In many cases, acceleration changes over time. You see this when a car that has been stopped at a traffic light speeds up to \(\mathbf{60}\,\textbf{km/h}\) after the light changes to green. Acceleration also varies throughout a rollercoaster ride. Both of these scenarios demonstrate non-constant (or non-uniform) acceleration. Use this resource to learn about non-constant acceleration, including how it applies to Aboriginal spear-throwing.
So far, you've used the constant acceleration equations because acceleration stayed fixed throughout the motion. But when acceleration itself changes, those equations no longer apply, and we need a more general approach.
Recall that velocity is the rate of change of displacement, and acceleration is the rate of change of velocity. Written using calculus notation, these are:
\[v=\dfrac{dx}{dt}\qquad a=\dfrac{dv}{dt}\]
This means we can find velocity and acceleration at any instant by differentiating displacement with respect to time – once, to get velocity, and again, to get acceleration.
Integration reverses differentiation, so we can also work backwards: starting from acceleration, we can integrate to find velocity, and integrate again to find displacement.
If you need a refresher on differentiation or integration before diving in, see the further resources at the end of this page.
Indigenous knowledges in physics
When a hunter throws a spear, it starts from rest and accelerates – but not at a constant rate. The force and geometry of the throw change continuously, meaning the spear's acceleration changes throughout the motion. This is non-constant acceleration in action.
Aboriginal peoples across Australia developed the woomera, a wooden spear-throwing tool, to maximise the velocity of the spear at the moment of release. The word comes from the Dharug language of the Eora people of the Sydney Basin. Across the continent, many nations developed their own designs with their own names: miru in Pitjantjatjara, spoken around Uluru; wamirri in Warumungu Country in the central Northern Territory; and meru in Noongar Country in South-West Western Australia, among many others.
Once airborne, the spear travels with constant horizontal velocity and is pulled back to Earth by the constant downward acceleration of gravity. The effectiveness of the woomera is a testament to tens of thousands of years of careful observation and refinement. Archaeological evidence suggests this technology may be over 20,000 years old, making it one of the earliest applications of acceleration physics in human history.
A toy car moves along a straight track. Its position is given by \(\mathbf{x=t^{3}-6t^{2}}\), where \(\mathbf{t}\) is in seconds and \(\mathbf{s}\) is in metres.
For \(v=0\), we simply substitute the value in and solve for \(t\).
\[\begin{align*} 0 & = 3t^{2}-12t \\[6pt]
0 & = 3t(t-4)\\[6pt]
0=3t\quad & \text{ or }\quad 0=t-4\\[6pt]
t=0\,\text{s}\quad & \text{ or }\quad t=4\,\text{s}\end{align*}\] Therefore, the toy car starts from rest and stops at \(4\,\text{s}\).
For \(a=0\), we substitute the value in and solve for \(t\).
\[\begin{align*} 0 & = 6t-12 \\[6pt]
12 & = 6t\\[6pt]
t & = \frac{12}{6}\\[6pt]
t & = 2\,\text{s}\end{align*}\]
We can find the total distance covered using displacement. We can start by finding the starting position, i.e. when \(t=0\,\text{s}\).
\[\begin{align*} x & = 0^{3}-6(0)^{2} \\[6pt]
& = 0\end{align*}\]
The car changes direction when \(v=0\,\text{m/s}\), i.e. at \(t=0\,\text{s}\) and \(t=4\,\text{s}\). We can then find the distance it travels before it changes direction at \(t=4\,\text{s}\).
\[\begin{align*} x & = (4)^{3}-6(4)^{2} \\[6pt]
& = 64-96\\[6pt]
& = -32\,\text{m}\end{align*}\]
From \(t=0\) to \(t=4\), the car travels \(32\,\text{m}\) in the negative direction.
We can now find the position of the car at \(t=5\,\text{s}\).
\[\begin{align*} x & = (5)^{3}-6(5)^{2} \\[6pt]
& = 125-150\\[6pt]
& = -25\,\text{m}\end{align*}\]
From \(t=4\) to \(t=5\), the car changes position from \(-32\,\text{m}\) to \(-25\,\text{m}\), which is \(7\,\text{m}\) in the positive direction.
The total distance travelled is the sum of these: \(32+7=39\,\text{m}\). This is different from the car's displacement, which would be the difference between the displacement at \(t=5\) and \(t=0\): \(-25\,\text{m}\).
To find velocity as function of time, we need to integrate \(a=-0.5v\) with respect to time, \(a=\dfrac{dv}{dt}\). First, we separate the variables, then integrate:
\[\begin{align*} \frac{dv}{dt} & = -0.5v \\[6pt]
\dfrac{1}{v}\,dv & = -0.5\,dt\\[6pt]
\displaystyle\int_{16}^{v} \dfrac{1}{v}\,dv & = \int_{0}^{t} -0.5\,dt\\[6pt]
[\ln v]_{16}^{v} & = [-0.5t]_{0}^{t}\\[6pt]
\ln v - \ln 16 & = -0.5t-(-0.5)(0)\\[6pt]
\ln\left(\dfrac{v}{16}\right) & = -0.5t\\[6pt]
\dfrac{v}{16} & = e^{-0.5t}\\[6pt]
v & = 16e^{-0.5t}\,\text{m}\,\text{s}^{-1}\end{align*}\]
The find position/displacement as a function of time, we integrate \(v=16e^{-0.5t}\) with respect to time, \(v=\dfrac{dx}{dt}\).
\[\begin{align*} \frac{dx}{dt} & = 16e^{-0.5t} \\[6pt]
\displaystyle\int_{0}^{s}\,dx & = 16\int_{0}^{t} e^{-0.5t}\,dt\\[6pt]
x & = 16(-2)\left[e^{-0.5t}\right]_{0}^{t}\\[6pt]
& = -32\left(e^{-0.5t} - 1\right)\\[6pt]
& = 32\left(1 - e^{-0.5t}\right)\end{align*}\]
To find the boat's acceleration at \(t=2\,\text{s}\), we simply substitute the function \(v\) into \(a=-0.5v\), then find \(a\) for \(t=2\).
\[\begin{align*} a & = -0.5\left(16e^{-0.5t}\right) \\[6pt]
& = -8e^{-0.5t}\\[6pt]
& = -8e^{-0.5(2)}\\[6pt]
& = -2.94\,\text{m}\,\text{s}^{-2}\end{align*}\]
Notice that the boat never fully stops in this model; its velocity keeps approaching zero but never quite reaches it. This is typical of resistance forces that depend on speed: as the boat slows, the resistance slowing it down also weakens.
\(12\,\text{m}\)
\(20\,\text{m}\,\text{s}^{-1}\)
\(97\,\text{m}\)
RMIT University acknowledges the people of the Woi wurrung and Boon wurrung language groups of the eastern Kulin Nation on whose unceded lands we conduct the business of the University. RMIT University respectfully acknowledges their Ancestors and Elders, past and present. RMIT also acknowledges the Traditional Custodians and their Ancestors of the lands and waters across Australia where we conduct our business - Artwork 'Sentient' by Hollie Johnson, Gunaikurnai and Monero Ngarigo.
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