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MathidiaMathematics Academy
Variable acceleration
Lesson 70 of 75
Review

Variable-acceleration models

Objective. Build and critique calculus-based motion models.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

Variable-acceleration models use calculus-based equations of motion, rather than constant-acceleration suvat formulas, to represent real situations where forces and hence acceleration change with time or position, such as resistive forces proportional to velocity. Building such a model means forming a differential equation from Newton's second law, F = ma, with a replaced by dv/dt or v dv/dx depending on context, then solving it using the given initial conditions, and critiquing the result against expected physical behaviour (e.g. checking velocity does not become negative or infinite inappropriately). For example, a model with resistance proportional to v gives a = −kv, leading to exponential decay of speed rather than constant deceleration.

Worked example

A particle experiences acceleration a = −2v m/s², where v is its velocity, with v = 10 m/s at t = 0. Critique this variable-acceleration model by finding dv/v in terms of dt and describing the resulting behaviour.

2100
Bar chart of the values 2, 10, 0
  1. 1From a = dv/dt = −2v, separate variables: dv/v = −2 dt
  2. 2Integrating gives ln v = −2t + c, so v = Ae^(−2t)
  3. 3At t = 0, v = 10, so A = 10, giving v = 10e^(−2t)
  4. 4This models exponential decay: velocity approaches zero but never becomes negative, which is physically sensible

Answer: v = 10e^(−2t); velocity decays exponentially towards zero, a physically realistic model

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Common mistake. Students apply suvat equations to situations explicitly described as having variable acceleration, producing an invalid model. Variable acceleration requires calculus methods (differentiation or integration), never constant-acceleration formulas.

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