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

• $x$-$t$ slope $\to v$-$t$ values.
• $v$-$t$ slope $\to a$-$t$ values.
• $a$-$t$ area $\to \Delta v$ added to $v$-$t$.
• $v$-$t$ 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.

Examples in context

Example 1. Bathurst 1000 pit-lane deceleration. A Supercars driver entering Bathurst's pit lane at $80$ km h$^{-1}$ ($22.2$ m s$^{-1}$) must reach the $40$ km h$^{-1}$ ($11.1$ m s$^{-1}$) limit by the timing line. On a velocity-time graph, the slope from $22.2$ to $11.1$ m s$^{-1}$ over a $2.0$ s braking interval has slope (acceleration) $a = (11.1 - 22.2)/2.0 = -5.55$ m s$^{-2}$. The area under that segment, $\tfrac{1}{2}(22.2 + 11.1) \times 2.0 = 33.3$ m, is the displacement during braking. Position-time graphs of the same event are concave-down because the car covers progressively less distance per second.

Example 2. CityLink approach to Burnley Tunnel. A driver decelerates from $100$ km h$^{-1}$ ($27.8$ m s$^{-1}$) to $80$ km h$^{-1}$ ($22.2$ m s$^{-1}$) over $3.0$ s on the descent into Burnley Tunnel, then holds steady speed. On a velocity-time graph, the first segment slopes negatively at $a = (22.2 - 27.8)/3.0 = -1.87$ m s$^{-2}$, then is horizontal. On an acceleration-time graph the same motion appears as a step from $-1.87$ to $0$ m s$^{-2}$. The displacement during deceleration equals the trapezoidal area $\tfrac{1}{2}(27.8 + 22.2) \times 3.0 = 75$ m, useful for designing the tunnel approach distance to the toll gantry.

Try this

Q1. Define instantaneous acceleration from a velocity-time graph. [2 marks]

• Cue. Slope of the tangent to the velocity-time curve at a given instant.

Q2. A cyclist travels at $5$ m s$^{-1}$ for $10$ s, then accelerates at $0.5$ m s$^{-2}$ for $20$ s. Calculate (a) the final velocity, and (b) the total displacement. [4 marks]

• Cue. (a) $v = 5 + 0.5 \times 20 = 15$ m s$^{-1}$. (b) $5\times 10 + \tfrac{1}{2}(5+15)\times 20 = 50 + 200 = 250$ m.

Q3. Refer to the Bathurst pit-lane example. (a) Calculate the deceleration. (b) Determine the braking distance. (c) Sketch the corresponding position-time graph and describe its curvature. [2+2+2 marks]

• Cue. (a) $-5.55$ m s$^{-2}$. (b) $33.3$ m. (c) Concave-down because slope (velocity) decreases over time.