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AS · Practice questions · Linear motion and graphs

Let the graph speak.

Six original Cambridge-style questions on distance and displacement, gradients and areas, the meaning of graph shapes, and average speed versus average velocity.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Gradient and area do the work.

01
Analysis
[2 marks]

A walker travels 6.0 km due north, then 8.0 km due east. State the total distance walked and find the magnitude of the displacement from the start.

  • Distance is the total path: 6.0 + 8.0 = 14 km. ✓
  • Displacement is the straight line: √(6.0² + 8.0²) = √100 = 10 km. ✓
02
Analysis
[2 marks]

A velocity-time graph for a train is a horizontal line at 25 m s⁻¹ lasting 40 s. Use the area to find the distance travelled, and state the acceleration.

  • Distance = area = 25 × 40 = 1000 m. ✓
  • The line is horizontal, so the gradient and hence the acceleration is zero. ✓
03
Analysis
[3 marks]

A velocity-time graph rises in a straight line from 0 to 18 m s⁻¹ over 6.0 s. Find the acceleration and the distance travelled.

  • Acceleration = gradient = 18 / 6.0 = 3.0 m s⁻². ✓
  • Distance = area of triangle = ½ × 6.0 × 18 = 54 m. ✓
04
Analysis
[2 marks]

Describe the motion shown by a displacement-time graph that is a horizontal straight line, and then by one that is a straight line sloping downward.

  • Horizontal line: the displacement is not changing, so the object is stationary. ✓
  • Straight line sloping downward: constant negative gradient, so constant velocity in the negative direction (moving back toward the origin). ✓
05
Analysis
[3 marks]

A car accelerates uniformly from 8.0 m s⁻¹ to 20 m s⁻¹ in 4.0 s. Sketch the shape of its velocity-time graph in words, and use the area to find the distance travelled.

  • The graph is a straight line sloping upward from 8.0 to 20 m s⁻¹ over 4.0 s. ✓
  • Area = a rectangle (8.0 × 4.0 = 32 m) plus a triangle (½ × 4.0 × 12 = 24 m). ✓
  • Total distance = 32 + 24 = 56 m. (Equivalently, average velocity 14 m s⁻¹ × 4.0 s.) ✓
06
Analysis
[3 marks]

A swimmer covers 50 m to the far wall in 25 s, then 50 m back in 25 s, finishing where they started. Find the average speed and the average velocity for the whole swim, and explain why they differ.

  • Average speed = total distance / time = 100 / 50 = 2.0 m s⁻¹. ✓
  • Average velocity = displacement / time = 0 / 50 = 0 m s⁻¹. ✓
  • They differ because speed uses the path length (a scalar) while velocity uses displacement (a vector), which is zero for a round trip. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.