Describing motion with quantities and graphs
Key idea: H2 Physics lessons on motion graphs, uniformly accelerated motion, momentum and Newton’s laws.
Continue where you stopped
The core idea
Build the idea
Learn the idea
Big question: What can the shape of a motion graph tell you without another equation?
Displacement locates an object relative to an origin; velocity is its rate of change, and acceleration is the rate of change of velocity. On a displacement–time graph the gradient is velocity. On a velocity–time graph the gradient is acceleration and signed area is displacement. These links let you move between words, graphs and constant-acceleration equations while keeping direction signs consistent.
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.
Key ideas to keep
- A horizontal displacement–time line means stationary, not constant non-zero speed.
- Area below the time axis is negative displacement.
- The constant-acceleration equations apply only over an interval with constant acceleration.
See 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.
Another worked model
Model 1
A velocity rises linearly from 4.0 to 16 m s⁻¹ in 6.0 s, then falls linearly to 10 m s⁻¹ over 3.0 s. Find the displacement and acceleration in each interval.
Check the worked solution
Displacement is velocity–time area: first interval = ½(4.0 + 16)(6.0) = 60 m; second = ½(16 + 10)(3.0) = 39 m; total = 99 m. Accelerations are (16 − 4)/6 = 2.0 m s⁻² and (10 − 16)/3 = −2.0 m s⁻².
Model 2
Derive v² = u² + 2as from constant-acceleration definitions.
Check the worked solution
Use a = (v − u)/t, so t = (v − u)/a. Constant acceleration gives average velocity (u + v)/2, hence s = (u + v)t/2. Substitution gives s = (u + v)(v − u)/(2a) = (v² − u²)/(2a), so v² = u² + 2as.
Use a hint if needed
Practise with support
Question 1
A cyclist’s position changes from −5 m to +19 m in 8.0 s along a straight axis. Find displacement and average velocity. Can distance be found from this information alone?
Hint: Use final position minus initial position.
Check your answer
Displacement = 19 − (−5) = +24 m; average velocity = +24/8.0 = +3.0 m s⁻¹. Distance cannot be found without knowing the path or whether direction changed.
Question 2
A stone is thrown vertically upward at 14 m s⁻¹. Neglect air resistance and use g = 9.8 m s⁻². Find the time to maximum height and that height above release.
Hint: Take upward positive, so a = −9.8 m s⁻² and v = 0 at the top.
Check your answer
0 = 14 − 9.8t gives t = 1.43 s. Using v² = u² + 2as: 0 = 14² − 2(9.8)s, so s = 10.0 m.
Now work without the hint
Practise independently
Question 1
Sketch in words a velocity–time graph for motion that starts at 2 m s⁻¹, accelerates non-uniformly to 8 m s⁻¹, then continues at 8 m s⁻¹. Explain how acceleration and displacement are obtained.
Check your answer
The first section is a curve rising from 2 to 8 m s⁻¹, with changing gradient; the second is horizontal at 8 m s⁻¹. Instantaneous acceleration is the tangent gradient and total displacement is the signed area under both sections.
Question 2
A train moving at 25 m s⁻¹ brakes uniformly to rest over 200 m. Find its acceleration and stopping time.
Check your answer
From v² = u² + 2as: 0 = 25² + 2a(200), so a = −1.5625 m s⁻². From v = u + at: t = (0 − 25)/(−1.5625) = 16 s.
Avoid these traps
Common mistakes
Common mistake
Distance and displacement, or speed and velocity, are interchangeable.
What is wrong with this reasoning?
Show better thinking
Distance and speed are scalar path measures. Displacement and velocity include a declared direction and may be zero even after motion.
Common mistake
The height of any motion graph gives the object’s acceleration.
What is wrong with this reasoning?
Show better thinking
Read the axis labels first. Acceleration is the velocity–time gradient; velocity is the position–time gradient. Velocity–time area gives displacement.
Common mistake
The uniformly accelerated equations apply whenever an object is moving.
What is wrong with this reasoning?
Show better thinking
They require constant acceleration over the chosen interval. For vertical motion without air resistance, choose a positive direction and use a constant acceleration of magnitude g with the correct sign.
Write for the examiner
Exam guidance
When asked to sketch, label axes and key values, then make each gradient and area match the described motion.
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.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
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.
Come back in three days
Check what stayed with you
Recall question 1
What does the gradient of a displacement–time graph give?
Check the answer
Velocity.
Recall question 2
What does signed area under a velocity–time graph give?
Check the answer
Displacement.
Recall question 3
When must the constant-acceleration equations not be used?
Check the answer
When acceleration changes during the interval being modelled.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. This topic connects descriptions of motion to the forces that cause acceleration. Read graph gradients and areas from their axes, and draw forces only on the chosen object before applying Newton's laws.
- GCE A-Level H2 PhysicsTopic 3(a) / Topic 3(b) / Topic 3(c) / Topic 3(d) / Topic 3(e) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027