Position–time & velocity–time graphs

The Language of Motion: Graphical Analysis

In Kinematics, graphs are not just pictures; they are a powerful language used to describe the story of an object's motion. For NEET, mastering the interpretation of Position–time (x-t) and Velocity–time (v-t) graphs is non-negotiable. Almost every year, questions are framed that test your ability to read slopes, calculate areas, and convert one type of graph into another.

This lesson will make you fluent in this graphical language. We will dissect what every line, curve, slope, and area means in the real world. By the end, you will be able to look at a graph and instantly visualize the object's journey—whether it's speeding up, slowing down, reversing, or standing still. The table above is your master key; we will unlock its full potential.

1. Position–Time (x-t) Graphs

A position-time graph plots the position (x) of an object on the y-axis against time (t) on the x-axis. It’s like a flipbook of where the object is at every instant. The single most important takeaway for x-t graphs is what its slope represents.

Slope is defined as the change in the y-axis divided by the change in the x-axis. Slope = Δy / Δx For an x-t graph, this becomes: Slope = Δx / Δt By definition, Δx / Δt is the average velocity. If we take the time interval Δt to be infinitesimally small (dt), the slope gives us the instantaneous velocity (v).

{KEY: type=definition | title=The Golden Rule of x-t Graphs | text=The slope of the position-time graph at any point gives the instantaneous velocity of the object at that point. A steeper slope means higher velocity. A flatter slope means lower velocity.}

Let's analyze the different scenarios you'll encounter.

Case A: Object at Rest

If an object is stationary, its position x does not change with time. It remains at a constant value, say x₀.

Plotting this gives a horizontal line. The slope of a horizontal line is zero (Δx = 0), which correctly tells us that the velocity is zero.

{{VISUAL: diagram: A position-time graph for a stationary object. The x-axis is labeled 'Time (t)' and the y-axis is 'Position (x)'. A horizontal line is drawn at a positive value of x, labeled 'x = x₀'. The slope is indicated as 'Slope = Δx/Δt = 0', hence 'v = 0'.}}

Case B: Object in Uniform Motion (Constant Velocity)

Uniform motion means the object covers equal distances in equal intervals of time. Its velocity is constant and non-zero.

Since velocity v = Δx/Δt is constant, the relationship between x and t is linear: x = x₀ + vt. This is the equation of a straight line (y = c + mx) with a constant slope m = v. Therefore, the x-t graph for uniform motion is a straight line with a constant, non-zero slope.

• A positive slope (line goes up from left to right) indicates positive velocity.
• A negative slope (line goes down from left to right) indicates negative velocity (moving in the opposite direction).
• The steepness of the slope corresponds to the magnitude of the velocity (speed). A steeper line means a higher speed.

{{VISUAL: diagram: Two position-time graphs for uniform motion on the same axes. Both are straight lines starting from the origin. Line A is steeper than Line B. A caption reads: 'Slope of A > Slope of B, so Velocity of A > Velocity of B'. Another graph shows a line with a negative slope starting from a positive x value.}}

Example 1: Interpreting an x-t Graph (Easy)

The position-time graph of a car is shown below. What is the velocity of the car?

{{VISUAL: chart: An x-t graph. The x-axis (time) goes from 0 to 5s. The y-axis (position) goes from 0 to 20m. A straight line passes through (0, 0) and (5s, 20m).}}

Given:

• Initial position (x₁) at t₁ = 0 s is 0 m.
• Final position (x₂) at t₂ = 5 s is 20 m.

To Find:

• The velocity (v) of the car.

Approach: The graph is a straight line, which signifies uniform motion (constant velocity). The velocity is equal to the slope of the line. We can calculate the slope using the formula: Slope = (x₂ - x₁) / (t₂ - t₁).

Working:

1. Identify two points on the line. Let's use (t₁, x₁) = (0 s, 0 m) and (t₂, x₂) = (5 s, 20 m).
2. Calculate the slope.

   v = Slope = (20 m - 0 m) / (5 s - 0 s)
   

   v = 20 m / 5 s
   

   v = 4 m/s
   

Final Answer: The velocity of the car is 4 m/s.

Case C: Object in Uniformly Accelerated Motion

When an object's velocity changes at a constant rate, it is in uniformly accelerated motion. Its velocity is no longer constant, meaning the slope of the x-t graph must change.

The equation of motion is x = x₀ + ut + ½at². This is a quadratic equation in t. The graph of a quadratic equation is a parabola.

• If acceleration a > 0 (speeding up in the positive direction), the parabola opens upwards. The slope continuously increases, indicating increasing velocity.
• If acceleration a < 0 (slowing down in the positive direction), the parabola opens downwards. The slope continuously decreases, indicating decreasing velocity.

