PhysicsISC Class 11 · Motion in a Straight Line

Motion Graphs and Calculus: Slope, Area and Variable Acceleration

Updated 17 October 2026 · Fact-checked

On an x-t graph the slope is velocity; on a v-t graph the slope is acceleration and the area under the graph is displacement; on an a-t graph the area is the change in velocity. When motion is described by a formula, velocity is dx/dt and acceleration is dv/dt = v dv/dx; integrate to go the other way.

Velocity-time graph with a rising line from v0 to v; yellow rectangle and green triangle under it give S = v0t + at²/2. — Notice that the area under the v-t line is the displacement: the yellow rectangle is v0·t and the green triangle is ½·at·t, which together give S = v0t + ½at².
Picture: Noname22124 · CC0 · Wikimedia Commons

Moving right, take the slope (d/dt); moving left, take the area (integrate): x-t to v-t to a-t, and back again.

Illustration drawn for QuestionBuddy.

Understand Motion Graphs and Calculus Methods

Reading graphs. The slope of the x-t graph at an instant is the instantaneous velocity; the slope of the chord between two points is the average velocity. The slope of the v-t graph is the acceleration. The area between the v-t graph and the time axis gives displacement, with area below the axis counted negative. For total distance, add the magnitudes of the areas. The area under an a-t graph is the change in velocity.

Shapes to remember. A straight line on x-t means constant velocity; a parabola means constant acceleration. A straight sloping line on v-t means constant acceleration; a horizontal line means constant velocity.

Calculus form. v = dx/dt and a = dv/dt = d²x/dt². Using the chain rule, a = v dv/dx, which is useful when the acceleration is given as a function of position. Reversing: v = ∫a dt + constant and x = ∫v dt + constant; the constants come from the initial conditions.

When is a particle at rest? Set v = 0 and solve for t. To find the distance (not displacement) over an interval where the particle reverses, find the instants at which v = 0 and add the distance of each leg.

Maximum or minimum speed. Set the derivative of v to zero, that is a = 0, and check that it is a maximum.

Motion Graphs and Calculus Methods: formulas and key facts

Velocity and acceleration: $v = d x / d t ∣ a = d v / d t = d^{2} x / d t^{2} ∣ a = v d v / d x$; The last form is used when a depends on x.
From acceleration to position: $v = \int a d t ∣ x = \int v d t$; Use the initial conditions to fix the constants.
Graph meaning: $s l o p eo f x - t = v ∣ s l o p eo f v - t = a ∣ a r e a u n d er v - t = d i s pl a ce m e n t$; Areas below the time axis are negative.

Motion Graphs and Calculus Methods in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A metro journey

Suppose a metro train accelerates uniformly from rest to 20 m s⁻¹ in 10 s, runs at that speed for 60 s, then brakes uniformly to rest in 20 s.

Find the distance between the two stations and the average speed.

Try it first, then show the solutionSolution

Distances from areas: ½ × 10 × 20 = 100 m; 60 × 20 = 1200 m; ½ × 20 × 20 = 200 m.
Total distance = 100 + 1200 + 200 = 1500 m.
Total time = 10 + 60 + 20 = 90 s, so average speed = 1500/90 = 16.7 m s⁻¹.

Answer: 1500 m; average speed about 16.7 m s⁻¹.

Memory hook: The area of the trapezium under the v-t graph is the distance.

Tricks to remember Motion Graphs and Calculus Methods

Short memory aids. Say each one aloud once, then use it on a question the same day.

Slope is rate, area is total
Slope gives how fast the quantity changes; area gives the total change of the quantity above it.
Use it when: Moving between x-t, v-t and a-t graphs.
Zero velocity means turning point
Set v = 0 to find where the motion reverses before computing distance.
Use it when: Distance versus displacement questions with a formula for x.

Motion Graphs and Calculus Methods: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Slope of x-t graph · 1 mark · MCQBoard level

The slope of the position-time graph of a particle gives its:
(a) acceleration
(b) velocity
(c) displacement
(d) jerk

Try it first, then show the solutionSolution

Velocity is dx/dt, which is the slope of x against t.

