Get Ready: Module 1 — Kinematics & Dynamics
Everyone starts somewhere. Before we resolve forces on ramps and chase blocks across velocity-time graphs, 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.
Skills to Build First
Five small building blocks. Master these and Module 1 will feel like the next natural step, not a wall.
Unit Conversion & SI Prefixes
What it is & why you need it: Physics answers carry units, and the formulas only work in the base SI units — meters (m), kilograms (kg), seconds (s). Before plugging in, you convert: km/h to m/s, grams to kilograms, centimeters to meters. The trick is to multiply by a fraction equal to \(1\) (like \(\tfrac{1000\,\text{m}}{1\,\text{km}}\)) so the unwanted unit cancels.
Worked example: convert \(72\) km/h to m/s
- Write what you have, with units: \(72\ \dfrac{\text{km}}{\text{h}}\).
- Convert kilometers to meters: \(\times \dfrac{1000\ \text{m}}{1\ \text{km}}\) — the "km" cancels.
- Convert hours to seconds: \(\times \dfrac{1\ \text{h}}{3600\ \text{s}}\) — the "h" cancels, leaving m/s.
- Multiply: \(72 \times \dfrac{1000}{3600} = 72 \times \dfrac{1}{3.6} = 20\) m/s.
Shortcut worth memorizing: to go from km/h to m/s, divide by \(3.6\). To go back, multiply by \(3.6\). SI prefixes you'll meet: kilo \(=10^3\), centi \(=10^{-2}\), milli \(=10^{-3}\), micro \(=10^{-6}\).
1. Convert \(90\) km/h to m/s.
2. A block has a mass of \(500\) g. Express it in kilograms (the SI unit for mass).
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2. Milli/centi/kilo: \(1\) kg \(= 1000\) g, so \(500\ \text{g} \times \dfrac{1\ \text{kg}}{1000\ \text{g}} = 0.5\) kg.
Solving Linear and Simple Quadratic Equations
What it is & why you need it: Almost every physics problem ends with "solve for the unknown." Linear equations (one power of the variable) use the balance rule — do the same thing to both sides. Simple quadratics like \(v^2 = 2ad\) need a square root. In Module 1 you'll rearrange \(v_f^2 = v_i^2 + 2ad\) and free-fall \(h = \tfrac12 g t^2\), which are exactly these two skills.
Worked example: solve \(\tfrac{1}{2}(9.8)t^2 = 19.6\) for \(t\)
- Simplify the constant: \(\tfrac{1}{2}(9.8) = 4.9\), so \(4.9\,t^2 = 19.6\).
- Isolate \(t^2\) by dividing both sides by \(4.9\): \(t^2 = \dfrac{19.6}{4.9} = 4\).
- Take the square root of both sides: \(t = \sqrt{4} = 2\) (take the positive root — time isn't negative).
- So \(t = 2\) s.
Undo operations in reverse: peel away addition/subtraction, then multiplication/division, then take a root last. A squared variable always means a square root is coming.
1. Solve the linear equation \(3x + 4 = 19\).
2. Solve \(v^2 = 144\) for \(v\) (take the positive root).
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2. Square-root both sides: \(v = \sqrt{144} = 12\).
Right-Triangle Trig: sin, cos, tan
What it is & why you need it: To split a force (or a velocity) into components, you use the sine and cosine of the angle. SOH-CAH-TOA: \(\sin\theta = \tfrac{\text{opp}}{\text{hyp}}\), \(\cos\theta = \tfrac{\text{adj}}{\text{hyp}}\), \(\tan\theta = \tfrac{\text{opp}}{\text{adj}}\). On a ramp, \(mg\sin\theta\) is the down-slope pull and \(mg\cos\theta\) is the press into the slope — this skill is the whole incline lab.
Worked example: find the components of a \(50\)-N force at \(30^\circ\)
- Draw the force as the hypotenuse of a right triangle, with the angle \(30^\circ\) at the base.
- The horizontal (adjacent) component uses cosine: \(F_x = 50\cos 30^\circ\).
- The vertical (opposite) component uses sine: \(F_y = 50\sin 30^\circ\).
- Evaluate: \(\cos 30^\circ \approx 0.866\) so \(F_x \approx 43.3\) N; \(\sin 30^\circ = 0.5\) so \(F_y = 25\) N.
Worth memorizing: \(\sin 30^\circ = 0.5\), \(\cos 60^\circ = 0.5\), \(\sin 45^\circ = \cos 45^\circ \approx 0.707\), \(\cos 30^\circ = \sin 60^\circ \approx 0.866\). To recover an angle from a ratio, use \(\tan^{-1}\) (arctan).
