Session 1 — Vectors from Scratch and Motion in a Straight Line: The Kinematic Equations

Duration: 60 minutes. It splits cleanly into two 30-minute halves: vectors, then kinematics.
Position in the course: the first Physics session. Everything after it builds on this one: Session 2’s projectiles, Session 4’s relative velocity and Sessions 5–6’s forces.
Grade-9 foundation: Class 9 Ch4 Describing Motion Around Us: distance vs displacement, speed vs velocity, acceleration, the equations of uniformly accelerated motion. The Class 9 PDF is on the Mac and wasn’t reachable today, so section numbers from it are left out.
Grade-11 depth: Glencoe Physics: Principles and Problems:

• Ch2 “Representing Motion”: position, displacement, velocity
• Ch3 “Accelerated Motion”: §3.1 Acceleration, §3.2 Motion with Constant Acceleration, §3.3 Free Fall
• Ch5 §5.1 “Vectors”: adding vectors, and components

Hewitt Conceptual Physics adds Ch2 §2.1–2.6 (motion is relative, speed, velocity, acceleration, free fall “how fast” and “how far”) and Ch6 §6.1–6.4 (vector and scalar quantities, adding vectors geometrically).
g = 9.81 m/s² throughout, matching the school.

Lesson Plan

Objectives. By the end of this session the student should be able to:

1. Tell scalars from vectors, add vectors tip-to-tail, and find a resultant’s size and direction with Pythagoras and tangent.
2. Resolve a vector into x- and y-components, and add vectors by components.
3. Tell distance from displacement and speed from velocity, and find average speed, average velocity and acceleration.
4. Derive the kinematic equations for constant acceleration and choose the right one for a problem.
5. Solve free-fall problems with a = −g.

Pacing note. If time is short, keep the derivations in Part 3 (they are the heart of the session) and set component addition (Worked Example 3) and free fall as reading plus homework.

Warm-up (5 min)

1. You walk 3 km east, then 4 km north. How far did you walk? How far are you from where you started? → You walked 7 km (distance), but you’re only 5 km from the start (displacement). The difference between “how far along the path” and “how far and which way” is the whole idea of vectors.

2. A car’s speedometer reads a steady 60 km/h as it goes around a bend. Is its velocity constant? → No. The speed is constant but the direction changes, so the velocity changes. Velocity is a vector.

Formula Sheet — Session 1

Symbols: u = initial velocity, v = final velocity, a = acceleration (constant), t = time, s = displacement.

Part 1 — Vectors from Scratch (20 min)

Scalars and vectors (Hewitt §6.1, Glencoe §5.1)

A scalar has only a size (magnitude) and a unit. A vector has a size and a direction.

How a vector is drawn: an arrow. Its length shows the size (to a chosen scale) and its direction shows the direction. In print a vector is written in bold (A) or with an arrow over the letter. Plain A means just its size.

Two useful facts:

• Two vectors are equal if they have the same size and the same direction, wherever they are drawn.
• −A is the same size as A but points the opposite way. In one dimension, direction is just a sign: choose a positive direction (east, or up) and make the opposite way negative.

Adding vectors in one dimension

Along a line, add the signs. Walk 5 m east, then 2 m west, with east positive: +5 + (−2) = +3 m, so the displacement is 3 m east.

Adding vectors in two dimensions: tip-to-tail (Hewitt §6.4)

To add A + B, draw A, then start B at the tip of A. The resultant R runs from the tail of A to the tip of B. The order doesn’t matter: A + B = B + A.

When A and B are perpendicular they form a right triangle with R as the hypotenuse:

• Size (Pythagoras): R = √(A² + B²)
• Direction: tanθ = B/A, so θ = tan⁻¹(B/A), measured from A.

Worked Example 1: the warm-up walk

3 km east then 4 km north.

• R = √(3² + 4²) = √25 = 5 km.
• θ = tan⁻¹(4/3) = 53.1°.

Answer: 5 km at 53.1° north of east. Distance walked = 7 km, but displacement = 5 km.

The range of possible answers. Two vectors of sizes A and B can give any resultant from A − B (pointing opposite ways) to A + B (the same way). For 3 and 4 the resultant can be anything from 1 to 7. It is 5 only when they are perpendicular.

