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Honors Physics: Mechanics, Unit 1: Measurement and motion in one dimension

Seven lessons including a lab. Use g = 9.8 m/s² unless told otherwise. Keep a lab notebook (paper or a doc).

Lesson 1: Units, significant figures and scalars vs. vectors

You will be able to:

Physics uses SI units: meters (m), kilograms (kg), seconds (s). Convert by multiplying by fractions equal to 1, like (1000 m / 1 km).

When multiplying or dividing, keep as many significant figures as the least precise measurement. A scalar has size only (distance, speed, mass, time). A vector has size and direction (displacement, velocity, acceleration, force).

Worked example. Convert 72 km/h to m/s.

72 km/h × (1000 m/1 km) × (1 h/3600 s) = 20 m/s.

Practice set (try each before opening the answers):

  1. Convert 15 m/s to km/h.
  2. A room is 3.0 m by 4.52 m. Area with correct significant figures?
  3. Convert 1.5 hours to seconds.
  4. Scalar or vector: 30 m north; 12 s; 9.8 m/s² down; 60 km/h.
  5. How many significant figures in 0.00450?
Answers and short solutions
  1. 15 × 3.6 = 54 km/h.
  2. 13.56 m² → 14 m² (2 significant figures).
  3. 5,400 s.
  4. Vector, scalar, vector, scalar (speed).
  5. 3.

Honors challenge: Use dimensional analysis to find which formula for a pendulum's period can be right: T = 2π√(L/g) or T = 2π√(g/L).

Challenge solution

L/g has units m ÷ (m/s²) = s², so √(L/g) is in seconds. g/L has units 1/s², giving 1/s. Only T = 2π√(L/g) has units of time.

Lesson 2: Position, displacement and velocity

You will be able to:

Pick an origin and a positive direction. Displacement Δx = xfinal − xinitial (a vector). Distance is the total path length (a scalar).

Average velocity = Δx / Δt. Average speed = distance / Δt. On an x–t graph, the slope is the velocity.

Worked example. You walk 30 m east, then 10 m west, in 20 s total.

Distance 40 m; displacement 20 m east. Average speed 40/20 = 2 m/s. Average velocity 20/20 = 1 m/s east.

Practice set (try each before opening the answers):

  1. A runner goes once around a 400 m track in 80 s. Average speed and average velocity?
  2. A car's position goes from x = −20 m to x = 60 m in 4 s. Average velocity?
  3. On an x–t graph, a straight line goes from (0 s, 10 m) to (5 s, 0 m). Velocity?
  4. You drive 60 km at 60 km/h and then 60 km at 30 km/h. Average speed?
  5. What does a horizontal line on an x–t graph mean?
Answers and short solutions
  1. 5 m/s; 0 m/s (back where she started).
  2. 80/4 = +20 m/s.
  3. −2 m/s.
  4. Times: 1 h + 2 h = 3 h; 120 km / 3 h = 40 km/h (not 45).
  5. The object is at rest (velocity 0).

Honors challenge: Prove that if you travel equal distances at speeds v₁ and v₂, your average speed is 2v₁v₂/(v₁ + v₂).

Challenge solution

Each leg is d, taking d/v₁ and d/v₂. Average speed = 2d/(d/v₁ + d/v₂) = 2/(1/v₁ + 1/v₂) = 2v₁v₂/(v₁ + v₂).

Lesson 3: Acceleration and motion graphs

You will be able to:

Average acceleration a = Δv / Δt (m/s²). On a v–t graph, the slope is the acceleration and the area under the line is the displacement.

Slowing down means velocity and acceleration point in opposite directions. It does not always mean a negative acceleration.

Worked example. A v–t graph rises from 0 to 10 m/s over 5 s, then stays at 10 m/s for 5 more s. Displacement?

Triangle ½ · 5 · 10 = 25 m, plus rectangle 5 · 10 = 50 m: 75 m in total.

Practice set (try each before opening the answers):

  1. A car goes from 0 to 24 m/s in 8.0 s. Average acceleration?
  2. A bike slows from 12 m/s to 4 m/s in 4 s (positive direction forward). Acceleration?
  3. Moving in the −x direction at −6 m/s, an object's acceleration is +2 m/s². Is it speeding up or slowing down?
  4. Displacement from a v–t graph that is a straight line from 4 m/s at t = 0 to 12 m/s at t = 4 s?
  5. Sketch in words the x–t graph for constant positive acceleration from rest.
Answers and short solutions
  1. 3.0 m/s².
  2. −2 m/s².
  3. Slowing down (opposite signs).
  4. Trapezoid: ½(4 + 12)(4) = 32 m.
  5. A curve that starts flat and gets steeper (an upward-opening parabola).

Honors challenge: A v–t graph is a straight line through (0, v₀) and (t, v). Use the area to derive Δx = v₀t + ½at².

Challenge solution

The area is a rectangle v₀t plus a triangle ½ · t · (v − v₀). Since v − v₀ = at, the triangle is ½at². So Δx = v₀t + ½at².

Lesson 4: The constant-acceleration equations

You will be able to:

For constant acceleration: v = v₀ + at; Δx = v₀t + ½at²; v² = v₀² + 2aΔx; Δx = ½(v₀ + v)t. List what you know (v₀, v, a, t, Δx) and pick the equation that leaves out the one you don't need.

Worked example. A car at 20 m/s brakes at −5.0 m/s². Stopping distance and time?

v² = v₀² + 2aΔx: 0 = 400 − 10Δx, so Δx = 40 m. t = (0 − 20)/(−5) = 4.0 s.

