UY1: Basics & Kinematics

This lesson is a UY1 primer for mechanics: how we model motion, choose conventions, and use kinematics cleanly in 1D and 2D (including projectiles and circular motion).

At a glance

• You will learn: how to define v and a from x(t); when constant-acceleration equations apply (and when they don’t); how to separate 2D motion into components; how to do relative-velocity addition; how to decompose curved motion into tangential/radial parts
• Key result: for constant acceleration, x(t) = x₀ + v₀t + (1/2)at².
• Common trap: writing equations before fixing axes/signs (especially the sign of g), or treating projectile motion as one coupled equation instead of two independent 1D motions linked by the same time t

Setup (models and conventions)

We usually model a moving object as a particle (point mass) unless its size/rotation matters. Kinematics describes motion without asking why it moves (forces come next).

Before you write component equations, choose:

• coordinates (e.g. x horizontal, y vertical),
• a sign convention (e.g. + y upward),
• what g means in your convention (if + y upward, then a_y = -g).

We use:

• In 1D, position is x(t), with v(t) = dx/dt, a(t) = dv/dt = d²x/dt².
• In 2D/3D, position is vector r(t), with vector v(t) = (d vector r)/dt, vector a(t) = (d vector v)/dt = (d² vector r)/dt².

Kinematics in 1D

Definitions (from calculus)

If x(t) is the position along an axis:

Displacement from (tᵢ,xᵢ) to (t_f,x_f): Δ x = x_f - xᵢ = x(t_f)-x(tᵢ).

Average velocity over Δ t = t_f-tᵢ: v_avg = (Δ x)/(Δ t) = (x(t_f)-x(tᵢ))/(t_f-tᵢ).

Instantaneous velocity (slope of the x–t graph): v(t) = lim _(Δ t → 0)(x(t + Δ t)-x(t))/(Δ t) = dx/dt.

Average acceleration: a_avg = (Δ v)/(Δ t) = (v(t_f)-v(tᵢ))/(t_f-tᵢ).

Instantaneous acceleration (slope of the v–t graph): a(t) = lim _(Δ t → 0)(v(t + Δ t)-v(t))/(Δ t) = dv/dt = d²x/dt².

Checks: [v] = m s⁻¹, [a] = m s⁻².

Constant-acceleration model (1D)

If acceleration is constant on the time interval of interest, integrate twice and use initial conditions (x₀,v₀) at t = 0:

These are valid only when acceleration is constant and motion is along one axis.

How to use the model (and not misuse it)

• If a is piecewise constant, apply these equations piece-by-piece and reset (x₀,v₀) at the boundary.
• If a depends on t, x, or v (e.g. drag), you generally need calculus rather than the constant-a formulas.

Checks: if a = 0, then v = v₀ and x = x₀ + v₀t (constant-velocity motion).

Free fall sign convention (1D)

Near Earth’s surface, neglecting air resistance, the acceleration is approximately constant with magnitude g ≈ 9.80 m s⁻².

If you choose + y upward, then a_y = -g.

Kinematics in 2D/3D

Vectors and components

In 2D/3D, use a position vector vector r(t) and work component-by-component once you have chosen axes.

Displacement: Δ vector r = vector r(t₂)- vector r(t₁).

Average velocity: vector v_avg = (Δ vector r)/(Δ t) = (vector r(t₂)- vector r(t₁))/(t₂-t₁).

Instantaneous velocity: vector v(t) = lim _(Δ t → 0)(vector r(t + Δ t)- vector r(t))/(Δ t) = (d vector r)/dt.

The direction of vector v is tangent to the trajectory.

Average acceleration: vector a_avg = (Δ vector v)/(Δ t) = (vector v(t₂)- vector v(t₁))/(t₂-t₁).

Instantaneous acceleration: vector a(t) = lim _(Δ t → 0)(vector v(t + Δ t)- vector v(t))/(Δ t) = (d vector v)/dt.

Scroll diagram horizontally to read all labels.

Constant-acceleration model (2D/3D)

If vector a is constant, you can integrate vector equations directly: vector v(t) = vector v₀ + vector at. Then integrate again: vector r(t) = vector r₀ + vector v₀t + (1/2) vector at².

In components, each axis obeys the 1D constant-acceleration equations. For 2D with vector a = (aₓ,a_y):

Key idea (when components decouple): if aₓ depends only on x/t and a_y depends only on y/t (e.g. aₓ = 0, a_y = -g), then you can solve x and y as two separate 1D problems linked only by the same time t. With drag or constraints, components can couple through vector v and/or vector r.

Relative motion (Galilean addition)

Relative motion is mostly bookkeeping with vectors. If an object has velocity vector v_(A/B) “of A relative to B”, then: vector v_(A/C) = vector v_(A/B) + vector v_(B/C).

