W1. Kinematics

1. Theory

Week 1 builds kinematics from zero: first motion along a straight line (lecture) — position, velocity, acceleration and their graphs — then the geometric methods of the tutorial: constraints, relative motion, Mach cones, pursuit and curved trajectories. The lab adds the working notation.

1.1 Position, Displacement, Velocity and Speed

To locate an object give its position relative to an origin; a change is the displacement (a vector: magnitude plus sign on an axis). Average velocity is the slope of the secant on the graph; average speed drops the direction. They coincide only for one-directional motion without turnarounds.

• Worked check (the uphill riddle): 30 km/h up and 60 km/h back over the same distance : km/h — not 45, and no extra data needed. Distance and time both cancel.
• Worked check (truck): 8.4 km at 70 km/h (7.2 min), then 2.0 km on foot (30 min) to the station: displacement km over min gives km/h; the round trip adds 2.0 km and 45 min for km/h.
• Notation used everywhere: position, distance, velocity, acceleration; metres, seconds, km/h; subscripts for initial values.

1.2 Instantaneous Velocity and Acceleration

Shrink : (instantaneous) velocity is the slope of the tangent to ; speed is what the speedometer shows. Likewise acceleration is the slope of — the elevator’s graph is read straight off the kinks of its graph ( m/s² speeding up, braking, cruising). Same signs of and mean speeding up; opposite signs mean slowing down. Large accelerations come in ’s ( m/s²; a roller coaster hits ).

• Worked check: for , , , : constant velocity holds for (1) and (4) — zero counts as constant; negative-direction motion for (2) and (3).

1.3 Constant Acceleration

When is constant, averages equal instants and everything integrates in closed form: from , and from integrating once more. The parabola, linear and flat graphs are the signatures to recognise.

1.4 Non-Constant Acceleration

When depends on or , integrate — and when it depends on position, trade for via the chain rule . For the tram with between stops ( at both ends): gives , and at the far stop yields (the root is the start).

• Key Pitfall: problems punish blind substitution — that formula is constant- only.

1.5 Constrained Motion

Rigid bodies couple coordinates. A rod of length with ends (floor) and (wall) obeys ; a rod rising on a moving wedge obeys . Write the geometric constraint, differentiate in time, solve for the unknown rate.

• Worked check: starts at moving right with : — accelerates downward while is uniform; wedge rise speed .
• Key Pitfall: coupled components are never independent — freezing one coordinate while differentiating silently breaks the constraint.

1.6 Relative Motion and Collision

Uniform particles collide iff for some : in particle 1’s frame this means the relative motion aims at the origin. The trains-and-bird classic dissolves this way: trains at 30 km/h each from 60 km meet in 1 h, so the 60 km/h bird covers 60 km — no infinite series needed. Pursuit variants (cheetah–antelope with a head start, police with delayed constant acceleration) are the same equation with careful clocks: the cheetah closes m at km/h m/s in s ( m, inside its s budget; that budget caps any head start at m); the officer’s gives s and m/s at the catch.

• Key Pitfall: delayed starts shift one clock — always write both motions against a single .

1.7 Mach Cone of Supersonic Motion

A source faster than the waves () outruns its own sound; wavefronts pile into a cone of half-angle . For height and delay after flyover, .

• Worked check: km, s, m/s gives m/s — consistently supersonic.

1.8 Pursuit Curves

Points perpetually heading at each other preserve symmetry: three points on a triangle keep it equilateral while it shrinks and spins. Project onto the shrinking side: chaser plus target’s give , meeting at . Never integrate the curved paths — project velocities onto the line that matters.

1.9 Curvilinear Motion and Curvature

Through the unit tangent, and : tangential part brakes or pushes, normal part turns, with curvature radius ( on straights, on circles).

• Worked check: from rest with , : converts time to distance, gives and .