Resolving a vector into components

Any vector can be replaced by two perpendicular vectors, its components, that add up to it. This is the reverse of adding. If A makes angle θ with the x-axis, then from the right triangle:

• cosθ = adjacent/hypotenuse = Aₓ/A, so Aₓ = A cosθ
• sinθ = opposite/hypotenuse = A_y/A, so A_y = A sinθ

Check: Aₓ² + A_y² = A²(cos²θ + sin²θ) = A² ✓

Worked Example 2. A ball is kicked at 20 m/s at 30° above the horizontal. Find the components of its velocity.

• vₓ = 20 cos30° = 20 × 0.866 = 17.3 m/s (horizontal)
• v_y = 20 sin30° = 20 × 0.500 = 10.0 m/s (vertical)

This is exactly how Session 2 splits a projectile’s launch velocity.

Adding any vectors by components

For vectors that are not perpendicular, don’t draw a messy triangle. Instead:

1. Resolve each vector into x and y components.
2. Add all the x’s, and add all the y’s: Rₓ = Aₓ + Bₓ, R_y = A_y + B_y.
3. Recombine: R = √(Rₓ² + R_y²) and θ = tan⁻¹(R_y/Rₓ).

Worked Example 3. A boat moves 10 m east, then 6 m at 60° north of east. Find the total displacement.

• A: Aₓ = 10, A_y = 0.
• B: Bₓ = 6 cos60° = 3.0, B_y = 6 sin60° = 5.20.
• Rₓ = 13.0 m, R_y = 5.20 m.
• R = √(13.0² + 5.20²) = √(169 + 27) = √196 = 14.0 m.
• θ = tan⁻¹(5.20/13.0) = 21.8°.

Answer: 14.0 m at 21.8° north of east.

Part 2 — Describing Motion (10 min)

Motion is relative (Hewitt §2.1)

Every position and velocity is measured from a reference point. You sit still in a bus, but you are moving at 60 km/h relative to the road. Session 4 returns to this idea.

Distance vs displacement

• Distance is the length of the path actually traveled. It is a scalar and never negative.
• Displacement is the straight-line change in position, from start to finish. It is a vector: Δx = x_final − x_initial.

The size of the displacement is never more than the distance. The two are equal only for a straight-line trip in one direction. A runner who completes one lap of a 400 m track has run 400 m but has a displacement of zero.

Speed vs velocity (Hewitt §2.2–2.3)

• Average speed = total distance / total time (scalar).
• Average velocity = displacement / time, v_avg = Δx/Δt (vector).
• Instantaneous speed or velocity is the value at one moment: what the speedometer shows.

Units: m/s (SI), km/h, mph. To convert km/h → m/s, divide by 3.6 (because 1 km/h = 1000 m / 3600 s). For example, 90 km/h = 25 m/s.

Worked Example 4. A runner completes one lap of a 400 m track in 80 s. Find the average speed and average velocity.

• Average speed = 400/80 = 5.0 m/s.
• Average velocity = 0/80 = 0 m/s: back at the start, so the displacement is zero.

Acceleration (Hewitt §2.4, Glencoe §3.1)

Acceleration is the rate of change of velocity:

a = Δv/Δt = (v − u)/t, unit m/s² (“meters per second, every second”)

Because velocity is a vector, you accelerate when you speed up, slow down or change direction.

Sign rule. If a and v point the same way, the object speeds up. If they point opposite ways, it slows down. “Negative acceleration” means “acceleration in the negative direction”. It only means slowing down if the object is moving in the positive direction.

Part 3 — The Kinematic Equations (17 min)

These hold only when the acceleration is constant. Each is derived from the one before, so the student never has to memorize them blindly.

S1 from the definition of acceleration

Start from a = (v − u)/t. Multiply both sides by t: at = v − u. So:

v = u + at (S1)

The velocity changes by the same amount, a, every second, so v grows linearly with t.

S0: distance = average velocity × time

For any motion, displacement = average velocity × time. When a is constant, v rises in a straight line from u to v, so its average is exactly halfway:

v_avg = (u + v)/2, so s = ½(u + v)t (S0)

This is the equation the student used first: “distance = average velocity × time”.

