Interpret and construct position-time, velocity-time and acceleration-time graphs for one-dimensional motion, including reading slope (instantaneous rates) and area (displacement and change in velocity)

What this dot point is asking

VCAA wants you to read motion graphs fluently and to convert between $x$-$t$, $v$-$t$ and $a$-$t$ using slope and area.

Three motion graphs

Position-time ($x$-$t$).

• Slope = instantaneous velocity.
• Horizontal line = stationary.
• Straight slope = constant velocity.
• Curved = changing velocity (acceleration).

Velocity-time ($v$-$t$).

• Slope = instantaneous acceleration.
• Area = displacement (with sign).
• Horizontal line = constant velocity.
• Straight slope = constant acceleration.

Acceleration-time ($a$-$t$).

• Area = change in velocity $\Delta v$.

Reading between graphs

• IMATH_13 -$t$ slope $\to v$-$t$ values.
• IMATH_17 -$t$ slope $\to a$-$t$ values.
• IMATH_21 -$t$ area $\to \Delta v$ added to $v$-$t$.
• IMATH_26 -$t$ area $\to \Delta x$ added to $x$-$t$.

For uniformly accelerated motion, $x$-$t$ is parabolic, $v$-$t$ is linear, $a$-$t$ is constant.

Sign of area

Area above the time axis is positive displacement; below is negative. A round-trip object has zero net displacement but positive total distance (sum of absolute areas).

Worked example

A ball thrown straight up at $19.6$ m s$^{-1}$ returns to the launcher.

$v$-$t$ graph: straight line from $(0, +19.6)$ with slope $-9.8$ m s$^{-2}$. Reaches zero at $t = 2.0$ s (peak). Continues to $(4.0, -19.6)$ at return.

Displacement: triangle above (area $19.6$ m, going up) + triangle below (area $-19.6$ m, returning) = $0$ net.

Distance: $39.2$ m total.

Common traps

Reading $x$-$t$ slope as displacement. Slope is velocity. Displacement is read off the vertical axis.

Treating area on $x$-$t$ as meaningful. Only $v$-$t$ and $a$-$t$ areas matter.

Ignoring sign. Negative area on $v$-$t$ reduces net displacement.

Confusing straight $v$-$t$ with constant velocity. A straight $v$-$t$ line means constant acceleration (zero acceleration if horizontal).

In one sentence

The slope of $x$-$t$ is velocity, the slope of $v$-$t$ is acceleration, the area under $v$-$t$ is displacement (signed), and the area under $a$-$t$ is $\Delta v$, which lets you convert between the three motion graphs for any one-dimensional journey.