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QLDPhysicsSyllabus dot point

Topic 1: Linear motion and force

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

A focused answer to the QCE Physics Unit 2 dot point on the equations of uniformly accelerated motion. Lists the four QCAA-formulae-sheet suvat equations, the conditions under which they apply, and works the free-fall standard question that recurs in IA1 and EA Paper 1.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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Jump to a section
  1. What this dot point is asking
  2. The four equations
  3. When the equations apply
  4. Sign conventions
  5. Free fall
  6. Examples in context
  7. Try this

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=9.8g = 9.8 m s2^{-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 uu for initial velocity, vv for final velocity, aa for acceleration, ss for displacement and tt for time:

v=u+atv = u + at

s=ut+12at2s = ut + \tfrac{1}{2} a t^2

v2=u2+2asv^2 = u^2 + 2 a s

s=12(u+v)ts = \tfrac{1}{2}(u + v) t

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 gg).
  • Motion is from t=0t = 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=9.8a = -g = -9.8 m s2^{-2}) or down positive (then a=+g=+9.8a = +g = +9.8 m s2^{-2}).

Stick to one convention through the whole problem.

Free fall

Air resistance is ignored unless the problem states otherwise. The acceleration is g=9.8g = 9.8 m s2^{-2} directed downward.

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

At the highest point of flight, v=0v = 0. Use this to find time to peak (t=u/gt = u/g) and peak height (s=u2/(2g)s = u^2 / (2g)).

Examples in context

Example 1. A Bremer River bridge safety design tests a 1500 kg1500 \text{ kg} crash barrier with an instrumented sled released from 4.0 m4.0 \text{ m} height. Vertical SUVAT: v2=2gsv^2 = 2gs gives v=8.85 m s1v = 8.85 \text{ m s}^{-1} at impact. With crumple distance 0.30 m0.30 \text{ m}, deceleration is a=v2/(2s)=131 m s2a = v^2/(2s) = 131 \text{ m s}^{-2}, force F=ma=1.96×105 NF = 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 m s21.4 \text{ m s}^{-2} from 14 m s114 \text{ m s}^{-1}. Stopping distance s=v2/(2a)=196/2.8=70.0 ms = v^2/(2a) = 196/2.8 = 70.0 \text{ m}. Time to stop t=v/a=10.0 st = 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+atv = u + at; s=ut+12at2s = ut + \tfrac{1}{2}at^2; v2=u2+2asv^2 = u^2 + 2as; s=12(u+v)ts = \tfrac{1}{2}(u+v)t; constant acceleration.

Q2. A stone is dropped from a 45 m45 \text{ m} cliff. Calculate the time to reach the ground and the impact velocity, using g=9.8 m s2g = 9.8 \text{ m s}^{-2}. [3 marks]

  • Cue. s=12gt2s = \tfrac{1}{2}gt^2 gives t=3.03 st = 3.03 \text{ s}; v=gt=29.7 m s1v = gt = 29.7 \text{ m s}^{-1}.

Q3. A Cairns tram brakes from 14 m s114 \text{ m s}^{-1} at 1.4 m s21.4 \text{ m s}^{-2}. (a) Calculate the stopping distance and time. (b) Determine the velocity after 5.0 s5.0 \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 m70 \text{ m}, 10 s10 \text{ s}; (b) v=141.4×5=7.0 m s1v = 14 - 1.4 \times 5 = 7.0 \text{ m s}^{-1}; (c) braking is not perfectly uniform - blends regenerative and friction.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Year 11 SAC4 marksA stone is dropped from rest from the top of a tall building and takes 3.23.2 s to reach the ground. Ignoring air resistance and using g=9.8g = 9.8 m s2^{-2}, calculate (a) the height of the building and (b) the speed of the stone when it lands.
Show worked answer →

Take down as positive. Initial velocity u=0u = 0, acceleration a=g=9.8a = g = 9.8 m s2^{-2}, time t=3.2t = 3.2 s.

(a) Height. Use s=ut+12at2s = ut + \frac{1}{2} a t^2.

s=0+12(9.8)(3.2)2=12(9.8)(10.24)=50.2s = 0 + \frac{1}{2}(9.8)(3.2)^2 = \frac{1}{2}(9.8)(10.24) = 50.2 m.

(b) Landing speed. Use v=u+atv = u + at.

v=0+(9.8)(3.2)=31.4v = 0 + (9.8)(3.2) = 31.4 m s1^{-1} downward.

Markers reward the stated sign convention, the explicit identification of uu, aa and tt, and units on both answers.

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