Two stages of motion reconstruction
Acceleration determines velocity up to one constant. Specific velocity determines position up to another
\[a\longrightarrow v+C_v\longrightarrow s+C_s\]
- 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
\[a(t)=6t,\quad v(1)=5,\quad s(2)=13\]
- \(\displaystyle v(t)=3t^2+C_v,\quad3+C_v=5\)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 recovers velocity and acceleration; both measured values are satisfied
Common mistakes
Using one condition twice
✕Use the position value to fix both constants✓Use one velocity condition and one position condition
The two stages introduce independent constantsSkipping a stage
✕Anti-differentiate acceleration once and call it position✓Recover velocity first, then position
Acceleration is the second derivative of positionDifficulty levels
introConstant positive acceleration and braking with initial conditions
standardLinear acceleration and conditions at different times
challengeExact endpoint or double-root stops within the model
examDistance with positive velocity and signed displacement from two reconstructed rules
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 11, VCE Units 1 & 2, Mathematical Methods, chapter 14: Anti-differentiation and integrals