In VCE Specialist Mathematics, kinematics involves the study of the motion of points, objects, or systems without considering the forces that cause them to move. When an object moves with constant acceleration (also known as uniform acceleration), its motion can be described using a specific set of algebraic formulas known as the SUVAT equations.
To use the equations of motion, we define five key variables. It is essential to maintain consistent units (typically SI units: metres and seconds).
| Symbol | Quantity | Unit | Description |
|---|---|---|---|
| $s$ | Displacement | $m$ | The change in position relative to a starting point. |
| $u$ | Initial Velocity | $ms^{-1}$ | The velocity at time $t = 0$. |
| $v$ | Final Velocity | $ms^{-1}$ | The velocity at time $t$. |
| $a$ | Acceleration | $ms^{-2}$ | The constant rate of change of velocity. |
| $t$ | Time | $s$ | The time interval over which the motion occurs. |
KEY TAKEAWAY: These equations are only valid when acceleration ($a$) is constant. If acceleration is a function of time, displacement, or velocity (e.g., $a = -x$ or $a = 2v$), you must use calculus (integration) instead of these formulas.
For a particle moving in a straight line with constant acceleration $a$, the following formulas apply:
The equations are derived by anti-differentiating the definition of acceleration:
* Starting with $\frac{dv}{dt} = a$:
$$\int du = \int a \, dt \implies v = at + c$$
Since $v = u$ when $t = 0$, then $c = u$, giving $v = u + at$.
* Starting with $\frac{dx}{dt} = v = u + at$:
$$\int ds = \int (u + at) \, dt \implies s = ut + \frac{1}{2}at^2 + d$$
Since $s = 0$ when $t = 0$, then $d = 0$, giving $s = ut + \frac{1}{2}at^2$.
EXAM TIP: When solving problems, always list your “knowns” ($s, u, v, a, t$) first. Identify which variable is missing and which variable you need to find, then select the equation that links those four specific variables.
In two dimensions, motion is often expressed using vector notation ($\mathbf{i}$ and $\mathbf{j}$ components). The constant acceleration equations extend directly to vectors:
Where:
* $\mathbf{r}_0$ is the initial position vector.
* $\mathbf{r}$ is the position vector at time $t$.
* $\mathbf{s} = \mathbf{r} - \mathbf{r}_0$ is the displacement vector.
In 2D motion, the horizontal ($\mathbf{i}$) and vertical ($\mathbf{j}$) components are independent. You can solve for the $\mathbf{i}$ and $\mathbf{j}$ directions separately using the 1D equations and then combine them.
COMMON MISTAKE: Do not use $v^2 = u^2 + 2as$ with vector quantities directly. The $v^2$ and $u^2$ terms in that equation refer to the magnitudes (speeds) squared, and $s$ refers to the distance or displacement in the direction of acceleration. It is safer to work with components or use the vector dot product forms if necessary.
A common application of constant acceleration is an object falling or thrown near the Earth’s surface.
* Acceleration: $a = -g \approx -9.8 \, ms^{-2}$ (assuming upwards is positive).
* At the maximum height of a projectile, the vertical component of velocity is zero ($v_y = 0$).
* The time taken to reach maximum height is half the total time of flight (if the launch and landing heights are the same).
| Scenario | Velocity ($v$) | Acceleration ($a$) |
|---|---|---|
| Object rising | Positive (+) | $-9.8$ |
| At peak height | Zero (0) | $-9.8$ |
| Object falling | Negative (-) | $-9.8$ |
VCAA FOCUS: VCAA often tests the distinction between distance and displacement. Displacement is the vector change in position, while distance is the total path length. For an object that changes direction (like a ball thrown up and caught), you must calculate the path segments separately to find the total distance.
STUDY HINT: Sketching a Velocity-Time ($v-t$) graph is an excellent way to visualize motion.
* The gradient of a $v-t$ graph represents the acceleration.
* The area under a $v-t$ graph represents the displacement.
* The total area (treating negative areas as positive) represents the distance travelled.