To find the instantaneous velocity at a specific time t, you must find the slope of the tangent drawn to the curve at that point.

{{VISUAL: chart: An x-t graph showing a parabola opening upwards, starting from the origin. Tangents are drawn at two points, t₁ and t₂ (where t₂ > t₁). The tangent at t₂ is steeper than at t₁. A caption reads: 'Slope at t₂ > Slope at t₁, so v(t₂) > v(t₁). This indicates positive acceleration.'}}

{FORMULA: expr=v_inst = dx/dt | symbols=v_inst: Instantaneous velocity, dx/dt: Derivative of position with respect to time (slope of the tangent to the x-t curve)}

2. Velocity–Time (v-t) Graphs

A velocity-time graph plots the velocity (v) of an object on the y-axis against time (t) on the x-axis. This is arguably the most important and information-rich graph in kinematics. It tells you not just where an object is going, but how its motion is changing.

A v-t graph provides two crucial pieces of information: its slope and the area under it.

{CALLOUT: type=tip | text=For v-t graphs, remember SAD: Slope is Acceleration, Area is Displacement. This is a vital memory aid for NEET.}

The Meaning of Slope: Acceleration

Slope = Δy / Δx For a v-t graph, this becomes: Slope = Δv / Δt By definition, Δv / Δt is the average acceleration. If we take the time interval Δt to be infinitesimally small (dt), the slope of the tangent gives us the instantaneous acceleration (a).

The Meaning of Area: Displacement

From the definition of velocity, v = Δx / Δt, we can rearrange to get Δx = v × Δt. If velocity is constant, the displacement is the product of velocity and time. On a v-t graph, this corresponds to the area of a rectangle. If velocity is changing, we can think of the motion as being made up of many tiny intervals dt where the velocity v is almost constant. The total displacement is the sum (integral) of all the small v × dt areas.

{KEY: type=definition | title=The Two Golden Rules of v-t Graphs | text=1. The slope of the velocity-time graph gives the instantaneous acceleration. <br> 2. The area under the velocity-time graph gives the displacement. }

Let's explore the common cases.

Case A: Uniform Motion (Constant Velocity)

If velocity is constant, say v = v₀, the v-t graph is a horizontal line.

• Slope: The slope is zero (Δv = 0), which correctly tells us the acceleration is zero.
• Area: The area under the graph from t=0 to t=T is a rectangle with height v₀ and width T. The area is v₀ × T, which is the displacement, Δx.

{{VISUAL: chart: A velocity-time graph for an object in uniform motion. The x-axis is 'Time (t)' and y-axis is 'Velocity (v)'. A horizontal line is at a positive value v₀. The area under the line from t=0 to t=T is shaded and labeled 'Area = v₀ × T = Displacement'. The slope is marked as 'Slope = 0, so a = 0'.}}

Case B: Uniformly Accelerated Motion

If an object has constant acceleration a, its velocity changes linearly with time: v = u + at. This is the equation of a straight line (y = c + mx) with slope m = a.

• A positive slope indicates positive constant acceleration.
• A negative slope indicates negative constant acceleration (deceleration).
• A zero slope (horizontal line) indicates zero acceleration.

The area under this graph is a trapezium (or a triangle if u=0). The area of a trapezium is ½ × (sum of parallel sides) × height. Area = ½ × (u + v) × t From v = u + at, we have t = (v-u)/a. Substituting this into the area formula gives Area = ½ (u+v)(v-u)/a = (v²-u²)/2a. So, Δx = (v²-u²)/2a, which rearranges to v² = u² + 2aΔx, the third equation of motion! This shows how the graphical concepts directly lead to the kinematic equations.

Example 2: Finding Displacement and Acceleration from a v-t graph (Medium)

A body starts from rest and its velocity-time graph is shown. Find (i) the acceleration in parts OA, AB, and BC, and (ii) the total displacement in 6 seconds.

{{VISUAL: chart: A v-t graph. Axes are v (m/s) and t (s).

• From t=0 to t=2s (Part OA), velocity increases linearly from 0 to 10 m/s.
• From t=2s to t=4s (Part AB), velocity is constant at 10 m/s.
• From t=4s to t=6s (Part BC), velocity decreases linearly from 10 m/s to 0 m/s.}}

This is a classic problem structure. Let's solve it on the whiteboard.

