For motion in a straight line, the area between the velocity–time curve and the time axis, between two instants, equals the displacement of the object over that interval. This geometric interpretation is the basis for graphical derivations of the kinematic equations.
-- NCERT Class 11 Physics, Ch. 2, p. 17Position Time Graph
8 MCQs2 revision cards9-step worked example
Source: NCERT KinematicsOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Position Time Graph, explained for NEET
Position-time graphs encode an object's entire motion history in a single picture. The one trap that costs marks on this topic: confusing trig functions when extracting velocity from the graph's angle.
A position-time (x-t) graph plots position on the vertical axis and time on the horizontal axis (NCERT Class 11 Physics Chapter 2, page 17). Each point on the curve tells you where the object is at that instant. The slope of the graph at any point gives the instantaneous velocity.
For a straight-line x-t graph, the slope is constant — the object moves with uniform velocity. The velocity equals the tangent of the angle the line makes with the time axis:
v = slope = tan θ (where θ is measured from the t-axis)
Here is the high-frequency trap: when a question gives you the angle, you must use tan, not sin or cos. Under exam pressure, students reach for sin θ or cos θ out of reflex. The slope of any graph is rise/run = Δx/Δt, and the trigonometric ratio that equals opposite/adjacent for an angle in a right triangle is the tangent.
A horizontal line on an x-t graph means zero slope — the object is at rest. A steeper line means higher velocity. If two lines have different angles (say 30° and 60°), the velocity ratio is tan 30° : tan 60° = (1/√3) : √3 = 1 : 3.
For a curved x-t graph, the slope changes with time. The tangent drawn at any point gives the instantaneous velocity at that moment. A parabolic x-t curve (x = x₀ + v₀t + ½at²) indicates uniformly accelerated motion — the slope increases (or decreases) steadily.
Watch out: the angle the tangent line makes with the t-axis gives velocity via tan, not via sin or cos. This holds for curved graphs too — draw the tangent, measure the angle from the t-axis, and take the tangent of that angle.
Can you answer these Position Time Graph MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The position-time graph of a particle moving along a straight line is a straight line making an angle of 45° with the time axis. What is the velocity of the particle?
Show answer and why every option is right or wrong
Answer: A. The velocity is the slope of the x-t graph, which equals tan θ where θ is the angle with the time axis. tan 45° = 1 m/s (NCERT Class 11 Physics Chapter 2, page 17).
Why B is wrong: B is wrong because cos 45° = 1/√2 is the ratio of the run along the time axis to the length of the line, not rise over run; the slope of a graph is tan θ (trap: trig-function confusion on graph slopes).
Why C is wrong: C is wrong because sin 45° = 1/√2; the slope of a graph is rise/run = tan θ, not sin θ (trap: trig-function confusion on graph slopes).
Why D is wrong: D is wrong because sec 45° = √2 is the ratio of the length of the line to its run along the time axis, not rise over run; velocity is the slope, tan θ (trap: trig-function confusion on graph slopes).
MCQ 2Direct ApplicationPractice
Two particles A and B have straight-line position-time graphs making angles of 30° and 60° respectively with the time axis. What is the ratio of velocity of A to velocity of B?
Show answer and why every option is right or wrong
Answer: D. v_A/v_B = tan 30°/tan 60° = (1/√3)/(√3) = 1/3, so the ratio is 1 : 3 (NCERT Class 11 Physics Chapter 2, page 17).
Why A is wrong: A is wrong because 3 : 1 is the inverse of the correct ratio. This would result from computing v_B/v_A instead of v_A/v_B (trap: ratio inversion).
Why B is wrong: B is wrong because 1 : √3 would be the ratio sin 30° : sin 60° = 0.5 : (√3/2). Using sin instead of tan to extract velocity from the slope is incorrect (trap: using sin instead of tan for graph slope).
Why C is wrong: C is wrong because √3 : 1 is the ratio cos 30° : cos 60° = (√3/2) : (1/2). Using cos instead of tan to read velocity from the slope is incorrect; 60° is the steeper line, so B is the faster particle (trap: trig-function confusion on graph slopes).
MCQ 3Easy RecallPractice
The position-time graph of an object is a horizontal straight line at x = 5 m. Which statement is correct?
Show answer and why every option is right or wrong
Answer: B. A horizontal line on the x-t graph has zero slope. Slope = velocity, so velocity = 0. The object is stationary at x = 5 m (NCERT Class 11 Physics Chapter 2, page 17).
Why A is wrong: A is wrong because a horizontal line has zero slope (Δx/Δt = 0). The value x = 5 m is the position, not the velocity (trap: confusing the y-intercept value with the slope).
Why C is wrong: C is wrong because neither velocity nor acceleration can be read as the y-value of the graph. The graph shows position, not velocity or acceleration. Since the line is flat, velocity is zero and acceleration is zero (trap: confusing position value with kinematic quantities).
Why D is wrong: D is wrong because the slope of a horizontal line is constant (zero), so velocity is constant at zero — it is not decreasing (trap: confusing position value with velocity behaviour).
