Straight-line kinematics
Use signed position relative to a chosen origin. Differentiate it once for velocity and twice for acceleration, then use velocity signs and test for direction reversals
\[v(t)=x'(t),\quad a(t)=v'(t)=x''(t)\]
- Differentiate displacement once Velocity is the derivative of displacement with respect to time \(\displaystyle x(t)=t^{3}-6t^{2}+9t+2\quad\Rightarrow\quad v(t)=3t^{2}-12t+9\)
- Differentiate velocity Acceleration is the derivative of velocity with respect to time \(\displaystyle v(t)=3t^{2}-12t+9\quad\Rightarrow\quad a(t)=6t-12\)
- Use velocity for rest and direction Rest at an instant means velocity is zero. The velocity sign gives the direction of motion. A rest time is a direction reversal only if velocity changes sign across it \(\displaystyle v(t)=0\quad\Rightarrow\quad 3(t-1)(t-3)=0\quad\Rightarrow\quad t=1,3\)
- Add path lengths for distance For the cubic model above on 0 ≤ t ≤ 4 seconds, positions at times 0, 1, 3 and 4 are 2, 6, 2 and 6 metres. Split at every velocity zero inside the interval and add the absolute position changes; a split without reversal is harmless \(\displaystyle D=|6-2|+|2-6|+|6-2|=12\text{ m}\)
Worked example
A particle has the position model below, measured in metres, for the first 6 seconds. Find when it is at rest and the distance travelled in this interval
\[x(t)=t^{2}-6t+8\]
- \(\displaystyle v(t)=x'(t)=2t-6\)Differentiate displacement to obtain velocity
- \(\displaystyle v(t)=0\quad\Rightarrow\quad 2t-6=0\quad\Rightarrow\quad t=3\)The particle is at rest when its velocity is zero
- \(\displaystyle x(0)=8,\quad x(3)=-1,\quad x(6)=8\)Evaluate displacement at the endpoints and at the rest time
- \(\displaystyle D=|-1-8|+|8-(-1)|=18\text{ m}\)Add the absolute changes in displacement on the two parts of the path
Common mistakes
Using displacement for rest
✕\(\displaystyle x(t)=0\quad\text{for rest}\)✓\(\displaystyle v(t)=0\quad\text{for rest}\)
Rest at an instant means v(t)=0. The condition x(t)=0 locates the origin instead; a zero at one instant does not establish rest over an intervalCalling displacement velocity
✕\(\displaystyle v(t)=x(t)\)✓\(\displaystyle v(t)=x'(t)\)
Velocity is the rate of change of displacement, so it is found by differentiating x(t)Using displacement for average speed
✕\(\displaystyle \text{average speed}=\frac{x(4)-x(0)}{4}=1\text{ m/s}\)✓\(\displaystyle \text{average speed}=\frac{D}{4}=3\text{ m/s}\)
For the cubic model on 0 ≤ t ≤ 4 seconds, distance is 12 m and displacement change is 4 m. Average speed uses distance divided by elapsed time; average velocity uses displacement changeDifficulty levels
introFind velocity or acceleration from a displacement function and evaluate one time
standardFind rest times and direction changes for a polynomial displacement function
challengeFind distance, average velocity or average speed by splitting the path at rest times
examSolve a mixed kinematics question involving derivatives, signs and motion over an interval
Questions about this topic
How are velocity and acceleration related to displacement?+
Velocity is the first derivative of displacement. Acceleration is the second derivative of displacement, or the derivative of velocity
Why can distance and displacement be different?+
The coordinate x(t) is signed position relative to the chosen origin. Displacement over an interval is the signed change x(b)−x(a). Distance adds the lengths of all travelled segments and is never negative
How is the direction of motion found?+
Check the sign of velocity. Positive velocity means the displacement is increasing, and negative velocity means it is decreasing
Does every rest time reverse direction?+
No. Suppose displacement is the cube of the difference between time and one, from zero to two seconds. Velocity is three times the square of that difference. It is positive on both sides of one second, so there is momentary rest without reversal
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 6: Applications of differentiation
In this chapter
Differentiate polynomialsChallenge enrichment: rational powers and derivative domainsDifferentiate natural exponential functions and solve simple derivative conditionsDifferentiate sine, cosine and tangent functions and solve simple derivative conditionsDifferentiate a composite function by the chain ruleDifferentiate a product or quotient of two functionsEvaluate and determine function rules6B Differentiation rules: Find the maximal domain of an algebraic functionExpand and factor polynomial expressionsSimplify and combine algebraic fractions6B Differentiation rules: Find the equation of a line from its graph or from its intercept and slopeAnalyse polynomial curves, stationary points and extremaUse a derivative to analyse and recognise the graph of a function6C Tangents, normals and curve sketching: Evaluate a polynomial or recover a coefficient parameter6C Tangents, normals and curve sketching: Move between roots, factors, polynomial rules and equationsRecognise and classify a polynomialConnect a factorised higher-degree rule and its graphAnalyse a cubic graph or determine its ruleWork with sets and interval notation6C Tangents, normals and curve sketching: Find the maximal domain of an algebraic function6C Tangents, normals and curve sketching: Find the equation of a line from its graph or from its intercept and slopeForm an equation of a parallel or perpendicular lineSolve a maximum or minimum problem by differentiationSet up and solve a practical linear equationRearrange literal equations and recognise equivalent line forms6D Optimisation: Find the maximal domain of an algebraic function6D Optimisation: Evaluate a polynomial or recover a coefficient parameter6D Optimisation: Move between roots, factors, polynomial rules and equationsCalculate and interpret average and instantaneous rates from a modelVolume of cones, pyramids and spheresVolume of prisms and cylindersPerimeter and area of standard shapesSimilar figures and scale factorsFind a side of a right-angled triangleAnalyse straight-line motion through displacement, velocity and acceleration6F Kinematics: Evaluate a polynomial or recover a coefficient parameter