{{SOLVE: {"problem":"From the given v-t graph, find: (i) Acceleration in parts OA, AB, and BC. (ii) Total displacement in 6 seconds. The graph shows velocity rising from 0 to 10m/s in 2s, staying at 10m/s for 2s, and falling back to 0 in the next 2s.","type":"numerical","subject":"physics","intro":"Yeh ek classic NEET level ka sawal hai. Chalo, isse whiteboard pe step-by-step solve karte hain.","outro":"Toh humne slope se acceleration aur area se displacement nikal liya. Bas yahi concept hai!","steps":[{"explanation":"First, let's find the acceleration for part OA. Acceleration is the slope of the v-t graph, which is (change in velocity) / (change in time).","write":"(i) Acceleration (a) = Slope = Δv / Δt","tough":false},{"explanation":"For part OA, the velocity changes from 0 to 10 m/s in 2 seconds.","write":"a_OA = (10 - 0) m/s / (2 - 0) s = 10/2 m/s² = 5 m/s²","tough":false},{"explanation":"For part AB, the graph is a horizontal line. The velocity is constant, so the change in velocity is zero.","write":"a_AB = (10 - 10) m/s / (4 - 2) s = 0/2 m/s² = 0 m/s²","tough":false},{"explanation":"For part BC, the velocity decreases from 10 m/s to 0 m/s over 2 seconds. The slope will be negative.","write":"a_BC = (0 - 10) m/s / (6 - 4) s = -10/2 m/s² = -5 m/s²","tough":true,"alt_explanation":"Remember, final velocity is 0 and initial velocity for this part is 10. So it's (final - initial) which is (0 - 10), giving a negative result. This means deceleration."},{"explanation":"Now for part (ii), total displacement. This is the total area under the v-t graph. The shape is a trapezium.","write":"(ii) Displacement (Δx) = Total Area under v-t graph","tough":false},{"explanation":"The total shape is a trapezium with parallel sides of length (4-2)=2s and 6s, and height 10 m/s. Or, we can add the areas of the triangle, rectangle, and the second triangle. Let's do that.","write":"Area = Area(ΔOAE) + Area(⎕ABFE) + Area(ΔBFC)","tough":false},{"explanation":"Area of the first triangle (OA) is half base times height.","write":"Area_OA = ½ × base × height = ½ × 2s × 10m/s = 10 m","tough":false},{"explanation":"Area of the rectangle (AB) is length times width.","write":"Area_AB = length × width = (4-2)s × 10m/s = 2s × 10m/s = 20 m","tough":false},{"explanation":"Area of the second triangle (BC) is again half base times height.","write":"Area_BC = ½ × base × height = ½ × (6-4)s × 10m/s = ½ × 2s × 10m/s = 10 m","tough":false},{"explanation":"Finally, the total displacement is the sum of these three areas.","write":"Total Displacement = 10 m + 20 m + 10 m = 40 m","tough":false}]}}}

Dealing with Negative Velocity

What if the v-t graph goes below the t-axis? This simply means the velocity is negative, i.e., the object is moving in the negative direction.

• The area calculated for the portion below the axis will be negative. This represents negative displacement.
• Displacement is the net area (Area above axis - Area below axis).
• Distance is the total path length, so you must add the magnitudes of the areas. Distance = |Area above axis| + |Area below axis|.

{FLASHCARD: q=For a v-t graph, which quantity can be negative: Distance or Displacement? | a=Displacement can be negative (if area under the t-axis is greater), but distance is a scalar and is always positive.}

Example 3: Distance vs. Displacement (Hard)

The velocity-time graph for a particle is shown. Find the total distance travelled and the net displacement after 8 seconds.

{{VISUAL: chart: A v-t graph.

• From t=0 to t=4s, velocity increases linearly from 0 to 5 m/s.
• From t=4s to t=8s, velocity decreases linearly from 5 m/s to -5 m/s, crossing the t-axis at t=6s.}}

Given:

• v-t graph with key points: (0,0), (4,5), (6,0), (8,-5).

To Find:

1. Total distance travelled.
2. Net displacement.

Approach: Displacement is the net area under the curve (area above the t-axis is positive, area below is negative). Distance is the sum of the absolute values of these areas. The graph consists of two triangles.

Working:

1. Calculate Area 1 (A₁): from t=0 to t=6s (above axis). This shape is a triangle with base = 6 s and height = 5 m/s.

   A₁ = ½ × base × height = ½ × 6 s × 5 m/s = 15 m
   

2. Calculate Area 2 (A₂): from t=6s to t=8s (below axis). This is a triangle with base = (8 - 6) s = 2 s and height = -5 m/s.

   A₂ = ½ × base × height = ½ × 2 s × (-5 m/s) = -5 m
   

3. Calculate Net Displacement. Displacement = A₁ + A₂

   Displacement = 15 m + (-5 m) = 10 m
   

4. Calculate Total Distance. Distance = |A₁| + |A₂|

   Distance = |15 m| + |-5 m| = 15 m + 5 m = 20 m
   

Final Answer: The net displacement is 10 m. The total distance travelled is 20 m. This is a very common trap in NEET!

3. Common Graphical Traps and Misconceptions

Students often make predictable errors when interpreting graphs under exam pressure. Here is a table to help you avoid them.

