Two stages of motion reconstruction
Acceleration determines velocity up to one constant. Specific velocity determines position up to another
- Match each condition Use velocity data for the first constant and position data for the second. Their numerical values may happen to be equal, but the constants are independent
- Retain the first constant After it is fixed, integrate the entire velocity. A constant velocity term becomes a linear position term
- Solve stops with the domain Find velocity zeros exactly and reject roots outside the model. An endpoint stop does not justify a prediction after the domain
- Prove the distance condition Use position increase for distance only when velocity is nonnegative throughout the requested interval
Worked example
Acceleration is 6t m/s². Given v(1) = 5 m/s and s(2) = 13 m, recover velocity and position
- \(\displaystyle v(t)=3t^2+C_v,\quad3+C_v=5\)Apply the velocity condition
- \(\displaystyle v(t)=3t^2+2\)The first constant is 2.
- \(\displaystyle s(t)=t^3+2t+C_s,\quad8+4+C_s=13\)Integrate the specific velocity and apply the position condition
- \(\displaystyle s(t)=t^3+2t+1\)Differentiating gives v = 3t² + 2 and a = 6t; both measured values are satisfied
Common mistakes
Difficulty levels
Questions about this topic
Are sign-changing distance calculations included?+
No. Distance is only requested when velocity is proved nonnegative throughout the requested interval
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 8: Integration