MCQ 4Direct ApplicationPractice
A particle's position-time graph is a straight line passing through the origin with a positive slope. If the angle this line makes with the time axis is θ, which expression gives the position of the particle at time t?
Show answer and why every option is right or wrong
Answer: C. The line passes through the origin, so x = vt with zero intercept. The velocity v = slope = tan θ. Therefore x = t tan θ (NCERT Class 11 Physics Chapter 2, page 17).
Why A is wrong: A is wrong because the slope of the x-t graph is tan θ, not sin θ. Using sin gives x = t sin θ, which underestimates position for angles less than 90° (trap: trig-function confusion on graph slopes).
Why B is wrong: B is wrong because cos θ is the adjacent/hypotenuse ratio, not the slope. The slope requires the opposite/adjacent ratio, which is tan θ (trap: trig-function confusion on graph slopes).
Why D is wrong: D is wrong because t/tan θ = t cot θ, which is the reciprocal of the correct expression. This would give position as though the axes were swapped (trap: inverting the slope ratio).
MCQ 5Easy RecallPractice
The position-time graph of a moving car is a parabola opening upward. This indicates that the car is:
Show answer and why every option is right or wrong
Answer: A. An upward-opening parabola has the form x = x₀ + v₀t + ½at² (a > 0). The slope increases with time, meaning velocity increases uniformly — the car has constant positive acceleration (NCERT Class 11 Physics Chapter 2, page 17).
Why B is wrong: B is wrong because uniform velocity produces a straight line on the x-t graph (constant slope), not a parabola (trap: confusing straight-line graphs with curved graphs).
Why C is wrong: C is wrong because an upward-opening parabola has an increasing slope (steeper tangent lines at later times), meaning the speed is increasing, not decreasing (trap: misreading curvature direction).
Why D is wrong: D is wrong because a stationary object produces a horizontal line on the x-t graph, not a curve. The parabolic shape shows the position is changing with time (trap: confusing a flat line with a parabola).
MCQ 6CalculationPractice
A particle's x-t graph consists of two straight-line segments. From t = 0 s to t = 4 s, the position rises linearly from x = 0 m to x = 8 m. From t = 4 s to t = 10 s, the position rises linearly from x = 8 m to x = 26 m. What is the particle's average velocity over the interval t = 1 s to t = 7 s?
Show answer and why every option is right or wrong
Answer: B. In the first segment the slope is (8 − 0)/(4 − 0) = 2 m/s, so the position at t = 1 s is x(1) = 2 × 1 = 2 m. In the second segment the slope is (26 − 8)/(10 − 4) = 3 m/s, so the position at t = 7 s is x(7) = 8 + 3 × (7 − 4) = 17 m. The displacement over the interval is 17 − 2 = 15 m in a time of 7 − 1 = 6 s, giving average velocity = 15/6 = 2.5 m/s (NCERT Class 11 Physics Chapter 2, page 17).
Why A is wrong: A is wrong because 2 m/s is only the slope of the first segment (8/4); it ignores that t = 7 s lies in the second segment, which has a different (steeper) slope (trap: applying one segment's slope to the whole interval).
Why C is wrong: C is wrong because 3 m/s is only the slope of the second segment (18/6); it ignores that t = 1 s lies in the first segment, which has a shallower slope, so it cannot be the average velocity over the full interval (trap: applying one segment's slope to the whole interval).
Why D is wrong: D is wrong because 17/7 divides the position at t = 7 s by the elapsed time measured from t = 0, i.e. treating it as if the interval started at the origin, instead of dividing the actual displacement (x(7) − x(1)) by the actual interval duration (7 − 1) (trap: confusing position-over-total-time with displacement-over-interval).
MCQ 7Direct ApplicationPractice
The position-time graph of an object consists of two connected straight-line segments: the first segment rises steeply from t = 0 to t = 5 s, and the second segment is horizontal from t = 5 s to t = 10 s. During the interval t = 5 s to t = 10 s, the object:
Show answer and why every option is right or wrong
Answer: D. A horizontal segment on the x-t graph has zero slope, which means velocity = 0. The object is stationary during that interval (NCERT Class 11 Physics Chapter 2, page 17).
Why A is wrong: A is wrong because acceleration would appear as a curved (parabolic) segment on the x-t graph, not a horizontal line. The horizontal segment indicates zero velocity, not changing velocity (trap: confusing graph shapes with kinematic states).
Why B is wrong: B is wrong because deceleration implies the object is still moving but slowing down, which would appear as a curve with decreasing slope — not a flat horizontal line (trap: confusing a flat line with deceleration).
Why C is wrong: C is wrong because negative velocity would produce a line with a negative (downward) slope. A horizontal line has zero slope, not negative slope (trap: confusing zero slope with negative slope).
MCQ 8Concept TrapPractice
The position-time graphs of two objects A and B are straight lines that intersect at a point (t₁, x₁). Which statement is necessarily true at t = t₁?
Show answer and why every option is right or wrong
Answer: B. The intersection of two x-t graphs means both objects have the same position coordinate at that instant. Their slopes (velocities) can be different — the lines can cross at any angle. Starting points (at t = 0) are determined by the y-intercepts, which are generally different (NCERT Class 11 Physics Chapter 2, page 14).
