Graphs — JEE Main Physics MCQs with Solutions
Free JEE Main 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. Velocity ✓ Correct
B. Force
C. Acceleration
D. Distance
Solution: Slope $= \frac{dx}{dt} = v$.
Q2 — Graphs · easy · theory
The slope of a velocity–time ($v$–$t$) graph gives
A. Acceleration ✓ Correct
B. Position
C. Velocity
D. Displacement
Solution: Slope $= \frac{dv}{dt} = a$.
Q3 — Graphs · easy · theory
The area under a velocity–time graph (with sign) gives
A. Average speed
B. Displacement ✓ Correct
C. Force
D. Acceleration
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. Uniform velocity ✓ Correct
B. Uniform acceleration
C. Rest
D. Increasing acceleration
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. Moving uniformly
C. At rest ✓ Correct
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. Rest
B. Decreasing speed always
C. Uniform acceleration ✓ Correct
D. Uniform velocity
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. Infinite acceleration
B. Zero acceleration (uniform velocity) ✓ Correct
C. The body is at rest
D. Uniform acceleration
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. Rest
B. Uniform acceleration starting from rest ✓ Correct
C. Uniform velocity
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 position
B. Negative time
C. Infinite velocity ✓ Correct
D. Zero velocity
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. Acceleration
B. Speed
C. Velocity
D. Position (they meet) ✓ Correct
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. Motion in the negative direction ✓ Correct
B. Rest
C. Deceleration
D. Negative acceleration
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. Rest
B. Deceleration only
C. Impossible motion
D. Motion in the opposite (negative) direction ✓ Correct
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 parabola
B. An exponential curve
C. A horizontal line
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 parabola
B. A V-shaped line
C. A horizontal line
D. A straight line of negative slope crossing the time axis at the top ✓ Correct
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 V shape (decreasing to zero, then increasing) ✓ Correct
B. A single straight line
C. A horizontal line
D. A circle
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 total area, counting parts below the axis as positive ✓ Correct
B. The signed area
C. The maximum velocity
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. At rest
B. Decelerating
C. Moving uniformly
D. Accelerating ✓ Correct
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. Both have equal speed
B. B is faster
C. A represents the faster body ✓ Correct
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. $9$ m ✓ Correct
C. $18$ m
D. $6$ m
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. $50$ m ✓ Correct
C. $2$ m
D. $15$ m
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. $20$ m
C. $100$ m ✓ Correct
D. $50$ 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 vertical line
B. A straight line with negative slope ✓ Correct
C. A horizontal line
D. A parabola opening up
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 positions
B. Different velocities
C. Equal velocities and never meet ✓ Correct
D. Equal accelerations but meet once
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. Uniform velocity
D. Rest
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. Time cannot repeat
C. A particle cannot be in two places at once ✓ Correct
D. Position cannot be negative
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 area under the curve
B. The slope of the tangent at the first point
C. The maximum slope
D. The slope of the straight line (chord) joining the two points ✓ Correct
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. $20$ m ✓ Correct
B. $60$ m
C. Zero
D. $40$ m
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. Both graphs rising
B. $x$–$t$ horizontal line and $v$–$t$ along the time axis ✓ Correct
C. $x$–$t$ on the time axis necessarily
D. $x$–$t$ rising line and $v$–$t$ horizontal
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. $30$ m
B. $60$ m ✓ Correct
C. $40$ m
D. $80$ m
Solution: Trapezium area $= \frac{(10+20)}{2}\times4 = 60$ m.