WEEK 2 TUE

Administrative Announcements

• Lecture: SLE123 – Physics for the Life Sciences, Lecture 5 (Motion – Part 2).

• Seminar this week reviews last week’s material (distance, displacement, speed, velocity, acceleration).
  • Bring a calculator and any questions.

• Experimental-demonstration assessment:
  • Group formation opens tomorrow 04:00; close Thursday 31st, 20:00.
  • Join via Tools → Groups (1 – 200). Groups of ≤4.
  • Read the posted instructions & rubric, start brainstorming immediately.

• Seminar slide decks are uploaded at the end of each week.

Review of Motion Concepts

• Speed (scalar): magnitude only.

• Velocity (vector): magnitude and direction.
  $v = \frac{\Delta x}{\Delta t}, \qquad \Delta x = x<em>f - x</em>i$

• Acceleration (vector):

  

• SI units: velocity $\text{m}\,\text{s}^{-1}$, acceleration $\text{m}\,\text{s}^{-2}$.

Kinematic Equations (Uniform Acceleration, 1-D)

1. $v<em>f = v</em>i + a t$ ( no $\Delta x$ )

2. $\Delta x = \frac{v<em>f + v</em>i}{2} t$ (mean-speed theorem, no $a$)

3. $\Delta x = v<em>i t + \frac{1}{2} a t^2$ (no $v</em>f$)

4. $v<em>f^2 = v</em>i^2 + 2 a \Delta x$ (no $t$)

Missing-variable trick:

• If a problem lacks time ⇒ use (4).

• If it lacks displacement ⇒ use (1).

• If it lacks acceleration ⇒ use (2).

• If it lacks final velocity ⇒ use (3).

Conventions & scope for this unit:

• Acceleration is constant and motion is straight-line.

• Horizontal: use $x$; vertical (free-fall, next lecture): swap $x \rightarrow y$ and $a \rightarrow g = 9.8\,\text{m}\,\text{s}^{-2}$ (downward).

Problem-Solving Strategy

1. Sketch / axis diagram of the motion.

2. List knowns & unknowns (identify variables, note units & sign).

3. Choose the kinematic equation containing the unknown but not containing any additional unknowns.

4. Convert to SI units before substituting (e.g. $\text{km}\,\text{h}^{-1} \rightarrow \text{m}\,\text{s}^{-1}$, minutes → seconds).

5. Solve algebraically, keep full calculator precision until the final step, then round.

6. State answer with units & direction (or “magnitude only” if specified).

Key words:

• “from rest” ⇒ $v_i = 0$.

• “comes to rest” ⇒ $v_f = 0$.

• “free-fall” ⇒ $a = g$ (downwards ⇒ negative in up-positive axes).

Significant Figures & Algebra Tips

• Retain all digits internally; round the displayed answer.

• Square-rooting a negative indicates a sign or algebra error.

• Dozens of older notations ($u, v, s$) exist; symbols do not change the physics, but course uses $v<em>i, v</em>f, a, t, \Delta x$.

Example Problems Walk-Through

(Equations chosen via the missing-variable rule; answers rounded appropriately.)

1 Airplane Take-Off Roll

Given: $v_i = 0$, $a = 3.2\,\text{m}\,\text{s}^{-2}$, $t = 32.8\,\text{s}$.
Use (3): $\Delta x = 0\cdot t + \tfrac12 a t^2 = \tfrac12(3.2)(32.8)^2 = 1.72\times10^3\,\text{m}$.

2 Race Car

(a) Acceleration from $18.5$ to $46.1\,\text{m}\,\text{s}^{-1}$ in $2.47\,\text{s}$.
• (1): $a = \dfrac{46.1 - 18.5}{2.47} = 11.2\,\text{m}\,\text{s}^{-2}$.
(b) Distance covered in that interval.
• (4): $\Delta x = \dfrac{46.1^2 - 18.5^2}{2\,\times 11.2} = 79.8\,\text{m}$.