S2: substitute S1 into S0

Replace v in S0 with u + at:

s = ½(u + u + at)t = ½(2u + at)t = ut + ½at²

s = ut + ½at² (S2)

The same result from the graph. The displacement is the area under the v–t graph. That area splits into a rectangle (u × t) and a triangle (½ × t × at):

S3: eliminate time

Some problems give no time at all. From S1, t = (v − u)/a. Substitute into S0:

s = ½(u + v) × (v − u)/a = (v² − u²)/(2a)

Multiply both sides by 2a:

v² = u² + 2as (S3)

Choosing the equation

Each equation leaves out one of the five quantities. Pick the one that leaves out the quantity you are neither given nor asked for.

Method: (1) choose a positive direction; (2) list u, v, a, s, t with signs, writing “?” for the unknown; (3) pick the equation; (4) substitute and solve; (5) check the units and whether the answer is sensible.

Worked Example 5: speeding up

A car goes from rest to 25 m/s in 5.0 s. Find its acceleration and the distance covered.

• Known: u = 0, v = 25 m/s, t = 5.0 s.
• S1: a = (25 − 0)/5.0 = 5.0 m/s².
• S0: s = ½(0 + 25)(5.0) = 62.5 m. Check with S2: 0 + ½ × 5.0 × 25 = 62.5 m ✓

Answer: 5.0 m/s²; 62.5 m.

Worked Example 6: braking distance

A car moving at 30 m/s brakes with an acceleration of −6.0 m/s². How far does it travel before stopping, and how long does it take?

• Forward is positive. Known: u = 30, v = 0, a = −6.0. Asked: s and t.
• S3: 0 = 30² + 2(−6.0)s, so s = 900/12 = 75 m.
• S1: 0 = 30 + (−6.0)t, so t = 5.0 s.

Answer: 75 m; 5.0 s. Double the speed and the stopping distance goes up four times (s ∝ u²). That is why speed limits matter.

Part 4 — Free Fall (5 min)

Free fall (Hewitt §2.5–2.6, Glencoe §3.3)

Without air resistance, every object near Earth’s surface accelerates downward at g = 9.81 m/s², whatever its mass. Take up as positive, so a = −9.81 m/s² in every free-fall problem, whether the object is going up, at the top, or coming down. The four equations are unchanged.

Useful results for an object thrown straight up at speed u:

• At the top, v = 0. From S1: t_top = u/g.
• From S3: 0 = u² − 2g·h, so h_max = u²/(2g).
• By symmetry it takes as long to fall as to rise, and it returns to your hand at the same speed it left with.

Worked Example 7. A ball is thrown straight up at 15 m/s. Find (a) the maximum height, (b) the time to reach it, (c) the total time in the air.

• (a) h = u²/(2g) = 15²/(2 × 9.81) = 225/19.62 = 11.5 m.
• (b) t = u/g = 15/9.81 = 1.53 s.
• (c) By symmetry, 2 × 1.53 = 3.06 s.

Exit Ticket (3 min)

1. Two forces of 6 N and 8 N act on an object. What are the largest and smallest possible resultants? → 14 N (same direction) and 2 N (opposite directions). 10 N only if they are perpendicular.
2. Which equation would you use to find how fast a car is going after 100 m, given u and a but not t? → S3, v² = u² + 2as (no time).
3. At the top of its flight, what are a ball’s velocity and acceleration? → v = 0, but a = 9.81 m/s² downward. It never stops accelerating.

Practice Problems

Use g = 9.81 m/s². Conceptual questions C1–C7, numerical questions N1–N12. Do C3, N3 and N8 in class if time allows; the rest are homework.

Conceptual

C1. Sort into scalars and vectors: mass, displacement, speed, velocity, time, acceleration, force, distance, temperature.

C2. (a) Can the size of an object’s displacement ever be greater than the distance it traveled? (b) Can the distance be non-zero while the displacement is zero? Give an example.

C3. Two vectors have sizes 6 units and 8 units. (a) What is the largest possible resultant? (b) The smallest? (c) At what angle between them is the resultant 10 units?

C4. Can an object have zero velocity and still be accelerating? Give an example.

C5. A car travels around a circular track at a steady 60 km/h. Is its (a) speed constant, (b) velocity constant, (c) acceleration zero? Explain.

C6. “A negative acceleration always means the object is slowing down.” True or false? Give an example.

C7. Explain why the displacement in the equations of motion equals the area under the velocity–time graph.

Numerical

N1. A hiker walks 5 m east and then 12 m north. Find the size and direction of her displacement.

