Recall and apply the equations for uniformly accelerated motion to one-dimensional problems, including problems involving free fall under gravity

What this dot point is asking

QCAA wants you to apply the four standard equations of uniformly accelerated motion to one-dimensional problems, including objects in free fall (acceleration $g = 9.8$ m s$^{-2}$ downward). The equations only apply when acceleration is constant. The IA1 and EA both reward fluent identification of the known and unknown quantities and clean substitution.

The four equations

Using $u$ for initial velocity, $v$ for final velocity, $a$ for acceleration, $s$ for displacement and $t$ for time:

DMATH_0
DMATH_1
DMATH_2

Each equation has four of the five variables. Identify what you know and what you want, then pick the equation that contains those four.

When the equations apply

• Motion is along a straight line.
• Acceleration is constant (including zero or $g$).
• Motion is from $t = 0$ with the stated initial conditions.

If acceleration changes during the motion, split the journey into segments and apply the equations to each segment separately. Do not use suvat across a segment where the acceleration jumps.

Sign conventions

Pick a positive direction at the start and apply it consistently. Common choices:

• Horizontal motion: positive in the direction of initial motion.
• Vertical motion: up positive (then $a = -g = -9.8$ m s$^{-2}$) or down positive (then $a = +g = +9.8$ m s$^{-2}$).

Stick to one convention through the whole problem.

Free fall

Air resistance is ignored unless the problem states otherwise. The acceleration is $g = 9.8$ m s$^{-2}$ directed downward.

For an object dropped from rest, $u = 0$; for an object thrown upward, $u$ is positive (up positive) and the velocity becomes negative after the peak.

At the highest point of flight, $v = 0$. Use this to find time to peak ($t = u/g$) and peak height ($s = u^2 / (2g)$).

Worked example

A car accelerates uniformly from $5.0$ m s$^{-1}$ to $25$ m s$^{-1}$ over a distance of $90$ m. Find the acceleration and the time taken.

Knowns: $u = 5.0$, $v = 25$, $s = 90$.

Acceleration from $v^2 = u^2 + 2as$:

$25^2 = 5.0^2 + 2 a (90)$

$625 = 25 + 180 a$

$a = 600 / 180 = 3.33$ m s$^{-2}$.

Time from $v = u + at$:

$25 = 5.0 + 3.33 t$

$t = 20 / 3.33 = 6.0$ s.

Common traps

Applying suvat across a change in acceleration. If a car accelerates then brakes, you cannot use one equation across both phases. Split the problem.

Mixing sign conventions. Choosing up as positive then writing $a = +9.8$ in a free-fall problem is the most common QCAA marking deduction.

Using $g = 10$ when the question says $9.8$. QCAA writes problems with $g = 9.8$ m s$^{-2}$ unless otherwise stated. Read the question.

Forgetting a negative root. $v^2 = u^2 + 2as$ gives two solutions for $v$. Pick the physically sensible one (direction of motion).

Try it: Projectile motion calculator handles two-dimensional cases; for one-dimensional motion the same equations apply with $\theta = 0$ or $\theta = 90°$.

In one sentence

The four constant-acceleration equations ($v = u + at$, $s = ut + \frac{1}{2}at^2$, $v^2 = u^2 + 2as$, $s = \frac{1}{2}(u+v)t$) solve any one-dimensional motion (including free fall with $a = g = 9.8$ m s$^{-2}$ downward) provided acceleration is constant and a consistent positive direction is chosen.