Practice set (try each before opening the answers):

  1. From rest, a = 2.0 m/s² for 6.0 s. Final speed and distance?
  2. A plane needs 80 m/s to take off and accelerates at 4.0 m/s². Minimum runway?
  3. A train slows from 30 m/s to 10 m/s over 200 m. Acceleration?
  4. A sprinter accelerates from rest to 10 m/s in 2.0 s, then holds that speed. How far in the first 6.0 s?
  5. A car at 25 m/s; the driver reacts after 0.80 s, then brakes at −6.25 m/s². Total stopping distance?
Answers and short solutions
  1. 12 m/s; 36 m.
  2. 6400/8 = 800 m.
  3. (100 − 900)/400 = −2.0 m/s².
  4. 10 m (first 2 s) + 40 m = 50 m.
  5. Reaction: 20 m. Braking: 625/12.5 = 50 m. Total 70 m.

Honors challenge: Derive v² = v₀² + 2aΔx by eliminating t from the first two equations.

Challenge solution

t = (v − v₀)/a. Substitute: Δx = v₀(v − v₀)/a + ½a(v − v₀)²/a². Multiply by 2a: 2aΔx = 2v₀v − 2v₀² + v² − 2v₀v + v₀² = v² − v₀². ∎

Lesson 5: Free fall

You will be able to:

Near Earth's surface, ignoring air resistance, every object falls with the same acceleration: 9.8 m/s² downward. At the top of its path a thrown ball has velocity 0, but its acceleration is still 9.8 m/s² down.

Worked example. A ball is thrown straight up at 14.7 m/s. Time to the top, maximum height and total time in the air?

Top: t = 14.7/9.8 = 1.5 s. Height: 14.7²/(2 · 9.8) = 11.0 m. Total time: 3.0 s (it's symmetric).

Practice set (try each before opening the answers):

  1. A stone is dropped from 45 m. Time to fall and impact speed?
  2. A ball is dropped. How far does it fall in the first 2.0 s?
  3. What is the acceleration at the top of a vertical throw?
  4. A ball is thrown up at 20 m/s. Velocity after 3.0 s?
  5. From a 20 m roof, a ball is thrown up at 15 m/s. When does it hit the ground?
Answers and short solutions
  1. t = √(2 · 45/9.8) ≈ 3.0 s; v = 9.8 · 3.03 ≈ 29.7 m/s.
  2. ½ · 9.8 · 4 = 19.6 m.
  3. 9.8 m/s² downward. (Velocity is 0, acceleration is not.)
  4. 20 − 29.4 = −9.4 m/s (moving down).
  5. −20 = 15t − 4.9t², so 4.9t² − 15t − 20 = 0; t = (15 + √(225 + 392))/9.8 ≈ 4.1 s.

Honors challenge: Show that a ball thrown up at speed v₀ returns to the launch height with speed v₀ (ignoring air resistance).

Challenge solution

Using v² = v₀² + 2aΔx with Δx = 0 gives v² = v₀², so |v| = v₀; it moves downward, so v = −v₀.

Lesson 6: Lab: measure g with a drop

You will be able to:

You need: a tape measure, a small heavy object (a coin or a ball), a phone with a slow-motion video or stopwatch app, and an adult spotter. Drop from a safe height (about 2 m, standing on the floor and holding the object high, never from a balcony).

Steps: (1) Measure the drop height h. (2) Record a slow-motion video of 5 drops and read the fall time t from the frames. (3) For each trial compute g = 2h/t². (4) Average the trials. (5) Percent error = |g − 9.8|/9.8 × 100%.

Write-up (CER): Claim (your value of g), Evidence (data table, mean, spread), Reasoning (why Δx = ½gt² applies; the biggest error sources, such as reaction time, frame rate and air resistance, and which way each one pushes your answer).

Worked example. Sample data: h = 2.00 m, t = 0.64 s. Find g and percent error.

g = 4.00/0.4096 ≈ 9.77 m/s²; error ≈ 0.3%.

Practice set (try each before opening the answers):

  1. If your timer starts 0.05 s late on a true 0.64 s fall, what g do you get?
  2. Why does a crumpled paper ball give a smaller g than a coin?
  3. Why does averaging 5 trials help?
Answers and short solutions
  1. t = 0.59 s → g = 4.00/0.3481 ≈ 11.5 m/s² (too high).
  2. Air resistance is a larger share of the force on a light, wide object.
  3. Random timing errors partly cancel; the mean is more reliable than one trial.

Lesson 7: Unit 1 test

You will be able to:

Allow 50 minutes. Show the equation, substitution and units for every answer.

Practice set (try each before opening the answers):

  1. Convert 90 km/h to m/s.
  2. A cyclist goes 3 km east then 1 km west in 0.25 h. Average speed and average velocity?
  3. A car accelerates from 5 m/s to 25 m/s in 4.0 s. Acceleration and distance?
  4. A ball is thrown up at 9.8 m/s. Maximum height?
  5. Two cars start together. Car A moves at a steady 20 m/s; car B starts from rest at 4.0 m/s². When does B catch A?
  6. On a v–t graph, what do slope and area mean?
Answers and short solutions
  1. 25 m/s.
  2. 16 km/h; 8 km/h east.
  3. 5.0 m/s²; ½(5 + 25)(4) = 60 m.
  4. 9.8²/19.6 = 4.9 m.
  5. 20t = 2t², so t = 10 s (at 200 m).
  6. Slope = acceleration; area = displacement.

AP® Physics 1 exam prep (optional)

AP® Physics 1 is algebra-based and asks you to explain your reasoning, not only compute. For each test question above, write one sentence explaining why your answer makes physical sense (sign, size and units). Graph-reading questions (slope and area) appear often.

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