NEET MCQ Practice Bank

Question 1: The position-time (x-t) graph for a particle in one-dimensional motion is shown. At which point (P, Q, R, or S) is the instantaneous velocity of the particle negative? (A diagram shows a sinusoidal-like x-t curve. P is at a crest, Q is on the downward slope, R is at a trough, S is on the upward slope.)

(a) P (b) Q (c) R (d) S

💡 Solution: (b) Explanation: Instantaneous velocity is the slope of the x-t graph.

• At P (crest), the tangent is horizontal, so slope ≈ 0.
• At Q, the graph is sloping downwards, so the slope is negative.
• At R (trough), the tangent is horizontal, so slope ≈ 0.
• At S, the graph is sloping upwards, so the slope is positive. Therefore, the velocity is negative at point Q.

Question 2: A car starts from rest and accelerates uniformly for 10 s to a velocity of 8 m/s. It then runs at a constant velocity and is finally brought to rest in 64 m with a constant retardation. The total distance covered by the car is 584 m. Find the value of acceleration, retardation and total time taken. The correct v-t graph is: (Four options showing different shaped v-t graphs)

💡 Solution: (The correct graph would be a triangle, followed by a rectangle, followed by another triangle going down to v=0) Explanation: This is a multi-step problem, but recognizing the shape of the graph is the first step.

1. "accelerates uniformly" → Straight line with positive slope from origin.
2. "runs at a constant velocity" → Horizontal line.
3. "constant retardation" → Straight line with negative slope, ending at v=0. Only one graph shape will match this description.

Question 3: The velocity of a particle is v = v₀ + gt + ft². If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is: (a) v₀ + g/2 + f (b) v₀ + 2g + 3f (c) v₀ + g/2 + f/3 (d) v₀ + g + f

💡 Solution: (c) Explanation: We are given velocity v as a function of time t. To find displacement x, we need to integrate velocity with respect to time. v = dx/dt. dx = v dt = (v₀ + gt + ft²) dt ∫dx = ∫(v₀ + gt + ft²) dt x = v₀t + g(t²/2) + f(t³/3) + C Since x=0 at t=0, the integration constant C=0. Now, substitute t=1: x(1) = v₀(1) + g(1²/2) + f(1³/3) = v₀ + g/2 + f/3.

Question 4: A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height d/2. Neglecting subsequent motion and air resistance, its velocity v varies with height h above the ground as: (Four options showing different v-h graph shapes)

💡 Solution: (The correct graph is a parabola opening to the right for the downward journey and a smaller parabola opening to the right for the upward journey). Explanation: From the third equation of motion, v² = u² + 2as. For downward motion: u=0, a=g, s = d-h. So, v² = 2g(d-h). Since velocity is downwards, v = -√(2g(d-h)). For upward motion: final velocity is 0 at h=d/2. Let velocity at ground be v'. 0² = (v')² - 2g(d/2). So v' = √gd. At any height h during upward motion, v² = (v')² - 2gh = gd - 2gh. So v = √(g(d-2h)). Both relations show that v² is proportional to h, which represents a parabola on a v-h graph.

Numerical Practice Set

1. The position of a particle moving along the x-axis is given by x = 10t - 2t² meters. Find the velocity of the particle at t = 2 s.

   💡 Answer: 2 m/s

2. A v-t graph is a straight line passing through (0, 5 m/s) and (10 s, 25 m/s). What is the acceleration of the body and what is the displacement in 10 seconds?

   💡 Answer: a = 2 m/s², Δx = 150 m

3. The velocity-time graph of a body moving in a straight line is shown. Find the displacement and distance traveled by the body in 6 seconds. The graph goes from (0,0) to (2,4), then to (4,-4), and finally to (6,-4).

   💡 Answer: Displacement = -4 m, Distance = 12 m

4. From the v-t plot shown, find the distance travelled by the particle during the first 40 seconds. Also find the average velocity during this period. The graph is a trapezium with vertices (0,0), (10,5), (30,5), (40,0).

   💡 Answer: Distance = 175 m, Average velocity = 4.375 m/s

{KEY: type=points | title=Quick Recap: Interpreting Kinematic Graphs | text=

• Position-Time (x-t) Graph:
  • Slope = Instantaneous Velocity (v)
  • Horizontal line → v = 0 (at rest)
  • Straight line with slope → constant v
  • Curved line → acceleration is present
• Velocity-Time (v-t) Graph:
  • Slope = Instantaneous Acceleration (a)
  • Area = Displacement (Δx)
  • Horizontal line → a = 0 (constant v)
  • Straight line with slope → constant a
  • Area below t-axis is negative displacement.
  • Total distance = sum of magnitudes of all areas. }