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Get Ready: Module 1 — Kinematics Foundations

Everyone starts somewhere. Before we launch projectiles and track motion in two dimensions, let's build the floor before the building — one small skill at a time. Go at your own pace; there's no rush and nothing to prove here.

Feeling shaky? That's completely okay — it just means you're about to fill in a gap that will make the whole module easier. Read the plain-language explanation, follow the worked example step by step, then try one yourself. The answers are hidden so you can think first — click Show answer whenever you're ready. Remember: this physics is enrichment, not a graded course — it's a place to stretch. You can do this.
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Skills to Build First

Four small building blocks from your math classes. Master these and Module 1 will feel like the next natural step, not a wall.

Skill 01

Solving a Two-Step Linear Equation

What it is & why you need it: Almost every kinematics formula is a one-line equation you must rearrange to solve for the unknown. The golden rule: whatever you do to one side, do to the other — like keeping a balance scale even. Solving \(v = v_0 + at\) for the time \(t\) is exactly this skill.

Worked example: solve \(2x + 3 = 11\)

  1. Goal: get \(x\) alone. First undo the \(+3\) by subtracting 3 from both sides: \(2x + 3 - 3 = 11 - 3\).
  2. Simplify: \(2x = 8\).
  3. Now undo the \(\times 2\) by dividing both sides by 2: \(\dfrac{2x}{2} = \dfrac{8}{2}\).
  4. Simplify: \(x = 4\). Check: \(2(4) + 3 = 8 + 3 = 11\). True!

Undo operations in reverse order: peel off addition/subtraction first, then multiplication/division — the opposite of the order of operations.

Try it

1. Solve \(3x - 4 = 14\).

2. The velocity formula \(v = v_0 + at\) becomes \(20 = 4 + 2t\). Solve for \(t\).

Show answer
1. Add 4 to both sides: \(3x = 18\). Divide both sides by 3: \(x = 6\). Check: \(3(6) - 4 = 14\). True.

2. Subtract 4 from both sides: \(16 = 2t\). Divide by 2: \(t = 8\) s. (Same two steps — just with physics letters.)
Skill 02

Squares and Square Roots

What it is & why you need it: Squaring means multiplying a number by itself; a square root undoes that. Kinematics is full of them — \(\tfrac12 a t^2\) squares the time, and \(v^2 = v_0^2 + 2a\,\Delta x\) needs a square root to finish. Get comfortable here and those formulas stop being scary.

Worked example: compute \(6^2\) and \(\sqrt{49}\)

  1. Square it. \(6^2\) means \(6 \times 6 = 36\). (A square never means "times 2.")
  2. Read a root as a question. \(\sqrt{49}\) asks: what number times itself gives \(49\)?
  3. Find it. \(7 \times 7 = 49\), so \(\sqrt{49} = 7\).
  4. They undo each other. \(\sqrt{6^2} = 6\) and \((\sqrt{49})^2 = 49\).

Memorize the perfect squares to 15: \(1,4,9,16,25,36,49,64,81,100,121,144,169,196,225\). Their roots are just \(1\) through \(15\).

Try it

1. Compute \(9^2\) and \(\sqrt{81}\).

2. In \(v^2 = 100\), find \(v\) (take the positive root).

Show answer
1. \(9^2 = 9 \times 9 = 81\). And \(\sqrt{81} = 9\) because \(9 \times 9 = 81\).

2. Take the square root of both sides: \(v = \sqrt{100} = 10\) m/s.
Skill 03

Sine and Cosine of \(0°,\,30°,\,45°,\,60°,\,90°\)

What it is & why you need it: Every 2-D launch splits into \(v_x = v\cos\theta\) and \(v_y = v\sin\theta\). To resolve those components you need the sine and cosine of the common launch angles at your fingertips. Knowing the five special angles makes most projectile problems pure arithmetic.

Worked example: the special-angle table

  1. Sine climbs \(0\to 1\). \(\sin 0° = 0,\ \sin 30° = \tfrac12,\ \sin 45° = \tfrac{\sqrt2}{2}\approx 0.707,\ \sin 60° = \tfrac{\sqrt3}{2}\approx 0.866,\ \sin 90° = 1\).
  2. Cosine falls \(1\to 0\). \(\cos 0° = 1,\ \cos 30° = \tfrac{\sqrt3}{2},\ \cos 45° = \tfrac{\sqrt2}{2},\ \cos 60° = \tfrac12,\ \cos 90° = 0\).
  3. They mirror. \(\sin\) and \(\cos\) swap between complementary angles: \(\sin 30° = \cos 60°\).
  4. Use it. For \(v=20\) m/s at \(30°\): \(v_x = 20\cos 30° \approx 17.3\) m/s, \(v_y = 20\sin 30° = 10\) m/s.

