Velocity–time graphs | CAIE 9709 Mechanics Notes

Mechanics

Mathematics9709

Use gradient for acceleration

At time $t$ , the graph height $v$ is the signed velocity. Above the time axis the particle moves in the positive direction; below it the particle moves in the negative direction. Its speed is $|v|$ , the distance from the time axis. The bars mean magnitude, so a negative velocity becomes a positive speed.

a=\dfrac{\Delta v}{\Delta t}

Symbols

$a$ = acceleration over a straight graph segment (m s⁻²)
$\Delta v$ = signed change in velocity (m s⁻¹)
$\Delta t$ = change in time (s)

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Above the time axis, area contributes positive displacement. Below it, area contributes negative displacement. Total distance adds the magnitudes of both regions.

Key idea

A downward graph segment has negative acceleration. Whether the particle slows down depends on whether that segment lies above or below the time axis.

Use signed area for displacement

Displacement is the signed area between the graph and the time axis.

Area above the axis is positive displacement.
Area below the axis is negative displacement.
Total distance adds the magnitudes of all separate areas.

The area unit is $(\text{m s}^{-1})(\text{s})=\text{m}$ , so an area represents a displacement, not a velocity.

Forward peak velocity8 m s⁻¹Return peak speed3 m s⁻¹

displacement = 10 m

distance = 22 m

The return region reduces displacement, but it still increases total distance.

Examiner note

A 2024 graph question awarded the method mark for using 68 m as the total area, including the region below the axis as positive distance. (9709/42/O/N/24 Q1)

Treat every axis crossing as a decision point

At $v=0$ the particle is instantaneously at rest. It changes direction only if the graph crosses the axis and the sign of velocity changes. Merely touching the axis does not prove a reversal.

Mark every time when $v=0$ .
Split the graph into regions with one velocity sign.
Add signed areas for displacement.
Add absolute areas for total distance.

Your turnIndependent check [4 marks]

A velocity–time graph is made of straight segments through $(0,0)$ , $(4,10)$ , $(10,0)$ , $(14,-3)$ and $(18,0)$ . Find the initial acceleration, the displacement and the total distance travelled.

Show worked answer

a=\dfrac{10-0}{4-0}=2.5\text{ m s}^{-2}

A_+=\tfrac12(10)(10)=50\text{ m}

A_-=\tfrac12(8)(3)=12\text{ m}

50-12=38\text{ m}

\text{distance}=50+12=62\text{ m}

Answer

2.5\text{ m s}^{-2}

38\text{ m}

62\text{ m}