---

2. Definitions

• Radius vector: Vector from the chosen origin to the point’s position at time .
• Material point: Idealised body whose size and rotation are ignored; only its position matters.
• Velocity: Time derivative of the radius vector, , tangent to the trajectory.
• Speed: Magnitude of the velocity, .
• Acceleration: Time derivative of velocity, .
• Geometric constraint: Rigid-body relation between coordinates (e.g. ) that couples their rates.
• Relative velocity: Velocity of one body as seen from another’s frame, .
• Mach cone: Envelope of wavefronts from a supersonic source, half-angle .
• Unit tangent vector: Unit vector along the trajectory in the direction of motion.
• Tangential acceleration: Component changing the speed.
• Normal acceleration: Component changing the direction of motion.
• Curvature radius: Radius of the osculating circle; inverse of the trajectory’s curvature.

---

3. Formulas

• Velocity: , tangent to the trajectory; speed .
• Acceleration: .
• Rigid-rod constraint: , differentiate for coupled rates.
• Mach angle: ; delay-height-speed .
• Pursuit closing rate: along the chased side.
• Tangential-normal split: .
• Averages: , , .
• Constant acceleration: , .
• General motion by integration: , .
• Position-dependent acceleration: .

---

4. Practice

4.1. Trains and a Bird (Lab 1, Task 1)

Two trains approach each other at km/h each from km apart; a bird at km/h shuttles between them until collision. What total distance does the bird cover?

Click to see the solution

Key Concept: Total time decides everything; the zigzag is a distraction.

1. Find the meeting time, multiply by bird speed.
   • Closing speed km/h over km gives h; bird distance .
   • Answer: km.

4.2. Cheetah Pursuit (Lab 1, Task 2a)

Cheetah km/h versus antelope km/h same direction, m head start. How long to catch, and how far does the cheetah run?

Click to see the solution

Key Concept: Work in relative motion; convert km/h to m/s first.

1. Close the gap at relative speed.
   • km/h m/s; s; cheetah distance m — inside the s stamina budget.
   • Answer: s and m.

4.3. Maximum Head Start (Lab 1, Task 2b)

Same speeds; the cheetah holds top speed only s. What is the largest catchable head start?

Click to see the solution

Key Concept: Invert the pursuit: gap equals relative speed times available time.

1. Multiply the budget.
   • m.
   • Answer: m.

4.4. Police Pursuit (Lab 1, Task 3)

A car at m/s passes a stationary officer, who starts s later at m/s². How fast is the officer at the overtake? Sketch both .

Click to see the solution

Key Concept: One clock for both motions; delayed start shifts the parabola.

1. Equate positions with at the pass.
   • gives , valid root s; officer speed m/s.
   • Answer: m/s; is a line versus a delayed parabola.

4.5. Decaying Acceleration (Lab 1, Task 4)

A particle has with . Find maximum speed and .

Click to see the solution

Key Concept: Integrate forward; the exponential dies, the distance keeps what it earned.

1. Integrate twice.
   • , so ; with .
   • Answer: , .

4.6. Reading an S–t Graph (Lab 1, Task 5)

From the distance-time graph (sigmoid to m over s): average velocity, maximum velocity, and the time where instantaneous velocity equals the mean over .

Click to see the solution

Key Concept: Secant slopes are averages, tangent slopes are instants.

1. Read slopes off the grid.
   • Average m/s; steepest tangent (inflection, –) m/s; is where the tangent passes through the origin, s — read yours to grid precision.
   • Answer: m/s average, m/s max, s.

4.7. Falling Bolt in an Elevator (Lab 1, Task 6)

Elevator cabin m tall ascends at m/s²; s in, a bolt drops from the ceiling. Find its fall time, displacement and distance in the shaft frame.

Click to see the solution

Key Concept: Share one clock; the drop starts with the cabin’s velocity.

1. Meet the accelerating floor.
   • At drop, m/s up; floor: ; bolt: ; gives , s — the terms cancel.
2. Track the bolt’s path.
   • It rises m first, then falls to m below the drop point: displacement m, distance m.
   • Answer: s fall; displacement m; distance m.

4.8. Motion from Position (Lab 1, Task 7)

, where is in metres and in seconds. At s determine (a) position, (b) velocity, (c) acceleration. For position determine (d) the maximum positive coordinate and (e) when it is reached. For velocity determine (f) the maximum positive velocity, (g) when it is reached, and (h) the acceleration at the nonzero time when the particle is not moving. Over s determine (i) the average velocity.

