Use velocity time gradients and signed areas
Sketch piecewise velocity graphs, calculate acceleration from slope and separate signed displacement from total distance.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.2. Motion is restricted to one dimension. Choose a positive direction; distance and speed are scalars, while displacement, velocity and acceleration carry a direction. Deceleration means decreasing speed, not necessarily negative acceleration. Calculus uses Paper 1 techniques only. Constant-acceleration formulae require constant acceleration on the stated interval; use a new interval and its initial velocity when conditions change. Where vertical motion occurs, use the supplied g = 10 m s⁻² and the stated no-air-resistance model. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Calculate triangle and trapezium areas.
- Interpret signed velocities and solve a linear equation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Velocity changes from 2 to 8 m s⁻¹ over 3 s on a straight graph. What is acceleration?
What you will learn
- Calculate acceleration from a velocity–time gradient.
- Use positive and negative areas to distinguish displacement and distance.
Slope and area have different meanings
The vertical coordinate of a velocity–time graph is directed velocity. Its gradient gives acceleration, measured in m s⁻² when the axes use seconds and m s⁻¹. A horizontal segment means constant velocity and zero acceleration, not necessarily rest. A straight sloping segment models constant acceleration; a curved velocity graph has changing acceleration and needs a tangent to find its instantaneous value.
The signed area between the graph and time axis gives displacement. An area below the axis counts negatively. To find distance, add the magnitudes of all areas and split a segment where it crosses v = 0. A triangle or trapezium can evaluate a straight segment exactly; no calculus is needed for that geometry. A time-axis crossing may indicate a reversal, whereas a turning point of the velocity graph concerns acceleration and need not reverse the motion.
Label endpoints, crossing times and the full interval
Consider straight segments through (0, 0), (2, 6), (5, 6) and (9, −2). Accelerations are +3, 0 and −2 m s⁻² on their open intervals. The last line reaches v = 0 at t = 8 s: it takes 3 s to reduce velocity from 6 to zero after t = 5. Use that crossing to divide the last trapezium into positive area 9 m and negative area −1 m.
The first triangle gives 6 m, and the constant-velocity rectangle gives 18 m. Therefore total displacement is 6 + 18 + 9 − 1 = 32 m, while distance is 6 + 18 + 9 + 1 = 34 m. Draw the final negative-velocity segment below the axis and keep the endpoint time at 9 s. At the joins velocity remains continuous, but acceleration changes abruptly; use the neighbouring interval slopes rather than claiming one common acceleration there.
Pause and explain
Positive velocity area is 9 m and negative area is −4 m. What are displacement and distance?
Worked example
For a velocity graph joining (0, 0), (2, 6), (5, 6) and (9, −2), find the accelerations, zero-velocity crossing, displacement and distance.
Show the worked solution
- Segment gradients are (6 − 0)/2 = 3, 0, and (−2 − 6)/4 = −2 m s⁻².
- On the final segment, zero velocity occurs at 5 + 6/2 = 8 s; signed areas are 6, 18, 9 and −1 m.
- Displacement is 32 m, while distance adds all area magnitudes to give 34 m.
Answer Accelerations +3, 0 and −2 m s⁻²; crossing 8 s; displacement 32 m; distance 34 m.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A velocity graph joins (0, 0), (4, 8) and (6, 8). Find both accelerations and displacement, using seconds and m s⁻¹.
Give me a hint
The first region is triangular and the second is rectangular.
Compare my reasoning
- First acceleration is 8/4 = 2 m s⁻²; second acceleration is zero.
- First area is ½ × 4 × 8 = 16 m; second area is 2 × 8 = 16 m.
- All velocities are non-negative, so displacement and distance are both 32 m.
Accelerations 2 and 0 m s⁻²; displacement and distance 32 m.
Look for these in your work
- Use segment duration rather than endpoint time for each gradient.
- Connect the non-negative areas to the equality of distance and displacement.
For v = 6 − 2t m s⁻¹ on 0 ≤ t ≤ 5 s, use the graph to find acceleration, reversal time, displacement and distance.
Give me a hint
Split the straight line where it crosses the time axis.
Compare my reasoning
- Acceleration is −2 m s⁻² and v = 0 at t = 3 s.
- The positive triangle has area ½ × 3 × 6 = 9 m; the final negative triangle has magnitude ½ × 2 × 4 = 4 m.
- Displacement is 9 − 4 = 5 m and distance is 9 + 4 = 13 m.
Acceleration −2 m s⁻²; reversal 3 s; displacement 5 m; distance 13 m.
Look for these in your work
- Draw the endpoint velocity −4 m s⁻¹ below the axis.
- Use signs for displacement and magnitudes for distance.
A velocity graph joins (0, 4), (2, 4), (4, 0) and (6, −4). Calculate accelerations, whole-journey displacement and distance, then average velocity and average speed.
Give me a hint
The last two segments lie on the same straight line but their areas have different signs.
Compare my reasoning
- Acceleration is zero for 0 < t < 2 and −2 m s⁻² for 2 < t < 6.
- Areas are +8, +4 and −4 m, so displacement is 8 m and distance is 16 m.
- Over 6 s, average velocity is 8/6 = 4/3 m s⁻¹ and average speed is 16/6 = 8/3 m s⁻¹.
Accelerations 0 then −2 m s⁻²; displacement 8 m; distance 16 m; average velocity 4/3 m s⁻¹; average speed 8/3 m s⁻¹.
Look for these in your work
- Separate the zero crossing from a change in acceleration.
- Divide both whole-journey quantities by the full 6 s.
Common mistakes
- A graph’s height is velocity, not acceleration.
- Below-axis area subtracts from displacement but adds to distance.
- Split an area at each zero-velocity crossing when finding distance.
What can you explain now?
What does a horizontal velocity graph at v = −4 m s⁻¹ mean?
Compare with the explanation
Constant negative velocity and zero acceleration.
The particle moves at speed 4 m s⁻¹ in the negative direction rather than remaining at rest.
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Make it stick
A velocity graph’s gradient gives acceleration and its signed area gives displacement; split at v = 0 for total distance.
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