Answer: (b) velocity

Example 2: Trapezium on v-t graph · 3 marks · Short answerBoard level

A body speeds up from rest to 20 m s⁻¹ uniformly in 4 s, moves at that speed for 6 s, then slows uniformly to rest in 2 s. Find the total distance and the average speed.

Try it first, then show the solutionSolution

Areas: ½ × 4 × 20 = 40 m; 6 × 20 = 120 m; ½ × 2 × 20 = 20 m. (1 mark)
Total distance = 180 m in 12 s. (1 mark)
Average speed = 180/12 = 15 m s⁻¹. (1 mark)

Answer: 180 m; average speed 15 m s⁻¹.

Example 3: Particle reversing · 4 marks · Short answerHarder board level

A cart moves along a rail so that its position is x = t³ − 9t² + 24t (x in metres, t in seconds). Find the instants at which it is momentarily at rest, its acceleration then, and the distance covered in the first 4 s.

Try it first, then show the solutionSolution

v = 3t² − 18t + 24 = 3(t − 2)(t − 4) = 0 gives t = 2 s and t = 4 s. (1 mark)
a = 6t − 18: a(2) = −6 m s⁻², a(4) = +6 m s⁻². (1 mark)
x(0) = 0, x(2) = 8 − 36 + 48 = 20 m, x(4) = 64 − 144 + 96 = 16 m. (1 mark)
Distance = 20 + (20 − 16) = 24 m, while the displacement is 16 m. (1 mark)

Answer: At rest at t = 2 s and 4 s; accelerations −6 and +6 m s⁻²; distance 24 m.

Example 4: Variable acceleration · 4 marks · Short answerHarder board level

A particle starts from rest at the origin with acceleration a = 6 − 2t (m s⁻²). Find its velocity and position as functions of time, when it next comes to rest, and its position then.

Try it first, then show the solutionSolution

v = ∫(6 − 2t)dt = 6t − t² (v = 0 at t = 0). (1 mark)
x = ∫(6t − t²)dt = 3t² − t³/3. (1 mark)
v = 0 → t(6 − t) = 0, so it next stops at t = 6 s. (1 mark)
x(6) = 3 × 36 − 216/3 = 108 − 72 = 36 m. (1 mark)

Answer: v = 6t − t², x = 3t² − t³/3; it stops at t = 6 s at x = 36 m.

Example 5: A bouncing ball · 5 marks · Long answerHOTS / competency

Suppose a ball is dropped from a height of 5 m (g = 10 m s⁻²). After each bounce it rebounds with 80% of the speed it had just before hitting the floor. Find (i) the speed before the first impact, (ii) the height reached after the first bounce, (iii) the height reached after the second bounce.

Try it first, then show the solutionSolution

v = √(2gh) = √(2 × 10 × 5) = 10 m s⁻¹. (1 mark)
Rebound speed = 0.8 × 10 = 8 m s⁻¹. (1 mark)
Height after the first bounce = 8²/(2 × 10) = 3.2 m. (1 mark)
Speed before the second impact = √(2 × 10 × 3.2) = 8 m s⁻¹; rebound 0.8 × 8 = 6.4 m s⁻¹. (1 mark)
Height after the second bounce = 6.4²/20 = 2.048 m ≈ 2.05 m. On the v-t graph (up positive) each fall is a straight line of slope −10 m s⁻² with a jump in velocity at each bounce. (1 mark)

Answer: 10 m s⁻¹; 3.2 m; about 2.05 m.

Where students lose marks in Motion Graphs and Calculus Methods

The slips examiners see most often, and what to write instead.

Mistake: Using the slope of v-t as velocity
Mixing which graph gives which quantity.
Do this: Slope of x-t is v; slope of v-t is a; area under v-t is displacement.
Usually costs: 1 to 2 marks.
Mistake: Forgetting the constant of integration
Writing v = 6t − t² + 0 without justifying it, or omitting u when it is not zero.
Do this: Use the initial conditions to find the constant, and show the step.
Usually costs: 1 mark.
Mistake: Adding signed areas for distance
Area below the axis cancels area above, giving displacement not distance.
Do this: Add the magnitudes of the areas for total distance.
Usually costs: 1 to 2 marks.