N2. A rope pulls a sled with a force of 50 N at 37° above the horizontal. Find the horizontal and vertical components of the force.

N3. A boat travels 10 m east and then 6 m at 60° north of east. Using components, find the resultant displacement. (Compare with Worked Example 3.)

N4. A family drives 120 km to a city in 1.5 h and returns home by the same road in 2.5 h. Find (a) the average speed for the round trip, (b) the average velocity for the round trip.

N5. Convert (a) 90 km/h to m/s, (b) 15 m/s to km/h.

N6. A cyclist speeds up uniformly from 4.0 m/s to 10 m/s in 3.0 s. Find (a) the acceleration, (b) the distance covered, using “distance = average velocity × time”.

N7. A plane starts from rest and accelerates at 3.2 m/s². It needs 80 m/s to take off. Find (a) the minimum runway length, (b) the time on the runway.

N8. A car moving at 20 m/s brakes at 4.0 m/s². Find (a) the time to stop, (b) the stopping distance, (c) the distance covered in the first 2.0 s of braking.

N9. A stone is dropped from rest off a 45 m cliff. Find (a) the time to reach the bottom, (b) its speed on impact.

N10. A ball is thrown straight up at 20 m/s. Find (a) the maximum height, (b) the time to reach it, (c) its velocity after 3.0 s, (d) its height after 3.0 s.

N11. A car accelerates from rest at 2.0 m/s² for 10 s, then travels at constant velocity for 20 s, then brakes uniformly to a stop in 5.0 s. Find (a) the total distance, (b) the average velocity for the whole trip.

N12. A driver moving at 25 m/s sees a hazard. Her reaction time is 0.80 s, and then the car brakes at 5.0 m/s². Find the total distance from seeing the hazard to stopping.

Answer Key — Full Step-by-Step Solutions

Conceptual

C1.

• Scalars: mass, speed, time, distance, temperature.
• Vectors: displacement, velocity, acceleration, force.

C2.

• (a) No. The straight line is the shortest path between two points, so the size of the displacement is at most the distance. They are equal only for straight-line motion in one direction.
• (b) Yes. One lap of a 400 m track: distance 400 m, displacement 0. The same goes for any round trip.

C3.

• (a) Same direction: 6 + 8 = 14 units.
• (b) Opposite directions: 8 − 6 = 2 units.
• (c) 6² + 8² = 36 + 64 = 100 = 10², so by Pythagoras they must be at 90°.

C4. Yes. A ball thrown straight up has v = 0 for an instant at the top, but its acceleration is still 9.81 m/s² downward. That is why it immediately starts to fall.

C5.

• (a) Yes, the speed is constant (60 km/h).
• (b) No. Its direction keeps changing, so its velocity (a vector) changes.
• (c) No. A changing velocity means a non-zero acceleration. Session 4 shows it is v²/r, toward the center.

C6. False. Negative acceleration means the acceleration points in the negative direction. A ball falling (taking up as positive) has negative velocity and negative acceleration, and it is speeding up. An object slows down only when a and v have opposite signs.

C7. Over a tiny time interval Δt the velocity is almost constant, so the displacement in that interval is v × Δt: the area of a thin strip under the graph. Adding up all the strips gives the total area, and the total displacement. For constant acceleration the area is a rectangle ut plus a triangle ½at², which is S2.

Numerical

N1.

• R = √(5² + 12²) = √(25 + 144) = √169 = 13 m.
• θ = tan⁻¹(12/5) = 67.4°.

Answer: 13 m at 67.4° north of east.

N2.

• Horizontal: Fₓ = 50 cos37° = 50 × 0.799 = 39.9 N.
• Vertical: F_y = 50 sin37° = 50 × 0.602 = 30.1 N.
• Check: √(39.9² + 30.1²) = √(1592 + 906) = √2498 = 50.0 N ✓

Answer: 39.9 N horizontal; 30.1 N vertical.

N3.

• First vector: (10, 0). Second: (6 cos60°, 6 sin60°) = (3.0, 5.20).
• Rₓ = 13.0 m, R_y = 5.20 m.
• R = √(13.0² + 5.20²) = √196 = 14.0 m; θ = tan⁻¹(5.20/13.0) = 21.8°.

Answer: 14.0 m at 21.8° north of east. It matches Worked Example 3, as it should.

