slope of a position–time graph is increasing with time, the body is
A. Accelerating ✓ Correct
B. Moving uniformly
C. At rest
D. Decelerating
Solution: Growing slope = growing velocity = acceleration.
Q19 — Graphs · easy · theory
Two position–time graphs pass through the origin; graph A is steeper than graph B. Then
A. A represents the faster body ✓ Correct
B. B is faster
C. Both have equal speed
D. A is accelerating
Solution: Steeper slope = larger velocity.
Q20 — Graphs · easy · theory
The velocity of a particle varies as $v = 2t$ (m/s, seconds). The displacement from $t=0$ to $t=3$ s is
A. $3$ m
B. $6$ m
C. $18$ m
D. $9$ m ✓ Correct
Solution: Area of the triangle under $v$–$t$: $\frac{1}{2}\times3\times6 = 9$ m.
Q21 — Graphs · easy · theory
A velocity–time graph is a horizontal line at $10$ m/s from $t=0$ to $t=5$ s. The displacement is
A. $10$ m
B. $2$ m
C. $15$ m
D. $50$ m ✓ Correct
Solution: Rectangle area: $10\times5 = 50$ m.
Q22 — Graphs · easy · theory
A velocity–time graph rises uniformly from $0$ to $20$ m/s over $10$ s. The displacement is
A. $200$ m
B. $100$ m ✓ Correct
C. $50$ m
D. $20$ m
Solution: Triangle area: $\frac{1}{2}\times10\times20 = 100$ m.
Q23 — Graphs · easy · theory
Uniform retardation appears on a velocity–time graph as
A. A straight line with negative slope ✓ Correct
B. A vertical line
C. A parabola opening up
D. A horizontal line
Solution: Constant negative $a$ = constant negative slope of $v$–$t$.
Q24 — Graphs · medium · theory
The position–time graphs of two bodies are parallel straight lines. The bodies have
A. Equal accelerations but meet once
B. Different velocities
C. Equal velocities and never meet ✓ Correct
D. Equal positions
Solution: Parallel lines = same slope (velocity) but constant separation — they never meet.
Q25 — Graphs · medium · theory
A horizontal line on an acceleration–time graph represents
A. Uniform acceleration ✓ Correct
B. Jerk
C. Rest
D. Uniform velocity
Solution: Acceleration constant in time — uniformly accelerated motion.
Q26 — Graphs · easy · theory
A graph showing TWO different positions for the same instant cannot be a real position–time graph because
A. Velocity would be zero
B. Position cannot be negative
C. A particle cannot be in two places at once ✓ Correct
D. Time cannot repeat
Solution: $x(t)$ must be single-valued: one position per instant.
Q27 — Graphs · medium · theory
On a position–time graph, the average velocity between two instants equals
A. The maximum slope
B. The slope of the straight line (chord) joining the two points ✓ Correct
C. The slope of the tangent at the first point
D. The area under the curve
Solution: $\bar v = \frac{\Delta x}{\Delta t}$ — geometrically the chord’s slope; the tangent gives instantaneous velocity.
Q28 — Graphs · easy · theory
A velocity–time graph shows $+20$ m/s for $2$ s, then $-10$ m/s for $2$ s. The net displacement is
A. Zero
B. $40$ m
C. $60$ m
D. $20$ m ✓ Correct
Solution: Signed areas: $+40$ then $-20$: net $= 20$ m. (Distance would be 60 m.)
Q29 — Graphs · easy · theory
For a body at rest, which pair of graphs is correct?
A. $x$–$t$ horizontal line and $v$–$t$ along the time axis ✓ Correct
B. $x$–$t$ rising line and $v$–$t$ horizontal
C. $x$–$t$ on the time axis necessarily
D. Both graphs rising
Solution: Rest: $x$ constant (flat line at its position) and $v = 0$ (line ON the time axis).
Q30 — Graphs · easy · theory
A velocity–time graph rises uniformly from $10$ m/s to $20$ m/s in $4$ s. The displacement in this interval is
A. $40$ m
B. $60$ m ✓ Correct
C. $30$ m
D. $80$ m
Solution: Trapezium area $= \frac{(10+20)}{2}\times4 = 60$ m.