Why A is wrong: A is wrong because equal velocity would require equal slopes at the intersection point. Two straight lines can intersect at any angle, so their slopes are generally different (trap: confusing same position with same velocity).
Why C is wrong: C is wrong because both lines are straight, meaning each has constant velocity and zero acceleration. While both accelerations happen to be zero here, that is a special case of straight-line graphs, not a consequence of the intersection (trap: over-reading the intersection point).
Why D is wrong: D is wrong because the starting point of each object is its position at t = 0, given by the y-intercept of its line. Two lines intersecting at (t₁, x₁) generally have different y-intercepts unless t₁ = 0 (trap: confusing intersection point with initial position).
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Position Time Graph: quick recall before you leave
How do you solve a Position Time Graph question? A worked example
Pattern: Position-time graph slope — extract velocity ratio from given angles (anchored to PYQ pattern NEET pattern: position time graph slope, observed 2022).
- 1
Given
The position-time graphs of two particles X and Y are straight lines making angles 30° and 45° with the time axis respectively.
- 2
Required
Find the ratio of the velocity of X to the velocity of Y (v_X : v_Y).
- 3
Concept
The velocity of an object is the slope of its position-time graph. For a straight-line x-t graph making angle θ with the time axis, the slope is tan θ.
- 4
Formula
v = tan θ
- 5
Substitution
v_X = tan 30° = 1/√3
v_Y = tan 45° = 1 - 6
Calculation
v_X / v_Y = (1/√3) / 1 = 1/√3
Note on exact values: 30° and 45° are exact angle specifications from the problem. The trigonometric values tan 30° = 1/√3 and tan 45° = 1 are exact mathematical constants. They do not limit significant figures. - 7
Final answer
v_X : v_Y = 1 : √3
- 8
Common trap
Using sin or cos instead of tan. If a student used sin: sin 30°/sin 45° = 0.5/0.707 ≈ 0.707, giving the wrong ratio 1 : √2. If they used cos: cos 30°/cos 45° = (√3/2)/(1/√2) = √6/2, also wrong. Only tan gives the slope.
- 9
Similar NEET-style question
The x-t graphs of cars A and B are straight lines through the origin, making angles 60° and 30° with the t-axis. Find the ratio of speeds v_A : v_B.
Answer: tan 60° / tan 30° = √3 / (1/√3) = 3. So v_A : v_B = 3 : 1.
---
What to remember before solving Position Time Graph questions
The slope of the position–time (x-t) graph at any instant gives the instantaneous velocity. A straight x-t line implies uniform velocity; a curved x-t line implies non-zero acceleration.
-- NCERT Class 11 Physics, Ch. 2, p. 14Which Position Time Graph formulas do you need for NEET?
2 formulas — click to collapse
First kinematic equation (uniform acceleration)
Final velocity equals initial velocity plus acceleration times the time elapsed, for motion under constant acceleration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v | Final velocity | m/s |
| v0 | Initial velocity | m/s |
| a | Constant (uniform) acceleration | m/s^2 |
| t | Elapsed time | s |
Valid when
- Acceleration a is CONSTANT (uniform) in both magnitude and direction
- All quantities measured in the same inertial reference frame
- Motion is along a straight line; signs encode direction along chosen axis
Do NOT use when
- Acceleration changes in magnitude or direction (use a(t) integration)
- Motion is uniformly circular at constant speed (a is centripetal, not tangential)
Second kinematic equation (displacement under uniform acceleration)
Displacement equals initial-velocity-times-time plus half of acceleration-times-time-squared. The (1/2) factor is the area of the triangle on the v-t graph.
| Symbol | Quantity | SI Unit |
|---|---|---|
| x | Final position | m |
| x0 | Initial position | m |
| v0 | Initial velocity | m/s |
| a | Constant acceleration | m/s^2 |
| t | Time elapsed | s |
Valid when
- Acceleration constant (magnitude and direction)
- Sign convention consistent across x, v, a (one chosen positive direction)
Where do students lose marks on Position Time Graph?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
1 item — click to collapse
Category: Graph Interpretation
Student uses sin or cos of the angle the line makes with the time axis, instead of tan, to extract velocity.
When it triggers
Question gives an angle the x-t line makes with the t-axis (often 30°, 45°, 60°) and asks for velocity or its ratio.
How to avoid
Velocity = dx/dt = slope of x-t line = tan(angle), where the angle is measured from the time axis. Always tan, not sin or cos.
More in Kinematics: 13 exam traps and mistakes · 4 formulas · 9 question patterns from its other lessons.
Position Time Graph questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
How does NEET ask about Position Time Graph?
1 recurring pattern from past papers — click to collapse
Two particles' position-time graphs are straight lines with given angles vs the time axis (e.g. 30° and 45°); find ratio of velocities. Velocity = slope = tan(angle). Common shape: ratio = tan(30°)/tan(45°) = 1/√3.
InterpretationEasy
Common distractors
uses sin or cos instead of tan
Trig confusion under time pressure
Sources
NCERT refs: Class 11 Physics Chapter 2, p.17
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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