Analyse the linear motion of an object using graphs of position, velocity and acceleration against time, interpreting slope and area under the graph

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.

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

Worked example

A ball is thrown straight up at $19.6$ m s$^{-1}$ from ground level. Sketch its velocity-time graph until it returns to the launcher.

Take up as positive. At $t = 0$, $v = +19.6$ m s$^{-1}$. The slope is $-g = -9.8$ m s$^{-2}$.

$v$ reaches zero at $t = 19.6 / 9.8 = 2.0$ s (peak of flight).

$v$ continues to decrease, reaching $-19.6$ m s$^{-1}$ at $t = 4.0$ s (back at launcher).

The graph is one straight line from $(0, +19.6)$ to $(4.0, -19.6)$. Area above the axis (a triangle of area $\frac{1}{2}(2.0)(19.6) = 19.6$ m) is the rise; area below (the same triangle reflected) is the descent. Net displacement = 0, total distance = $39.2$ m.

Common traps

Reading $x$-$t$ slope as displacement. $x$-$t$ slope is velocity, not displacement. Displacement is read directly off the vertical axis or as the change in $x$.

Treating area on $x$-$t$ as meaningful. It usually is not. Areas matter on $v$-$t$ and $a$-$t$ graphs, not on $x$-$t$.

Ignoring sign of area. Negative areas reduce net displacement. A round-trip object has zero net displacement but non-zero distance.

Confusing straight $v$-$t$ with constant velocity. A straight $v$-$t$ line means constant acceleration (which can be zero). Only a horizontal $v$-$t$ line means constant velocity.

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 change in velocity, which lets you convert between the three motion graphs for any one-dimensional journey.