Exam tips

Label the axes and units on every graph you sketch.
Break a v-t graph into triangles and rectangles and write each area separately.
Show d/dt or the integral explicitly before substituting numbers.
For distance when the particle reverses, find the stopping instants first.

Practice questions on Motion Graphs and Calculus Methods

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The v-t graph of a body is a horizontal line at 5 m s⁻¹. The displacement in 10 s is:
(a) 5 m
(b) 10 m
(c) 50 m
(d) 100 m
Hint
Area of the rectangle under the graph.
Answer
(c) 50 m
Q2 · 2 marks · Short answerBoard level
A trolley has x = 2t³ + 5t (SI). Differentiate to get its speed and acceleration at t = 3 s.
Hint
Differentiate once and twice.
Answer
v = 6t² + 5 = 59 m s⁻¹. a = 12t = 36 m s⁻².
Q3 · 3 marks · Short answerHarder board level
Given v = 3t² − 2t (SI) and x = 0 when t = 0, integrate to find x(3) and then a(3).
Hint
Integrate v for x; differentiate v for a.
Answer
x = t³ − t² = 27 − 9 = 18 m. a = 6t − 2 = 16 m s⁻².
Q4 · 3 marks · Short answerHarder board level
A body starts from rest with an acceleration of 4 m s⁻² for 5 s, moves with constant velocity for the next 5 s, then retards uniformly at 5 m s⁻² until it stops. Find the total distance and the average speed.
Hint
Use areas under the v-t graph.
Answer
v after 5 s = 20 m s⁻¹. Distances: ½ × 5 × 20 = 50 m; 20 × 5 = 100 m; the stop takes 4 s, so ½ × 4 × 20 = 40 m. Total 190 m in 14 s, average speed 13.6 m s⁻¹.
Q5 · 3 marks · Short answerHarder board level
A particle's speed depends on its position as v² = 36 − 9x (SI). Show that its acceleration is constant and find it.
Hint
Use a = v dv/dx, or differentiate v² with respect to x.
Answer
Differentiate: 2v (dv/dx) = −9, so a = v dv/dx = −4.5 m s⁻². It is constant and negative.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a metro train accelerates uniformly from rest to 20 m s⁻¹ in 10 s, runs at constant speed for 60 s, then retards uniformly to rest in 20 s.
(i) Find the acceleration in the first phase. (1 mark)
(ii) Find the retardation in the last phase. (1 mark)
(iii) Find the distance between the stations. (1 mark)
(iv) Find the average speed for the whole journey. (1 mark)
Hint
Split the trapezium into triangle, rectangle and triangle.
Answer
(i) 20/10 = 2 m s⁻². (ii) 20/20 = 1 m s⁻². (iii) 100 + 1200 + 200 = 1500 m. (iv) 1500/90 = 16.7 m s⁻¹.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What does the area under an a-t graph give?Show answer
The change in velocity.
2.Write a in terms of v and x.Show answer
a = v dv/dx.
3.What does a straight line on an x-t graph mean?Show answer
Constant velocity (zero acceleration).
4.How is total distance found from a v-t graph?Show answer
Add the magnitudes of the areas.

Motion Graphs and Calculus Methods: frequently asked questions

What is the difference between average and instantaneous velocity on an x-t graph?

Average velocity is the slope of the chord joining two points; instantaneous velocity is the slope of the tangent at one point.

Why is a = v dv/dx useful?

When acceleration is given in terms of position, integrating a with respect to t is awkward. Writing a = v dv/dx lets you integrate directly in x to get v.

Does a curved v-t graph mean non-uniform acceleration?

Yes. The slope changes from point to point, so the acceleration is not constant, and the equations of uniformly accelerated motion do not apply.