Kinematics Using Calculus: Velocity, Acceleration, Distance

When Acceleration Is Not Constant

The equations of motion, v = u + at and s = ut + ½at², only work when acceleration is constant. A car pulling away, a rocket burning fuel or a particle in a changing force field all have acceleration that varies with time. For these, you need calculus. This tutorial explains kinematics using calculus: how displacement, velocity and acceleration are linked by differentiation and integration, how to find when a particle is at rest or changes direction, the difference between displacement and distance travelled, and how to handle acceleration given as a function of velocity or displacement. It is aimed at Further Mathematics (mechanics) and first-year university students.

Kinematics Using Calculus: The Three Links

For motion in a straight line, with displacement s from a fixed point O at time t:

Velocity is the rate of change of displacement: v = ds/dt.
Acceleration is the rate of change of velocity: a = dv/dt = d²s/dt².

Going the other way, integrate:

v = ∫ a dt and s = ∫ v dt, each with a constant of integration found from the initial conditions.

A useful memory picture: differentiate going down the list s → v → a, integrate going up a → v → s.

Example 1: Differentiating Displacement

A particle moves so that its displacement in metres after t seconds is s = t³ − 6t² + 9t + 2.

(a) Velocity and acceleration. v = ds/dt = 3t² − 12t + 9 and a = dv/dt = 6t − 12.

(b) When is the particle at rest? At rest means v = 0: 3t² − 12t + 9 = 0, so t² − 4t + 3 = 0, (t − 1)(t − 3) = 0, giving t = 1 s and t = 3 s.

(c) Positions at those times. s(1) = 1 − 6 + 9 + 2 = 6 m and s(3) = 27 − 54 + 27 + 2 = 2 m. The initial position is s(0) = 2 m.

(d) Acceleration when at rest. a(1) = −6 m/s² and a(3) = 6 m/s².

(e) Direction of motion. v = 3(t − 1)(t − 3) is positive for 0 ≤ t < 1, negative for 1 < t < 3, and positive for t > 3. So the particle moves forwards, then backwards, then forwards again. It changes direction at t = 1 and t = 3.

Displacement Versus Distance Travelled

This is the most common source of lost marks. Displacement is the change in position (it can be negative); distance travelled is the total length of the path (always positive).

In Example 1, between t = 0 and t = 4:

Displacement = s(4) − s(0). s(4) = 64 − 96 + 36 + 2 = 6, so the displacement is 6 − 2 = 4 m.
For the distance, split the journey wherever the particle changes direction. From t = 0 to 1 it moves from 2 to 6 (4 m). From t = 1 to 3 it moves back from 6 to 2 (4 m). From t = 3 to 4 it moves from 2 to 6 (4 m). Total distance = 4 + 4 + 4 = 12 m.

In integral form: displacement = ∫ v dt, but distance = ∫ |v| dt. Integrating v straight from 0 to 4 gives only 4 m, because the backward motion cancels part of the forward motion.

Example 2: Integrating Acceleration

A particle starts from rest at the origin. Its acceleration is a = 6t − 4 m/s².

Velocity. v = ∫(6t − 4) dt = 3t² − 4t + C. At t = 0, v = 0, so C = 0: v = 3t² − 4t.

Displacement. s = ∫(3t² − 4t) dt = t³ − 2t² + D. At t = 0, s = 0, so D = 0: s = t³ − 2t².

When does it return to the origin? s = t²(t − 2) = 0 gives t = 0 or t = 2 s.

Its velocity then: v(2) = 12 − 8 = 4 m/s.

Greatest distance from O before returning: the particle is momentarily at rest when v = t(3t − 4) = 0, so at t = 4/3. Then s = (64/27) − 2(16/9) = 64/27 − 96/27 = −32/27 ≈ −1.19 m. So it moves 32/27 m in the negative direction before turning back.

Example 3: Constant Acceleration Recovered

Calculus gives back the familiar equations. If a is constant, then v = ∫ a dt = at + C, and v = u when t = 0, so v = u + at. Integrating again, s = ut + ½at² (taking s = 0 at t = 0). The SUVAT equations are a special case of the calculus approach.

Acceleration as a Function of Velocity or Displacement

Sometimes the acceleration depends on the velocity (as with air resistance) or on the position (as with a spring). Use the chain rule:

a = dv/dt = (dv/ds)(ds/dt) = v dv/ds.

Example 4 (a depends on s). A particle moves with a = −4s (simple harmonic motion). It passes through s = 0 with speed 6 m/s. Find v in terms of s.

v dv/ds = −4s. Separate the variables: ∫ v dv = ∫ −4s ds, so ½v² = −2s² + C. At s = 0, v = 6, so C = 18. Then v² = 36 − 4s². The particle stops when v = 0, at s = ±3 m: this is the amplitude of the oscillation.

Example 5 (a depends on v). A boat switches off its engine at 10 m/s and is slowed by water resistance so that a = −0.5v. Find v after t seconds.

dv/dt = −0.5v. Separate: ∫ (1/v) dv = ∫ −0.5 dt, so ln v = −0.5t + C. At t = 0, v = 10, so C = ln 10. Therefore v = 10e^(−0.5t). After 4 s, v = 10e^(−2) ≈ 1.35 m/s. The boat never quite stops in this model, but it slows quickly.

How far does it drift? s = ∫ 10e^(−0.5t) dt from 0 to ∞ = 10 × 2 = 20 m.

Reading Velocity–Time Graphs with Calculus

The gradient of a velocity–time graph is the acceleration (the derivative).
The area under a velocity–time graph is the displacement (the integral). Areas below the time axis count as negative displacement.
For distance travelled, add all the areas as positive values.

Practice Questions

s = 2t³ − 9t² + 12t. Find when the particle is at rest. Answer: v = 6t² − 18t + 12 = 6(t − 1)(t − 2) = 0, so t = 1 and t = 2.
For the same particle, find the distance travelled from t = 0 to t = 2. Answer: s(0) = 0, s(1) = 5, s(2) = 4; distance = 5 + 1 = 6 m.
a = 12 − 6t, and the particle starts from rest. Find its maximum velocity. Answer: v = 12t − 3t², maximum when a = 0 at t = 2: v = 24 − 12 = 12 m/s.

Summary

v = ds/dt and a = dv/dt; integrate to go back, using initial conditions for the constants.
"At rest" means v = 0; a particle changes direction where v changes sign.
Distance travelled is found by splitting the journey at every change of direction.
When a depends on s, use a = v dv/ds; when it depends on v, separate variables in a = dv/dt.
Constant-acceleration equations are a special case of these calculus methods.