Click to see the solution

Key Concept: Differentiate for velocity and acceleration; zeroes of derivatives locate extrema.

1. Read position, velocity and acceleration at s.
   • m; gives m/s; gives m/s².
   • Answer: (a) m; (b) m/s; (c) m/s².
2. Locate the maximum of .
   • gives , so or ; confirms a maximum, m, and as so it is global.
   • Answer: (d) m; (e) s.
3. Locate the maximum of and the standstill acceleration.
   • gives s with m/s; the particle stands still when (), so at the nonzero standstill s the acceleration is m/s².
   • Answer: (f) m/s; (g) s; (h) m/s².
4. Average over .
   • m/s.
   • Answer: (i) m/s.

4.9. Truck Trip Averages (Lecture 1, Task 1)

Drive km at km/h, run out of fuel, walk km more ( min) to the station, then walk back ( min). Find displacement to the station, time to the station, average velocity there, and average speed for the whole trip.

Click to see the solution

Key Concept: Displacement cares about endpoints; distance counts every step.

1. Tally distance and time separately.
   • Displacement km over min ( h) gives km/h; full trip km over min ( h) gives km/h.
   • Answer: km; h; km/h; km/h.

4.10. Constant-Velocity Quiz (Lecture 1, Task 2)

For , , , : in which is velocity constant, and in which is it in the negative direction?

Click to see the solution

Key Concept: Differentiate; zero is a constant too.

1. Read .
   • Velocities , , , : constant in (1) and (4); negative-direction in (2) and (3).
   • Answer: constant — (1), (4); negative — (2), (3).

4.11. Tram Between Stops (Lecture 1, Example 1)

A tram runs stop A to stop B with ( at both). Find .

Click to see the solution

Key Concept: Position-dependent acceleration wants .

1. Integrate over distance, apply both rest conditions.
   • ; at gives .
   • Answer: (the root is the start).

4.12. Rod on a Moving Incline (Tutorial 1, Example 1)

A rod slides vertically in guides AB while its lower end rests on a wedge of angle moving left with speed . Find the rod’s rise speed .

Click to see the solution

Key Concept: A rigid contact couples the coordinates — differentiate the geometric constraint.

1. Relate the displacements over .
   • Wedge shifts left; rod rises ; contact stays on the slope, so .
   • Answer: .

4.13. Ladder Sliding Down a Wall (Tutorial 1, Example 2)

A rod of length has end on the floor (initial coordinate , constant speed rightward) and end on the wall. Find ’s coordinate versus time.

Click to see the solution

Key Concept: Pythagoras as a time-dependent constraint.

1. Write the constraint and differentiate.
   • with , hence .
   • Answer: .

4.14. Collision Condition (Tutorial 1, Example 3)

Particles move uniformly with , from , . When do they collide?

Click to see the solution

Key Concept: Go to the frame of particle 1; collision means the relative motion aims at the origin.

1. Demand a common meeting time.
   • must hold for some , i.e.  is collinear with with the correct sign.
   • Answer: collision iff the initial separation is parallel to the relative velocity and closing.

4.15. Supersonic Aircraft Speed (Tutorial 1, Example 4)

An aircraft flies horizontally at km; sound reaches an observer s after flyover. Sound speed m/s. Find the aircraft speed .

Click to see the solution

Key Concept: Mach-cone geometry links delay, height and speed.

1. Read the cone triangle.
   • and horizontal leg ; with get .
   • Answer: m/s.

4.16. Three Pursuing Points (Tutorial 1, Example 5)

Three points at vertices of an equilateral triangle of side move with speed , each heading for the next. When do they meet?

Click to see the solution

Key Concept: Project all velocities onto the shrinking side.

1. Compute the closing rate of one side.
   • Chaser contributes ; target’s velocity projects along the side, so with .
   • Answer: .

4.17. Curvature Radius on a Spiral (Tutorial 1, Example 6)

A point starts from rest with , ( constants). Find curvature radius and total acceleration versus distance .

Click to see the solution

Key Concept: Trade time for distance via , then use .

1. Express time through distance.
   • , so and .
2. Insert into the normal law.
   • , hence ; .
   • Answer: , .