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MathidiaMathematics Academy
Variable acceleration
Lesson 69 of 75
Challenge

Acceleration as a function of displacement

Objective. Use v dv/dx to connect acceleration and position.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
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Learn: the key idea

Acceleration as a function of displacement uses the identity a = v(dv/dx), derived from the chain rule applied to dv/dt, to connect acceleration directly to position x rather than time, which is useful when acceleration is given in terms of x. Rearranging gives v dv = a dx, which can be integrated directly with respect to x to find v² in terms of x, avoiding the need to find t explicitly. For example, if a = 2x, then v dv = 2x dx, integrating gives v²/2 = x² + c, so v² = 2x² + 2c.

Worked example

A particle has acceleration a = 6x m/s², where x is displacement, and velocity v = 4 m/s when x = 0. Find v² as a function of displacement x.

6 … 4
Illustrated card showing 6 … 4
  1. 1Use a = v dv/dx, so v dv = 6x dx
  2. 2Integrate: v²/2 = 3x² + c
  3. 3At x = 0, v = 4, so 16/2 = c, giving c = 8
  4. 4v² = 6x² + 16

Answer: v² = 6x² + 16

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Common mistake. Students try to integrate a = v dv/dx with respect to t instead of separating variables and integrating with respect to x. Always rearrange to v dv = a dx first, then integrate both sides with respect to their own variable.

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