1. A \(20\)-N force points \(60^\circ\) above the horizontal. Find its vertical component.
2. A right triangle has opposite side \(3\) and adjacent side \(4\). Find \(\tan\theta\), then the angle \(\theta\).
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2. \(\tan\theta = \dfrac{3}{4} = 0.75\), so \(\theta = \tan^{-1}(0.75) \approx 36.87^\circ\).
Slope and Area of a Line Graph
What it is & why you need it: Graphs are physics shorthand. On a velocity-time graph, the slope (rise over run) is the acceleration, and the area under the line is the displacement. Both are skills you already have from algebra — you just read them off the picture instead of a formula.
Worked example: a v-t graph rises from \((0,\,4)\) to \((6,\,22)\) (s, m/s)
- Slope = acceleration. Rise \(= 22 - 4 = 18\) m/s; run \(= 6 - 0 = 6\) s; slope \(= \dfrac{18}{6} = 3\) m/s².
- Area = displacement. Under a sloped line the area is a trapezoid: \(\tfrac12(\text{base})(\text{height}_1 + \text{height}_2)\).
- Plug in: \(\tfrac12(6)(4 + 22) = \tfrac12(6)(26)\).
- Evaluate: \(= 3 \times 26 = 78\) m of displacement.
Rule of thumb: on a v-t graph, slope tells you how the speed is changing (acceleration) and area tells you how far it went (displacement). A flat line = constant speed = zero acceleration.
1. A v-t graph is a straight line from \((0,\,0)\) to \((4,\,16)\). What is the acceleration?
2. Find the displacement (area) under that same line from \(t = 0\) to \(t = 4\) s. (Hint: it's a triangle.)
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2. Triangle area \(= \tfrac12(\text{base})(\text{height}) = \tfrac12(4)(16) = 32\) m.
Scientific Notation
What it is & why you need it: Physics constants are huge or tiny — the gravitational constant is \(6.67\times10^{-11}\), the speed of light is \(3\times10^{8}\) m/s. Scientific notation writes a number as a digit between \(1\) and \(10\) times a power of ten. You'll plug these into formulas constantly, so reading and multiplying them has to be automatic.
Worked example: write \(245{,}000\) and \(0.0042\) in scientific notation
- For \(245{,}000\): move the decimal so one nonzero digit is in front — \(2.45\). You moved it \(5\) places left.
- Moving left makes the exponent positive: \(245{,}000 = 2.45\times10^{5}\).
- For \(0.0042\): move the decimal right to get \(4.2\). You moved it \(3\) places right.
- Moving right makes the exponent negative: \(0.0042 = 4.2\times10^{-3}\).
To multiply: multiply the front numbers and add the exponents. \((3\times10^{8})(2\times10^{-3}) = 6\times10^{5}\). On a calculator, type the exponent with the EE or EXP key, not "×10^".
1. Write \(0.00067\) in scientific notation.
2. Multiply \((2\times10^{4})(4\times10^{3})\) and give the answer in scientific notation.
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2. Multiply fronts: \(2 \times 4 = 8\). Add exponents: \(4 + 3 = 7\). Answer: \(8\times10^{7}\).
Quick Readiness Check
Six short questions across all five skills. Try each one first, then peek at the answer.
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Convert \(108\) km/h to m/s.
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Divide by \(3.6\): \(108 \div 3.6 = 30\) m/s. -
Solve \(4.9\,t^2 = 44.1\) for \(t\) (positive root).
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\(t^2 = \dfrac{44.1}{4.9} = 9\), so \(t = \sqrt{9} = 3\) s. -
A \(40\)-N force is directed \(30^\circ\) above the horizontal. Find its horizontal component.
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Horizontal uses cosine: \(40\cos 30^\circ = 40(0.866) \approx 34.6\) N. -
On a v-t graph the velocity rises from \(2\) m/s to \(14\) m/s over \(4\) s. Find the acceleration.
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Slope \(= \dfrac{14 - 2}{4} = \dfrac{12}{4} = 3\) m/s². -
Write \(0.00091\) in scientific notation.
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Move the decimal \(4\) places right: \(9.1\times10^{-4}\). -
Find the area (a triangle) under a v-t line from \((0,\,0)\) to \((5,\,10)\) s, m/s.
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\(\tfrac12(\text{base})(\text{height}) = \tfrac12(5)(10) = 25\) m.
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.
Practice More — Free
Want extra reps? These free sites let you drill each skill until it clicks. (Links open in a new tab.)
Unit conversion & SI prefixes
Linear & quadratic equations
Right-triangle trig (SOH-CAH-TOA)
Slope & area of a graph
Scientific notation
When You're Ready →
No rush. When the skills above feel solid, step into Module 1.
Instructor: Dr. Goodluck Ijezie-Desbois, PharmD · Beta Academy · Room: TBA
Foundations support for the AP Physics · enrichment course — supplemental,
college-level practice following the College Board AP Physics 1 framework. Not a TEKS-graded course.