Uniform acceleration equations (suvat) (QCE Physics Unit 2)

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:

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 ($s = u^2 / (2g)$).

Examples in context

Example 1. A Bremer River bridge safety design tests a $1500 \text{ kg}$ crash barrier with an instrumented sled released from $4.0 \text{ m}$ height. Vertical SUVAT: $v^2 = 2gs$ gives $v = 8.85 \text{ m s}^{-1}$ at impact. With crumple distance $0.30 \text{ m}$, deceleration is $a = v^2/(2s) = 131 \text{ m s}^{-2}$, force $F = ma = 1.96 \times 10^5 \text{ N}$. SUVAT plus Newton's second law are the QCAA Unit 2 staples for engineering-safety stimulus.

Example 2. A Cairns light-rail tram driver brakes at $1.4 \text{ m s}^{-2}$ from $14 \text{ m s}^{-1}$. Stopping distance $s = v^2/(2a) = 196/2.8 = 70.0 \text{ m}$. Time to stop $t = v/a = 10.0 \text{ s}$. QCAA EA Unit 2 frequently sets this with one piece of missing data, requiring the student to choose the correct SUVAT equation from the four-equation set on the QCAA formula sheet.

Try this

Q1. List the four SUVAT equations and state the conditions under which they apply. [2 marks]

• Cue. $v = u + at$; $s = ut + \tfrac{1}{2}at^2$; $v^2 = u^2 + 2as$; $s = \tfrac{1}{2}(u+v)t$; constant acceleration.

Q2. A stone is dropped from a $45 \text{ m}$ cliff. Calculate the time to reach the ground and the impact velocity, using $g = 9.8 \text{ m s}^{-2}$. [3 marks]

• Cue. $s = \tfrac{1}{2}gt^2$ gives $t = 3.03 \text{ s}$; $v = gt = 29.7 \text{ m s}^{-1}$.

Q3. A Cairns tram brakes from $14 \text{ m s}^{-1}$ at $1.4 \text{ m s}^{-2}$. (a) Calculate the stopping distance and time. (b) Determine the velocity after $5.0 \text{ s}$ of braking. (c) Discuss one assumption in applying constant-acceleration SUVAT to a real tram. [3+2+2 marks; ISMG: Analysis and interpretation, Evaluation]

• Cue. (a) $70 \text{ m}$, $10 \text{ s}$; (b) $v = 14 - 1.4 \times 5 = 7.0 \text{ m s}^{-1}$; (c) braking is not perfectly uniform - blends regenerative and friction.