Motion graphs: position, velocity and acceleration vs time (QCE Physics Unit 2)

What this dot point is asking

QCAA wants you to read motion graphs fluently. Three types appear: position-time ($x$-$t$), velocity-time ($v$-$t$) and acceleration-time ($a$-$t$). The relationships between them are slope and area, and the same data set will appear on all three graphs in different forms.

The three graphs and what they show

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

• Slope at a point = instantaneous velocity.
• Horizontal line = stationary object.
• Straight sloping line = constant velocity.
• Curved line = changing velocity (acceleration).

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

• Slope at a point = instantaneous acceleration.
• Horizontal line = constant velocity.
• Straight sloping line = constant acceleration.
• Area under the graph = displacement (with sign).

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

• Horizontal line = constant acceleration.
• Area under the graph = change in velocity, $\Delta v$.

Going between graphs

You can build each graph from the previous one by reading slopes and areas.

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

For uniformly accelerated motion, $x$-$t$ is a parabola, $v$-$t$ is a straight line and $a$-$t$ is constant. For free fall, $v$-$t$ has a constant slope of $-g$ if up is taken as positive.

Areas with sign

Area below the time axis is negative. A $v$-$t$ graph that goes positive then negative shows an object that moves forward, stops, and returns. The net displacement is the algebraic sum (forward area minus return area). Total distance is the sum of absolute areas.

Examples in context

Example 1. A Bremer River bridge load-test produces a velocity-time trace of a truck approaching, crossing and decelerating. The slope of the $v$-$t$ graph is acceleration; the area under is displacement. A run from $20 \text{ m s}^{-1}$ to $4 \text{ m s}^{-1}$ over $8.0 \text{ s}$ gives slope $-2.0 \text{ m s}^{-2}$ and area $\tfrac{1}{2}(20+4) \times 8 = 96 \text{ m}$. QCAA Unit 2 IA1 stimulus often delivers exactly this trace and asks for both interpretations.

Example 2. A Sunshine Coast tidal-study group plots position of a drifter against time and observes a curve. The instantaneous velocity at any point is the tangent slope, the acceleration is the slope of the derived $v$-$t$ graph. When the position-time curve is concave upward, acceleration is positive; concave downward, negative. The QCAA EA Unit 2 data-test commonly gives such an $x$-$t$ curve and asks for the maximum speed time and the average speed.

Try this

Q1. State what the slope and the area under a velocity-time graph represent. [2 marks]

• Cue. Slope = acceleration; area = displacement.

Q2. A car's $v$-$t$ graph shows a constant $20 \text{ m s}^{-1}$ for $10 \text{ s}$, then a linear decrease to zero over $8.0 \text{ s}$. Calculate the total displacement and the deceleration phase magnitude. [3 marks]

• Cue. Area $= 20 \times 10 + \tfrac{1}{2} \times 20 \times 8 = 280 \text{ m}$; $|a| = 20/8 = 2.5 \text{ m s}^{-2}$.

Q3. A truck on the Bremer bridge has a $v$-$t$ trace from $20 \text{ m s}^{-1}$ uniform for $4.0 \text{ s}$, then linear decrease to $4 \text{ m s}^{-1}$ over $8.0 \text{ s}$. (a) Sketch the corresponding $a$-$t$ graph. (b) Calculate the total displacement. (c) Discuss one source of uncertainty in such IA1 stimulus data. [3+3+2 marks; ISMG: Analysis and interpretation]

• Cue. (a) $a = 0$ then $-2.0 \text{ m s}^{-2}$; (b) $20 \times 4 + \tfrac{1}{2}(20+4) \times 8 = 176 \text{ m}$; (c) sensor sampling rate aliasing.