Trick: write \(\sin\) of \(0°,30°,45°,60°,90°\) as \(\tfrac{\sqrt0}{2},\tfrac{\sqrt1}{2},\tfrac{\sqrt2}{2},\tfrac{\sqrt3}{2},\tfrac{\sqrt4}{2}\). Reverse the list for cosine.

Try it

1. What are \(\sin 60°\) and \(\cos 60°\) (as decimals, 3 places)?

2. A ball is launched at \(10\) m/s, \(45°\). Find \(v_x\) and \(v_y\).

Show answer
1. \(\sin 60° = \tfrac{\sqrt3}{2} \approx 0.866\); \(\cos 60° = \tfrac12 = 0.500\).

2. At \(45°\) both are \(\tfrac{\sqrt2}{2}\approx 0.707\): \(v_x = v_y = 10(0.707) \approx 7.07\) m/s.
Skill 04

SI Units: m, s, m/s, and m/s²

What it is & why you need it: Physics keeps score in SI units. Distance is measured in metres (m), time in seconds (s), velocity in metres per second (m/s), and acceleration in metres per second squared (m/s²). Reading the units of a quantity tells you what it is — and catches mistakes before they cost you.

Worked example: read the units, then convert km/h to m/s

  1. Velocity is distance over time. "Metres per second" literally means \(\dfrac{\text{m}}{\text{s}}\) — how many metres covered each second.
  2. Acceleration is velocity-change over time. "Metres per second, per second" is \(\dfrac{\text{m/s}}{\text{s}} = \text{m/s}^2\).
  3. Convert speeds. To turn km/h into m/s, divide by \(3.6\) (since \(1\) m/s \(= 3.6\) km/h).
  4. Example. \(72\) km/h \(\div\, 3.6 = 20\) m/s. (And \(20\) m/s \(\times 3.6 = 72\) km/h to go back.)

Sanity check: a quantity's units must match the formula. If you compute a "distance" and the units come out as m/s, you made an algebra slip somewhere.

Try it

1. Convert \(90\) km/h to m/s.

2. A car's speed changes by \(6\) m/s every second. State that as an acceleration with its unit.

Show answer
1. \(90 \div 3.6 = 25\) m/s.

2. A change of \(6\) m/s each second is \(6\) m/s² (metres per second squared) — that's what acceleration means.
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Quick Readiness Check

Six short questions across all four skills. Try each one first, then peek at the answer.

  1. Solve \(5x + 2 = 17\).
    Show answer
    Subtract 2: \(5x = 15\). Divide by 5: \(x = 3\).
  2. Rearrange and solve: \(30 = 6 + 3t\) for \(t\).
    Show answer
    Subtract 6: \(24 = 3t\). Divide by 3: \(t = 8\) s.
  3. Compute \(8^2\) and \(\sqrt{144}\).
    Show answer
    \(8^2 = 64\). \(\sqrt{144} = 12\) because \(12 \times 12 = 144\).
  4. What is \(\sin 30°\), and what is \(\cos 0°\)?
    Show answer
    \(\sin 30° = \tfrac12 = 0.5\). \(\cos 0° = 1\).
  5. A launch is \(20\) m/s at \(60°\). Find \(v_x = 20\cos 60°\).
    Show answer
    \(\cos 60° = 0.5\), so \(v_x = 20(0.5) = 10\) m/s.
  6. Convert \(54\) km/h to m/s, and name the SI unit of acceleration.
    Show answer
    \(54 \div 3.6 = 15\) m/s. Acceleration is measured in \(\text{m/s}^2\) (metres per second squared).

If these feel comfortable, you're ready for the module. If a couple felt tricky, that's fine — revisit that skill card above and drill a little with the free links below. No shame, just steps.

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Practice More — Free

Want extra reps? These free sites let you drill each skill until it clicks. (Links open in a new tab.)

Two-step equations

Khan Academy IXL

Squares & square roots

Khan Academy IXL

Sine & cosine of special angles

Khan Academy IXL

SI units & unit conversion

Khan Academy IXL
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When You're Ready →

No rush. When the skills above feel solid, step into Module 1.

Module 1 Overview

The full picture of Kinematics — vocabulary, the ideas, and the interactive Projectile Sandbox.

Go to Module 1 →

The Projectile Sandbox

Set a speed, angle, and gravity, then watch the arc and read the range, height, and flight time live.

Open the Visual Lab

Student Support

Stuck? Here's how to get help, ask questions, and find more practice across the studio.

Student Support

Instructor: Dr. Goodluck Ijezie-Desbois, PharmD
Physics enrichment foundations — not a TEKS-graded course and not tied to a state End-of-Course exam. Optional stretch material for curious scholars. Non-commercial classroom use.