Kinematics, graphs and uniform acceleration

Key idea: Motion graphs are compact stories. Their coordinates, gradients and areas mean different physical quantities, so read the axes before using a method.

GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 1 of 3

Check your understanding

By the end of this lesson, you should be able to

Use position, distance, displacement, speed, velocity and acceleration precisely.
Interpret gradients and signed areas on motion graphs, including non-uniform acceleration.
Derive and apply the constant-acceleration equations only when acceleration is uniform.

Learn the idea

Big question: How can one graph tell you both how an object is moving and how far it has travelled?

Position, displacement and distance answer different questions

Position locates an object relative to a chosen origin. Displacement is the signed change from initial to final position. Distance is the total path length and cannot be negative. A journey that returns to its starting point has zero displacement but a non-zero distance.

Instantaneous velocity is the gradient of the tangent to a position–time graph. On a curved velocity–time graph, the tangent gradient similarly gives instantaneous acceleration; the total signed area still gives displacement.

Check your understanding: A runner completes one 400 m lap. What are distance and displacement?

Distance is 400 m; displacement is zero because the final position equals the initial position.

Derive before choosing an equation

For constant acceleration, a = (v − u)/t rearranges to v = u + at. The velocity–time graph is a straight line, so displacement is average velocity times time: s = ½(u + v)t. Substituting v = u + at gives s = ut + ½at²; eliminating t instead gives v² = u² + 2as.

These are one connected family, not unrelated formulas. Choose the equation containing the known quantities and omitting the unwanted one, and do not use the family across an interval where acceleration changes.

Check your understanding: Which assumption lets average velocity equal (u + v)/2?

Velocity must change uniformly, so acceleration is constant.

A distance–time gradient gives speed. On a speed–time graph, gradient gives acceleration and area gives distance travelled.

Key ideas

A negative velocity gives negative displacement area, not negative distance.
A horizontal displacement–time graph means rest; a horizontal velocity–time graph means zero acceleration.
Declare a positive direction before substituting signed quantities.

Relationships to know

v = u + at
s = (u + v)t/2
s = ut + ½at²
v² = u² + 2as

Follow the reasoning

Worked example

Separate displacement from distance

Question: Velocity increases uniformly from −2.0 m s⁻¹ to +6.0 m s⁻¹ in 4.0 s. Find acceleration, displacement and distance travelled.

Step 1: Read acceleration from gradient
Why: A straight velocity–time line has constant acceleration equal to its gradient.
Working: a = [6.0 − (−2.0)]/4.0 = 2.0 m s⁻².
Step 2: Find signed area for displacement
Why: Area below the time axis is negative and area above is positive.
Working: Velocity is zero at 1.0 s. Displacement = −½(1.0)(2.0) + ½(3.0)(6.0) = −1.0 + 9.0 = 8.0 m.
Step 3: Add area magnitudes for distance
Why: Distance counts path length regardless of direction.
Working: Distance = 1.0 + 9.0 = 10.0 m.

Answer: Acceleration = 2.0 m s⁻², displacement = +8.0 m and distance = 10.0 m.

Check: Distance is greater than displacement magnitude because the object first moves in the negative direction and reverses.

Now try it with support

Practise with support

A car starts at 4.0 m s⁻¹ and accelerates uniformly at 3.0 m s⁻² for 5.0 s. Find its final velocity and displacement.

Hints

Use v = u + at first.
Use s = ut + ½at² or the average velocity.

View the guided answer

v = 4.0 + 3.0(5.0) = 19 m s⁻¹. s = 4.0(5.0) + ½(3.0)(5.0²) = 57.5 m.

Your turn

Practise independently

A velocity–time line falls from 12 to −4 m s⁻¹ in 8 s. Determine acceleration, displacement and total distance.

Check your answer

a = (−4 − 12)/8 = −2.0 m s⁻². Displacement = ((12 + (−4))/2)(8) = 32 m. Velocity reaches zero after 6 s: distance = ½(6)(12) + ½(2)(4) = 40 m.

Common mistakes and exam guidance

Watch out for

Using total area as distance when part of a velocity graph lies below the axis.
Using constant-acceleration equations on a curved velocity–time graph without justification.

In an exam

Write the meaning and unit of a gradient or area before calculating it.
Use signs consistently; report distance as non-negative and displacement with direction or sign.

Put the ideas together

Exam-style practice [7 marks]

A car accelerates uniformly from rest to 12 m s⁻¹ in 4.0 s, travels at 12 m s⁻¹ for 3.0 s, then changes velocity uniformly to −4.0 m s⁻¹ in 4.0 s. Find the acceleration in the final stage, total displacement and total distance.

Plan before you answer

Sketch the three velocity–time sections.
Use signed areas for displacement.
Split the last section where velocity crosses zero for distance.

View the marking points and model answer

Marking points

Finds final-stage acceleration −4.0 m s⁻².
First area 24 m.
Second area 36 m.
Signed final area 16 m.
Total displacement 76 m.
Final-stage distance 20 m after splitting at zero.
Total distance 80 m.

Model answer

The final acceleration is (−4 − 12)/4 = −4.0 m s⁻². The signed areas are ½(4)(12) = 24 m, 3(12) = 36 m and ½(12 − 4)(4) = 16 m, so displacement is 76 m. In the final stage the car reaches zero after 3 s, giving distance ½(3)(12) + ½(1)(4) = 20 m. Total distance is 24 + 36 + 20 = 80 m.

Finish from memory

Three-question recap

What does the gradient of a displacement–time graph give?
Check
Velocity.
What does signed area under a velocity–time graph give?
Check
Displacement.
When must the constant-acceleration equations not be used?
Check
When acceleration changes during the interval being modelled.

Try this next

Sketch the graph for a journey that reverses direction, then solve one free-fall problem with upward positive.

Continue to Mass, inertia and linear momentum Return to Motion and Forces

Continue with the next resource in this course.

Course and syllabus information

Course: GCE A-Level H1 Physics
Edition: GCE A-Level H1 Physics 2027