Search topics

Search

Analyse straight-line motion through displacement, velocity and acceleration

Start a set of 8
Step 3 of 8

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

\[v(t)=x'(t),\quad a(t)=v'(t)=x''(t)\]
  1. 1Differentiate displacement once Velocity is the derivative of displacement with respect to time \(\displaystyle x(t)=t^{3}-6t^{2}+9t+2\quad\Rightarrow\quad v(t)=3t^{2}-12t+9\)
  2. 2Differentiate velocity Acceleration is the derivative of velocity with respect to time \(\displaystyle v(t)=3t^{2}-12t+9\quad\Rightarrow\quad a(t)=6t-12\)
  3. 3Use 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 \(\displaystyle v(t)=0\quad\Rightarrow\quad 3(t-1)(t-3)=0\quad\Rightarrow\quad t=1,3\)
  4. 4Add 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 \(\displaystyle D=|6-2|+|2-6|+|6-2|=12\text{ m}\)
Year 12, VCE Units 3 & 4, Chapter 6, 3 difficulty levels, No calculatorBuild a worksheet

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

\[v(t)=x'(t),\quad a(t)=v'(t)=x''(t)\]
  1. Differentiate displacement once Velocity is the derivative of displacement with respect to time \(\displaystyle x(t)=t^{3}-6t^{2}+9t+2\quad\Rightarrow\quad v(t)=3t^{2}-12t+9\)
  2. Differentiate velocity Acceleration is the derivative of velocity with respect to time \(\displaystyle v(t)=3t^{2}-12t+9\quad\Rightarrow\quad a(t)=6t-12\)
  3. 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 \(\displaystyle v(t)=0\quad\Rightarrow\quad 3(t-1)(t-3)=0\quad\Rightarrow\quad t=1,3\)
  4. 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 displacement rule shown below, in metres, from 0 to 6 seconds. Find when it is at rest and the distance travelled in this interval

\[x(t)=t^{2}-6t+8\]
  1. \(\displaystyle v(t)=x'(t)=2t-6\)Differentiate displacement to obtain velocity
  2. \(\displaystyle v(t)=0\quad\Rightarrow\quad 2t-6=0\quad\Rightarrow\quad t=3\)The particle is at rest when its velocity is zero
  3. \(\displaystyle x(0)=8,\quad x(3)=-1,\quad x(6)=8\)Evaluate displacement at the endpoints and at the rest time
  4. \(\displaystyle D=|-1-8|+|8-(-1)|=18\text{ m}\)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 interval
Calling 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 change

Difficulty 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

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. For x(t)=(t−1)³ on 0 ≤ t ≤ 2, velocity is 3(t−1)². It is positive on both sides of t=1, 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, Specialist Mathematics, chapter 6: Differential equations and rectilinear motion

In this chapter

6A Models, verification and direction fields: Form differential models, verify solutions and read direction fieldsExponential growth and decay rate modelsSeparate variables, impose initial values and retain solution domainsAnti-differentiate a power of a linear functionDifferentiate parametric curves and integrate vector motionSolve and interpret continuous logistic models6C Logistic growth and applications: Form differential models, verify solutions and read direction fieldsApproximate solutions with explicit Euler stepsBinary place value and algorithm tracesUse acceleration forms and differential equations in rectilinear motionAnalyse straight-line motion through displacement, velocity and acceleration