Motion graphs and acceleration (VCE Physics Unit 2)

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.