For two inertial frames S and S' in standard relative motion (constant frame velocity vector u): vector r = vector r ' + vector ut, vector v = vector v ' + vector u, vector a = vector a ' (accelerations are the same in all inertial frames in Newtonian mechanics).

Projectile motion (standard model)

Assumptions (typical first pass):

• aₓ = 0 (neglect air resistance)
• a_y = -g (uniform gravitational field)

If vector v₀ makes an angle α₀ above the horizontal and you choose x₀ = y₀ = 0 at t = 0:

Eliminating t gives the trajectory: y = (tan α₀) x-(g/(2 v₀² cos² α₀)) x²

Maximum height (at v_y = 0): h = (v₀² sin² α₀)/2g

Range on level ground ( y = 0 again): R = (v₀² sin(2 α₀))/g

Time of flight (same launch/landing height): T = (2v₀ sin α₀)/g.

General landing height y_f (use the component method, then solve a quadratic): y_f = y₀ + v_(0,y)t-(1/2)gt² ⇒ (1/2)gt²-v_(0,y)t-(y₀-y_f) = 0. Pick the physical (positive-time) root, then get the horizontal displacement from x_f-x₀ = v_(0,x)t.

Checks: h and R have units of metres; R → 0 as α₀ → 0.

Curvilinear motion (tangent–normal form) (optional but powerful)

For motion along any smooth curve, define:

• t hat: unit tangent (instantaneous direction of motion),
• n hat: unit normal (toward the center of curvature),
• v = ‖ vector v‖: speed,
• ρ: radius of curvature.

Then the kinematics decomposes into: vector v = v t hat vector a = dv/dt t hat + v²/ρ n hat .

For a circle, ρ = R and n hat points radially inward, giving the familiar aᵣ = v²/R.

Circular motion (uniform and non-uniform)

Circular motion is easiest when you separate radial (toward the center) and tangential (along the motion) directions.

Define unit vectors:

• r hat: radial inward (centripetal) direction,
• t hat: tangential direction (along the motion).

Then: vector a = aᵣr hat + aₜt hat, aᵣ = v²/R, aₜ = dv/dt.

Using angular variables: ω = dθ/dt, α = dω/dt. v = ω R, aᵣ = ω²R, aₜ = α R.

Uniform circular motion

Period: T = (2 π R)/v

Radial (centripetal) acceleration magnitude: aᵣ = v²/R = (4 π² R)/T²

Checks: if you double v at fixed R, then aᵣ increases by a factor of 4.

Non-uniform circular motion

If the speed changes, there is also tangential acceleration: aₜ = (d ‖ vector v ‖)/dt = dv/dt

The total acceleration decomposes into radial and tangential parts: vector a = aᵣr hat + aₜt hat aᵣ = v²/R

Worked examples

Example: 1D constant-acceleration stop

Problem

• Given: v₀ = 25 m s⁻¹, v = 0, Δ x = 80 m, constant a
• Find: a and the stopping time t

Approach

• Use the constant-acceleration model (1D) because the problem states constant a.
• Use v² = v₀² + 2aΔ x to find a, then use v = v₀ + at to find t.

Working Use v² = v₀² + 2aΔ x: 0 = 25² + 2a(80) ⇒ a = -25²/160 = -3.91 m s⁻².

Then use v = v₀ + at: 0 = 25 + (-3.91)t ⇒ t = 6.40 s.

Answer + checks

• a = -3.91 m s⁻², t = 6.40 s
• Units: a is in m s⁻²; sign: negative means acceleration opposite to the motion (slowing down)

Example: projectile range and peak height

Problem

• Given: v₀ = 20 m s⁻¹, α₀ = 30°, g = 9.80 m s⁻², launch and landing at the same height
• Find: peak height h and range R

Approach

• Use the component method: x-motion has aₓ = 0, y-motion has a_y = -g.
• Use standard results for same-height launch/landing: h = (v₀² sin² α₀)/2g, R = (v₀² sin(2α₀))/g.

Working Peak height: h = (v₀² sin² α₀)/2g = (400 (0.5)²)/19.6 = 5.10 m.

Range: R = (v₀² sin(2α₀))/g = (400 sin 60°)/9.80 = 35.3 m.