3 Bullet in Rifle Barrel

$v<em>i = 0$, $v</em>f = 521\,\text{m}\,\text{s}^{-1}$, $\Delta x = 0.84\,\text{m}$.
(4): $a = \dfrac{521^2}{2\times0.84} = 1.62\times10^5\,\text{m}\,\text{s}^{-2}$.

4 Rocket-Powered Sled

(a) $v<em>i = 0$ → $v</em>f = 444\,\text{m}\,\text{s}^{-1}$ in $1.83\,\text{s}$.
• (1): $a = 2.43\times10^2\,\text{m}\,\text{s}^{-2}$.
(b) Distance in that time.
• (3): $\Delta x = \tfrac12 a t^2 = 0.5(2.43\times10^2)(1.83)^2 = 4.06\times10^2\,\text{m} = 0.406\,\text{km}$.

5 Skidding Car to Rest

$v<em>i = 22.4\,\text{m}\,\text{s}^{-1}$, $v</em>f = 0$, $t = 2.55\,\text{s}$.
(2): $\Delta x = \tfrac{(0+22.4)}{2}(2.55) = 28.6\,\text{m}$.

6 Runway Design (Slowest Plane)

$v<em>i = 0$, $v</em>f = 65\,\text{m}\,\text{s}^{-1}$, $a = 3\,\text{m}\,\text{s}^{-2}$.
(4): $\Delta x = \dfrac{65^2}{2\times3} = 7.04\times10^2\,\text{m}$.

7 Estimating Speed from Skid Marks

Given skid $\Delta x = 290\,\text{m}$, $v<em>f = 0$, $a = -3.9\,\text{m}\,\text{s}^{-2}$. (4): $v</em>i = \sqrt{2(-3.9)(290)} = 47.6\,\text{m}\,\text{s}^{-1} \;(\approx 171\,\text{km}\,\text{h}^{-1})$.

8 Plane Reaching Take-Off Speed

(a) Acceleration:
$v_f = 88.3\,\text{m}\,\text{s}^{-1},\; \Delta x = 1.365\times10^3\,\text{m}$.
(4): $a = 2.86\,\text{m}\,\text{s}^{-2}$.
(b) Time required:
(1): $t = \dfrac{88.3}{2.86} = 30.9\,\text{s}$.

9 Object Known at Two Positions

At $x=6\,\text{m}$, $v=10\,\text{m}\,\text{s}^{-1}$; at $x=10\,\text{m}$, $v=15\,\text{m}\,\text{s}^{-1}$.
$\Delta x = 4\,\text{m}$.
(4): $a = \dfrac{15^2 - 10^2}{2\times4} = 15.6\;\text{m}\,\text{s}^{-2} \;(\approx 16)$.

10 Displacement in a Sub-Interval

Uniformly accelerated from $75$ to $135\,\text{m}\,\text{s}^{-1}$ in $10\,\text{s}$.

1. Acceleration: (1): $a = \dfrac{135-75}{10} = 6\,\text{m}\,\text{s}^{-2}$.

2. Velocity at $t = 2\,\text{s}$: $v = 75 + 6(2) = 87\,\text{m}\,\text{s}^{-1}$.

3. Displacement from $t=2$ to $t=4$ ((\Delta t = 2\,\text{s})):
   (3): $\Delta x = 87(2) + \tfrac12(6)(2)^2 = 186\,\text{m}$.

Experimental & Statistical Reminders

• Perform ≥3 replicates; report mean, standard deviation, error bars, $p$-values.

• Apply earlier statistics lecture to upcoming laboratory assessment.

Looking Ahead

• Next lecture: Free-fall (vertical motion) — same four equations, swap $x \rightarrow y$, $a \rightarrow -g$.

Practical Advice & Resources

• Keep practising algebra; use Deakin’s Maths Mentor pages if rusty.

• Review revision quizzes before graded quizzes.

• Maintain consistent sign convention; check direction keywords.

• Carry units throughout calculations; convert early, round late.