Calculus of motion | CAIE 9709 Mechanics Notes

Mechanics

Mathematics9709

Differentiate to move from position to motion

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Differentiate to move right through the chain. Integrate to move left, then use a stated value of velocity or position to determine the constant.

v=\dfrac{ds}{dt}

a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}

Worked example

Stationary does not automatically mean turning

If $s=t^3-6t^2+9t$ , then $v=3(t-1)(t-3)$ . The particle is stationary at $t=1$ and $t=3$ . For a sign chart, these times split the timeline into intervals; test one time in each interval. The signs are positive, then negative, then positive, so the particle changes direction at both times.

Common mistake

Solving

v=0

finds stationary times. It does not by itself prove a direction change; check the sign of velocity on both sides.

Integrate, then use a boundary value

Integration produces a family of functions that differ by a constant. A boundary value states the value of $v$ or $s$ at a named time and selects the one function that fits the motion.

v=\int a\,dt+C_1

s=\int v\,dt+C_2

Worked example

Use each initial condition at the correct level

A particle has $a=6t-4$ , $v=3$ when $t=0$ , and $s=2$ when $t=0$ .

v=3t^2-4t+C_1

C_1=3,\quad v=3t^2-4t+3

s=t^3-2t^2+3t+C_2

C_2=2

s=t^3-2t^2+3t+2

Examiner note

In a 2024 piecewise-acceleration question, the second velocity constant had to be found from the velocity carried across the boundary at

t=1

; substituting a made-up value earned no method mark. (9709/42/O/N/24 Q6)

Split total distance wherever velocity changes sign

Swipe left or right to see the whole diagram →

Solve v = 0 first. Those times are the only possible direction changes, so they are the cut points for total distance.

Solve $v(t)=0$ inside the stated time interval.
Use those times to split the integral or the displacement function.
Find each signed displacement separately.
Add their magnitudes for total distance.

Key idea

The direct integral

\int_a^b v\,dt

gives displacement. It gives total distance only when velocity keeps one sign throughout the interval.

Your turnIndependent check [6 marks]

A particle has velocity $v=t^2-6t+5$ for $0\le t\le6$ . Find its displacement and total distance travelled during the 6 s.

Show worked answer

v=(t-1)(t-5)=0

t=1,5

Let $S(t)$ be an antiderivative of $v(t)$ : this means $S'(t)=v(t)$ . Only differences such as $S(5)-S(1)$ are used, so its arbitrary constant cancels; choose it as zero.

S(t)=\tfrac13t^3-3t^2+5t

S(0)=0,\qquad S(1)=\tfrac73

S(5)=-\tfrac{25}{3},\qquad S(6)=-6

Displacement:

S(6)-S(0)=-6\text{ m}

Distance:

\tfrac73+\tfrac{32}{3}+\tfrac73=\tfrac{46}{3}\text{ m}

Answer

\text{displacement}=-6\text{ m}

\text{distance}=15.3\text{ m}

Differentiate to move from position to motion

Swipe left or right to see the whole diagram →

Differentiate to move right through the chain. Integrate to move left, then use a stated value of velocity or position to determine the constant.

v=\dfrac{ds}{dt}

a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}

Worked example

Stationary does not automatically mean turning

Common mistake

Solving

v=0

finds stationary times. It does not by itself prove a direction change; check the sign of velocity on both sides.

Integrate, then use a boundary value

Integration produces a family of functions that differ by a constant. A boundary value states the value of $v$ or $s$ at a named time and selects the one function that fits the motion.

v=\int a\,dt+C_1

s=\int v\,dt+C_2

Worked example

Use each initial condition at the correct level

A particle has $a=6t-4$ , $v=3$ when $t=0$ , and $s=2$ when $t=0$ .

v=3t^2-4t+C_1

C_1=3,\quad v=3t^2-4t+3

s=t^3-2t^2+3t+C_2

C_2=2

s=t^3-2t^2+3t+2

Examiner note

In a 2024 piecewise-acceleration question, the second velocity constant had to be found from the velocity carried across the boundary at

t=1

; substituting a made-up value earned no method mark. (9709/42/O/N/24 Q6)

Split total distance wherever velocity changes sign

Swipe left or right to see the whole diagram →

Solve v = 0 first. Those times are the only possible direction changes, so they are the cut points for total distance.

Solve $v(t)=0$ inside the stated time interval.
Use those times to split the integral or the displacement function.
Find each signed displacement separately.
Add their magnitudes for total distance.

Key idea

The direct integral

\int_a^b v\,dt

gives displacement. It gives total distance only when velocity keeps one sign throughout the interval.

Your turnIndependent check [6 marks]

A particle has velocity $v=t^2-6t+5$ for $0\le t\le6$ . Find its displacement and total distance travelled during the 6 s.

Show worked answer

v=(t-1)(t-5)=0

t=1,5

Let $S(t)$ be an antiderivative of $v(t)$ : this means $S'(t)=v(t)$ . Only differences such as $S(5)-S(1)$ are used, so its arbitrary constant cancels; choose it as zero.

S(t)=\tfrac13t^3-3t^2+5t

S(0)=0,\qquad S(1)=\tfrac73

S(5)=-\tfrac{25}{3},\qquad S(6)=-6

Displacement:

S(6)-S(0)=-6\text{ m}

Distance:

\tfrac73+\tfrac{32}{3}+\tfrac73=\tfrac{46}{3}\text{ m}

Answer

\text{displacement}=-6\text{ m}

\text{distance}=15.3\text{ m}