Answer + checks

• h = 5.10 m, R = 35.3 m
• Units: metres; limit: R → 0 as α₀ → 0

Practice set (with hints + answers)

1. A particle moves along x with x(t) = 2t²-3t + 5 (SI units). Find v(t) and a(t). Find the time when the particle is instantaneously at rest. Find the displacement and distance travelled from t = 0 to t = 3 s.
2. A particle has time-dependent acceleration a(t) = 4-6t (SI units). Given x(0) = 0 and v(0) = 2 m s⁻¹, find v(t) and x(t). When does it first momentarily stop?
3. Show that in 1D a = vdv/dx. A particle has a(x) = -kx with k = 2.0 s⁻². If x(0) = 0 and v(0) = 3.0 m s⁻¹, find v(x) and the turning point xₘₐₓ.
4. A particle moves in the plane with vector r(t) = (3t²)i hat + (2t³-t)j hat (m). Find vector v(t) and vector a(t). Evaluate the speed and direction of motion at t = 1 s.
5. A projectile is launched from a cliff of height y₀ = 20 m with speed v₀ = 18 m s⁻¹ at α₀ = 35° above the horizontal. Take g = 9.80 m s⁻² and neglect air resistance. Find the flight time to the ground and the horizontal range from the launch point.
6. (Relative motion) A river is 200 m wide. A boat moves at 1.5 m s⁻¹ relative to the water; the current is 0.8 m s⁻¹ downstream. What heading (angle upstream from straight across) makes the boat land directly opposite? How long does the crossing take?
7. A particle moves on a circle of radius R = 0.50 m with angular position θ(t) = 2t + 0.40t² (rad). Find ω(t), α(t), the speed v, and the radial/tangential accelerations at t = 3.0 s.
8. A particle has velocity v(t) = 6-4t for 0 ≤ t ≤ 3 s and x(0) = 0. Find x(t). Find the displacement and the distance travelled over the interval.

Hints

1. Differentiate: v = dx/dt, a = dv/dt. For distance travelled, split the time interval where v(t) = 0.
2. Integrate: v(t) = v(0) + ∫₀^t a(t') dt', then x(t) = x(0) + ∫₀^t v(t') dt'.
3. Use the chain rule dv/dt = (dv/dx)dx/dt. The turning point occurs when v = 0.
4. Differentiate vector components; direction angle satisfies tan φ = v_y/vₓ at that instant.
5. Solve y(t) = 0 using y(t) = y₀ + v_(0,y)t-(1/2)gt², keep the positive-time root, then use x = v_(0,x)t.
6. Use vector v_(boat/ground) = vector v_(boat/water) + vector v_(water/ground). “Directly opposite” means zero downstream component.
7. Use ω = dθ/dt, α = dω/dt, v = ω R, aᵣ = ω²R, aₜ = α R.
8. Integrate v(t) to get x(t). Distance travelled is ∫₀³ |v(t)| dt so split at v(t) = 0.

Answers

1. v(t) = 4t-3, a(t) = 4. Instantaneous rest at t = 3/4 s. Displacement: Δ x = x(3)-x(0) = 14-5 = 9 m. Distance travelled: 11.25 m.
2. v(t) = 2 + 4t-3t², x(t) = 2t + 2t²-t³. First stop at t = (2 + square root of 10)/3 ≈ 1.72 s.
3. a = vdv/dx. With a(x) = -kx and x(0) = 0, v(0) = v₀: v² = v₀²-kx² ⇒ v(x) = ± square root of (9-2x²) m s⁻¹. Turning point: xₘₐₓ = v₀/(square root of k) = 3/(square root of 2) = 2.12 m.
4. vector v(t) = (6t)i hat + (6t²-1)j hat, vector a(t) = 6i hat + (12t)j hat . At t = 1 s: speed = square root of 61 = 7.81 m s⁻¹, direction φ = tan⁻¹ (5/6) = 39.8° above + x.
5. Flight time t = 3.33 s, range R = 49.1 m. (to 3 s.f.)
6. Heading angle θ = sin⁻¹ (0.8/1.5) = 32.2° upstream from straight across; time t = 200/(1.5 cos θ) = 158 s (3 s.f.).
7. ω(t) = 2 + 0.80t, α = 0.80 rad s⁻². At t = 3.0 s: ω = 4.4 rad s⁻¹, v = 2.20 m s⁻¹, aᵣ = 9.68 m s⁻², aₜ = 0.40 m s⁻². Magnitude | vector a| ≈ 9.69 m s⁻².
8. x(t) = ∫₀^t(6-4t') dt' = 6t-2t². Displacement on [0,3]: x(3)-x(0) = 0. Distance travelled: 9.0 m.

Quiz (MCQ)

Quiz — Basics & Kinematics (UY1)

Summary + next steps

• Choose axes and a sign convention before writing equations (especially for g).

• In 1D: v = dx/dt, a = d²x/dt². Check units early.

• Constant-acceleration equations apply only when a is (approximately) constant on the interval.

• In 2D, solve x and y motions separately; connect them with the same time t.

• Relative velocity is vector addition: vector v_(A/C) = vector v_(A/B) + vector v_(B/C).

• For projectiles (no drag): aₓ = 0, a_y = -g; standard formulas follow from the component method.

• For curved motion: tangent–normal decomposition cleanly separates “change of speed” and “change of direction”.

• For circular motion: use radial/tangential decomposition; aᵣ = v²/R.

  Next: Circular Motion (UY1) Back To Mechanics (UY1)