A/L Combined Mathematics · Revision

Applied: Newtonian Dynamics and Motion

Revise Applied: Newtonian Dynamics and Motion fast with concise, exam-ready key points · Aligned with the NIE syllabus

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Grade 2 · Term 5

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Revise Applied: Newtonian Dynamics and Motion fast with concise, exam-ready key points

About A/L Combined Mathematics: Combined Mathematics consists of Pure Mathematics and Applied Mathematics across two years.

  • From t=0 to t=2s, displacement increases linearly from 0 to 10m. - From t=2s to t=5s, displacement remains constant at 10m. - From t=5s to t=8s, displacement decreases linearly from 10m to 0m
  • Identify initial displacement (at t=0s) = 0 m and final displacement (at t=2s) = 10 m. 2. The time interval is 2 s - 0 s = 2 s. 3. Average Velocity = (Change in Displacement) / (Change in Time) = (10 m - 0 m) / 2 s = 5 m/s. (b) Instantaneous velocity at t = 3 seconds:
  • From t=2s to t=5s, the displacement-time graph is a horizontal line, indicating constant displacement at 10 m. 2. The gradient of a horizontal line is zero. 3. Instantaneous Velocity = Gradient of the displacement-time graph = 0 m/s. The particle is stationary. (c) Average velocity for the entire 8-second journey:
  • Identify initial displacement (at t=0s) = 0 m and final displacement (at t=8s) = 0 m. 2. Total time = 8 s. 3. Average Velocity = (Total Displacement) / (Total Time) = (0 m - 0 m) / 8 s = 0 m/s. (d) Motion between t=5s and t=8s:
  • The graph shows a straight line with a negative gradient. 2. Initial displacement (at t=5s) = 10 m and final displacement (at t=8s) = 0 m. 3. Gradient = (0 m - 10 m) / (8 s - 5 s) = -10 m / 3 s ≈ -3.33 m/s. 4. This means the particle is moving with a constant velocity of approximately 3.33 m/s in the negative direction (returning to its starting point). ⚠ Tricky: Distinguishing between average and instantaneous velocity, especially when the velocity changes. ⚠ Tricky: Understanding that a horizontal line on a displacement-time graph signifies zero velocity (the object is at rest). ⚠ Tricky: Recognizing that a negative gradient on a displacement-time graph means motion in the opposite direction from the positive reference

Source: Idasara knowledge pack — english/AL_SCIENCE/combined_mathematics v13, short_notes: Lesson 20 - Competency 3 _ Applies the Newtonian model to describe the instantaneous of moti

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