Frame of reference: where are you measuring from?
Motion is always relative. To say where a thing is, we need a frame of reference: an origin (zero point), a chosen positive direction and a clock to read time.
A passenger sitting in a moving train is at rest compared to the train, but moving compared to the platform. Both answers are right; they use different frames.
In a straight line, one number x (with a sign) tells the position. Displacement Δx = x₂ − x₁ is the change of position; distance (path length) is the total ground covered, always positive.
Uniform and non-uniform motion
Uniform motion: equal displacements in equal time intervals, however small. The velocity is constant and the x–t graph is a straight line.
Non-uniform motion: unequal displacements in equal time intervals. The velocity changes and the x–t graph is curved.
Average and instantaneous speed and velocity
Average velocity = displacement ÷ time = Δx/Δt. Average speed = total path length ÷ time. If you go 100 m and come back in 50 s, average velocity is 0 but average speed is 4 m/s.
Instantaneous velocity is the velocity at one instant: v = lim(Δt→0) Δx/Δt = dx/dt. On the x–t graph it is the slope of the tangent. Instantaneous speed is the size of instantaneous velocity.
Average speed is always ≥ the size of average velocity; they are equal only if the object never turns back.
Simple differentiation and integration for motion
You only need a few rules: d/dt(tⁿ) = n tⁿ⁻¹, d/dt(constant) = 0, and ∫tⁿ dt = tⁿ⁺¹/(n + 1) + C.
- x = 5t² → v = dx/dt = 10t → a = dv/dt = 10 m/s².
- a = 4 → v = ∫4 dt = 4t + u → x = ∫v dt = 2t² + ut + x₀.
Differentiating goes x → v → a (slopes). Integrating goes a → v → x (areas), and the starting values u and x₀ fill in the constants.
Acceleration and uniformly accelerated motion
Acceleration is the rate of change of velocity: average a = Δv/Δt, instantaneous a = dv/dt (unit m/s²). When a is the same at every instant, the motion is uniformly accelerated; a freely falling body near the Earth is the best example (a = g ≈ 9.8 m/s², downward).
A negative a does not always mean slowing down. The object slows down only when v and a have opposite signs.
Position–time and velocity–time graphs
- x–t graph: slope = velocity. Straight line → uniform velocity; curve bending up → speeding up in + direction; horizontal line → at rest.
- v–t graph: slope = acceleration; area between the graph and the time axis = displacement (area below the axis counts as negative).
- a–t graph: area = change in velocity.
A graph can never show two positions at the same time, and speed can never be negative.
Equations of motion by graphs and calculus
Graph method (v–t straight line from u to v)
Slope: a = (v − u)/t → v = u + at. Area = rectangle ut + triangle ½·t·(at) → s = ut + ½at². Area as trapezium s = ½(u + v)t; put t = (v − u)/a → v² = u² + 2as.
Calculus method
dv/dt = a → ∫ᵤᵛ dv = ∫₀ᵗ a dt → v = u + at. dx/dt = u + at → ∫dx = ∫(u + at)dt → x − x₀ = ut + ½at². Also a = v dv/dx → ∫v dv = ∫a dx → v² − u² = 2a(x − x₀).
Distance in the nth second: sₙ = u + a(2n − 1)/2. These equations work only when a is constant.
Key formulas and definitions
- v = dx/dt, a = dv/dt = v·dv/dx
- v = u + at
- s = ut + ½at²
- v² = u² + 2as
- s = ½(u + v)t, sₙ = u + a(2n − 1)/2
- Average velocity = Δx/Δt; average speed = path length/time
Worked examples
1. A runner goes 400 m round a circular track and returns to the start in 80 s. Find average speed and average velocity.
Step 1: path length = 400 m, displacement = 0. Step 2: average speed = 400/80 = 5 m/s. Step 3: average velocity = 0/80 = 0.
2. The position of a particle is x = 3t² − 2t + 4 (x in m, t in s). Find velocity and acceleration at t = 2 s.
Step 1: v = dx/dt = 6t − 2. Step 2: at t = 2 s, v = 12 − 2 = 10 m/s. Step 3: a = dv/dt = 6 m/s² (constant).
3. A car starts from rest with a = 2 m/s². Find its velocity and distance after 5 s.
Step 1: u = 0, a = 2, t = 5. Step 2: v = u + at = 0 + 10 = 10 m/s. Step 3: s = ut + ½at² = 0 + ½ × 2 × 25 = 25 m.
4. A bike moving at 20 m/s brakes with a retardation of 5 m/s². How far does it go before stopping?
Step 1: u = 20, v = 0, a = −5. Step 2: v² = u² + 2as → 0 = 400 − 10s. Step 3: s = 40 m.
5. A ball is dropped from a 45 m high building (g = 10 m/s²). Find the time to reach the ground and the speed on hitting it.
Step 1: take downward positive: u = 0, a = 10, s = 45. Step 2: s = ½gt² → 45 = 5t² → t = 3 s. Step 3: v = gt = 30 m/s.
6. A v–t graph goes in a straight line from 4 m/s at t = 0 to 16 m/s at t = 6 s. Find the acceleration and the displacement.
Step 1: a = slope = (16 − 4)/6 = 2 m/s². Step 2: displacement = area of trapezium = ½(4 + 16) × 6 = 60 m. Step 3: check: s = ut + ½at² = 24 + 36 = 60 m ✓.
7. A body with u = 3 m/s and a = 2 m/s² moves in a straight line. Find the distance covered in the 5th second.
Step 1: sₙ = u + a(2n − 1)/2. Step 2: s₅ = 3 + 2 × 9/2 = 3 + 9 = 12 m. Step 3: check: s(5) − s(4) = (15 + 25) − (12 + 16) = 40 − 28 = 12 m ✓.
8. The acceleration of a particle is a = 6t m/s². It starts from rest at x = 0. Find v and x at t = 2 s.
Step 1: a is not constant, so use calculus, not v = u + at. Step 2: v = ∫₀ᵗ 6t dt = 3t² → v(2) = 12 m/s. Step 3: x = ∫₀ᵗ 3t² dt = t³ → x(2) = 8 m.
Common mistakes
- Using v = u + at or s = ut + ½at² when the acceleration changes with time. Use calculus instead.
- Treating distance and displacement as the same. They differ when the body turns back.
- Forgetting signs: once you choose up (or right) as positive, g is −9.8 m/s² for upward motion.
- Reading the area under an x–t graph as displacement. Only the area under a v–t graph gives displacement.