Velocity and acceleration as derivatives
Let a particle move on a straight line. Its position at time t is s(t) (metres). The average velocity between two times is the change in s divided by the change in t. If we make the time gap very small, we get the instantaneous velocity:
v(t) = ds/dt = s′(t)
In the same way acceleration is the rate of change of velocity:
a(t) = dv/dt = d²s/dt²
On the s–t graph, v is the slope of the tangent. On the v–t graph, a is the slope of the tangent. Units: s in m, v in m/s, a in m/s².
Signs: v > 0 means moving in the positive direction, v < 0 the negative direction, v = 0 means momentarily at rest. If v and a have the same sign, the particle speeds up; opposite signs, it slows down.
Rest, turning points and speed
A particle is at rest when v = 0. If v changes sign there, the particle turns round. In the 3D graph, this is where the s–t tangent is flat (a peak or a valley of s).
Speed is |v|, always positive. Velocity has a sign; speed does not. A car can have velocity −5 m/s (going backward) and speed 5 m/s.
Example pattern: s = t³ − 6t² + 9t. Then v = 3t² − 12t + 9 = 3(t − 1)(t − 3), so v = 0 at t = 1 and t = 3. The particle goes forward until t = 1 (s = 4), comes back until t = 3 (s = 0) and then goes forward again.
Displacement and distance by integration
Reverse the process. If v(t) is known, the displacement from t = a to t = b is the definite integral of v:
Displacement = ∫ₐᵇ v(t) dt = s(b) − s(a)
This is the signed area under the v–t graph: above the axis counts positive, below counts negative. So the displacement can be 0 even if the particle moved a lot.
The distance travelled counts every part as positive, so it is the integral of the speed:
Distance = ∫ₐᵇ |v(t)| dt
How to compute it: find where v = 0, split the interval at those times, integrate on each piece, take the absolute value of each piece and add.
Motion in a plane: parametric position
When a particle moves in a plane, its position is given by two functions: x(t) and y(t) (a parametric description). Differentiate each one:
- Velocity vector: (x′(t), y′(t))
- Acceleration vector: (x″(t), y″(t))
- Speed: √(x′(t)² + y′(t)²)
Example: x = 2t, y = t². Velocity = (2, 2t), acceleration = (0, 2), and at t = 1 the speed is √(4 + 4) = 2√2 ≈ 2.83 m/s. This is exactly the path of a thrown ball: steady motion sideways and accelerated motion up and down.
Distance along a curve (arc length)
If a particle runs along a curve, the distance it covers is the integral of its speed:
L = ∫ₐᵇ √(x′(t)² + y′(t)²) dt
Idea: in a tiny time dt the particle moves dx sideways and dy up; by Pythagoras the tiny step is √(dx² + dy²). Add all tiny steps by integration.
If the curve is given as y = f(x), the formula becomes L = ∫ₐᵇ √(1 + f′(x)²) dx.
Check with a circle of radius 3: x = 3cos t, y = 3 sin t gives speed 3, so the half circle (t from 0 to π) has length 3π ≈ 9.42, which is half of 2πr.
Key formulas and definitions
- v = ds/dt, a = dv/dt = d²s/dt²
- Displacement = ∫ₐᵇ v dt = s(b) − s(a)
- Distance = ∫ₐᵇ |v| dt (split where v = 0)
- Plane: v = (x′, y′), speed = √(x′² + y′²), a = (x″, y″)
- Curve length: L = ∫ √(x′² + y′²) dt or ∫ √(1 + f′(x)²) dx
Worked examples
1. A particle moves with s = 5t² + 2t (s in metres, t in seconds). Find its velocity and acceleration at t = 3.
v = ds/dt = 10t + 2, so v(3) = 32 m/s. a = dv/dt = 10 m/s² (constant).
2. For s = t³ − 6t² + 9t, find when the particle is at rest and its positions then.
v = 3t² − 12t + 9 = 3(t − 1)(t − 3). v = 0 at t = 1 and t = 3. s(1) = 1 − 6 + 9 = 4 m. s(3) = 27 − 54 + 27 = 0 m.
3. A particle has a = 6t, v(0) = 2 and s(0) = 1. Find v(t) and s(t), and v and s at t = 2.
v = ∫6t dt = 3t² + C. v(0) = 2 gives C = 2, so v = 3t² + 2. s = ∫(3t² + 2) dt = t³ + 2t + D; s(0) = 1 gives D = 1, so s = t³ + 2t + 1. At t = 2: v = 12 + 2 = 14 m/s, s = 8 + 4 + 1 = 13 m.
4. The velocity is v = 2t − 6 (m/s) for 0 ≤ t ≤ 5. Find the displacement and the distance travelled.
Displacement = ∫₀⁵ (2t − 6) dt = [t² − 6t]₀⁵ = 25 − 30 = −5 m. v = 0 at t = 3. On 0–3: [t² − 6t] = 9 − 18 = −9, so the distance is 9. On 3–5: (25 − 30) − (9 − 18) = 4, so the distance is 4. Total distance = 9 + 4 = 13 m.
5. A particle moves with x = 2t, y = t². Find its velocity vector, acceleration vector and speed at t = 1.
x′ = 2, y′ = 2t, so velocity = (2, 2t) = (2, 2) at t = 1. x″ = 0, y″ = 2, so acceleration = (0, 2). Speed = √(2² + 2²) = √8 ≈ 2.83 m/s.
6. A particle moves with v = t² − 4t + 3 for 0 ≤ t ≤ 4. Find the displacement and the total distance.
Antiderivative: s = t³/3 − 2t² + 3t. At 4: 64/3 − 32 + 12 = 4/3. At 0: 0. Displacement = 4/3 m ≈ 1.33 m. v = (t − 1)(t − 3) = 0 at t = 1 and 3. On 0–1: s(1) − s(0) = 1/3 − 2 + 3 = 4/3. On 1–3: s(3) − s(1) = 0 − 4/3 = −4/3, absolute 4/3. On 3–4: 4/3 − 0 = 4/3. Distance = 4/3 + 4/3 + 4/3 = 4 m.
7. Find the length of the curve y = (2/3)x^(3/2) from x = 0 to x = 3.
f′(x) = x^(1/2), so 1 + f′² = 1 + x. L = ∫₀³ √(1 + x) dx = (2/3)(1 + x)^(3/2) from 0 to 3 = (2/3)(8 − 1) = 14/3 ≈ 4.67 units.
Common mistakes
- Taking displacement as distance. Displacement is ∫v dt with signs; distance is ∫|v| dt.
- Not splitting at v = 0 when finding distance. A single integral lets the backward part cancel the forward part.
- Forgetting the constant when integrating a to get v. Use the starting values (initial conditions) to find it.
- Using speed = x′ + y′ in the plane. Speed is √(x′² + y′²), the length of the velocity vector.