N4.

• (a) Total distance = 120 + 120 = 240 km; total time = 1.5 + 2.5 = 4.0 h. Average speed = 240/4.0 = 60 km/h.
• (b) Back home, so the displacement is 0. Average velocity = 0.

Note that 60 km/h is not the average of 80 km/h and 48 km/h (64 km/h). The slower leg takes longer, so it counts for more.

N5.

• (a) 90/3.6 = 25 m/s.
• (b) 15 × 3.6 = 54 km/h.

N6.

• (a) a = (10 − 4.0)/3.0 = 2.0 m/s².
• (b) S0: s = ½(4.0 + 10)(3.0) = ½ × 14 × 3.0 = 21 m. Check with S2: 4.0 × 3.0 + ½ × 2.0 × 3.0² = 12 + 9 = 21 m ✓

Answer: 2.0 m/s²; 21 m.

N7.

• Known: u = 0, v = 80 m/s, a = 3.2 m/s².
• (a) S3: 80² = 0 + 2 × 3.2 × s, so s = 6400/6.4 = 1000 m (1.0 km).
• (b) S1: 80 = 0 + 3.2t, so t = 25 s.

Answer: 1000 m; 25 s.

N8.

• Forward is positive: u = 20, a = −4.0.
• (a) S1: 0 = 20 − 4.0t, so t = 5.0 s.
• (b) S3: 0 = 20² − 2 × 4.0 × s, so s = 400/8.0 = 50 m.
• (c) S2 with t = 2.0 s: s = 20 × 2.0 + ½ × (−4.0) × 2.0² = 40 − 8.0 = 32 m.

Answer: 5.0 s; 50 m; 32 m. More than half the distance is covered in the first 2 s, because the car is fastest then.

N9.

• Take down as positive to keep the signs simple: u = 0, a = +9.81, s = 45 m.
• (a) S2: 45 = 0 + ½ × 9.81 × t², so t² = 90/9.81 = 9.17 and t = 3.03 s.
• (b) S3: v² = 0 + 2 × 9.81 × 45 = 883, so v = 29.7 m/s. Check with S1: 9.81 × 3.03 = 29.7 ✓

Answer: 3.03 s; 29.7 m/s.

N10.

• Up is positive: u = 20, a = −9.81.
• (a) S3 with v = 0 at the top: 0 = 20² − 2 × 9.81 × h, so h = 400/19.62 = 20.4 m.
• (b) S1: 0 = 20 − 9.81t, so t = 2.04 s.
• (c) S1 at t = 3.0 s: v = 20 − 9.81 × 3.0 = 20 − 29.4 = −9.43 m/s. The minus sign means it is moving downward at 9.43 m/s.
• (d) S2: s = 20 × 3.0 − ½ × 9.81 × 3.0² = 60 − 44.1 = 15.9 m above the launch point.

Answer: 20.4 m; 2.04 s; 9.43 m/s downward; 15.9 m.

N11.

• Stage 1: u = 0, a = 2.0, t = 10. v = 20 m/s; s₁ = ½ × 2.0 × 10² = 100 m.
• Stage 2: constant 20 m/s for 20 s: s₂ = 20 × 20 = 400 m.
• Stage 3: 20 → 0 m/s in 5.0 s. S0: s₃ = ½(20 + 0)(5.0) = 50 m.
• (a) Total: 100 + 400 + 50 = 550 m.
• (b) It moves in one direction, so the displacement is 550 m. Total time: 10 + 20 + 5.0 = 35 s. v_avg = 550/35 = 15.7 m/s.

Answer: 550 m; 15.7 m/s.

N12.

• Reaction phase (constant speed, no braking yet): s₁ = 25 × 0.80 = 20 m.
• Braking phase: 0 = 25² − 2 × 5.0 × s₂, so s₂ = 625/10 = 62.5 m.
• Total: 20 + 62.5 = 82.5 m.

Answer: 82.5 m. The reaction distance alone is about a quarter of the total, which is why distracted driving is so dangerous.

Notes for next session

Session 2 takes these tools into two dimensions. A projectile’s launch velocity is resolved into components (V2). The horizontal motion is constant velocity, and the vertical motion is free fall using S1–S3 with a = −g. The key idea: the two directions are independent. The warm-up for Session 2 asks for these equations from memory.