Kinematics with Calculus

A motorised cart on a track follows x = 2.0t³ − 9.0t² + 12t, with x in metres and t in seconds. Differentiating once gives v = 6.0t² − 18t + 12 = 6.0(t − 1)(t − 2) m s⁻¹, and again gives a = 12t − 18 m s⁻². Going down the chain needs no initial condition, and the derivatives answer questions the formula for x hides. The cart starts at 12 m s⁻¹ with a = −18 m s⁻², so it is slowing. v = 0 at t = 1.0 s and 2.0 s, and v changes sign at both: those are turning points, at x = 5.0 m and x = 4.0 m, so the cart runs out, backs up a metre, then drives on. a = 0 at t = 1.5 s, where v = −1.5 m s⁻¹ is its fastest backward motion. Comparing signs gives the rest. From 1.0 to 1.5 s, v and a are both negative and the cart speeds up backwards; from 1.5 to 2.0 s they differ and it slows. Speeding up means v and a share a sign, not that a is positive.

Run the chain backwards. A cart has a = (6.0 m s⁻³)t. Every function 3.0t² + C has that derivative, so the acceleration alone describes a whole family of velocity graphs, each shifted vertically from the next. Knowing v = −4.0 m s⁻¹ at t = 0 selects one: v = 3.0t² − 4.0. Integrating again gives another family, x = 1.0t³ − 4.0t + C₂, and x = 2.0 m at t = 0 selects x = 1.0t³ − 4.0t + 2.0. At t = 2.0 s the cart is back at 2.0 m, moving at 8.0 m s⁻¹. Start it from the same 2.0 m at +1.0 m s⁻¹ instead, and by t = 2.0 s the same acceleration has carried it to 12 m. Writing the integrals as definite integrals, v(t) = v₀ + ∫ a dt′ and x(t) = x₀ + ∫ v dt′ taken from t₀ to t, builds each condition in at the instant it was measured, which starts to matter as soon as that instant is not t = 0.

An electric bus pulls away from rest while its controller tapers the push: a = A exp(−t/τ), with A = 1.8 m s⁻² and τ = 6.0 s. Integrating from t = 0 gives v = Aτ[1 − exp(−t/τ)], which starts at zero and climbs towards Aτ = 10.8 m s⁻¹. At t = 6.0 s, v = 10.8(1 − e⁻¹) = 6.8 m s⁻¹, and one more integration gives x = Aτ[t − τ(1 − exp(−t/τ))] = 23.8 m. Two warnings sit inside this example. The antiderivative −Aτ exp(−t/τ) is not zero at t = 0, so setting its constant equal to v₀ gives v(6.0 s) = −4.0 m s⁻¹, a bus rolling backwards; the definite integral avoids that automatically. And the constant-acceleration answer built on the initial 1.8 m s⁻² gives 10.8 m s⁻¹ and 32.4 m, 58% and 36% too high, because by t = 6.0 s the real acceleration has fallen to 0.66 m s⁻².

∫ v dt counts area above the time axis as positive and area below it as negative, so it returns displacement. Distance travelled is ∫ |v| dt, and the practical way to compute it is to find where v = 0, split the interval there, and add the magnitudes of the pieces. Return to the cart with x = 2.0t³ − 9.0t² + 12t, over 0 to 3.0 s. The displacement is x(3.0) − x(0) = 9.0 m. But v changes sign at 1.0 s and 2.0 s, so the path has three legs: out from 0 to 5.0 m, back to 4.0 m, then on to 9.0 m. The distance is 5.0 + 1.0 + 5.0 = 11.0 m. A single integral of v across the whole interval counts the metre driven backwards as −1.0 m, so it falls 2.0 m short of the distance. The roots of v are not an optional refinement; they are where the arithmetic has to change.

A magnetic launch track pushes a cart hardest at the start and fades linearly: a = a₀(1 − x/L), with a₀ = 12 m s⁻² and L = 30 m. You cannot integrate a dt, because you do not know when the cart reaches each position until the motion is solved. The chain rule removes time: a = dv/dt = (dv/dx)(dx/dt) = v dv/dx, so a dx = v dv. Integrating from rest at x = 0 gives ½v² = a₀(x − x²/(2L)). At the end of the track ½v² = a₀L/2 = 180 m² s⁻², so v = √360 = 19 m s⁻¹. Treating a as a constant 12 m s⁻² would give √720 = 27 m s⁻¹, too fast by a factor of √2. This method returns speed as a function of position, which is often all a problem asks for. If you also need the time, you must still evaluate t = ∫ dx/v(x), which is usually the harder integral.

A canoe coasting after the paddling stops has an acceleration a = −bv, with b = 0.25 s⁻¹ for this hull. Because a depends on v rather than t, write dv/dt = −bv and separate: dv/v = −b dt. Integrating from v₀ at t = 0 gives ln(v/v₀) = −bt, so v = v₀ exp(−bt). From v₀ = 2.0 m s⁻¹, the canoe is down to 0.74 m s⁻¹ after 4.0 s. Integrating v gives x = (v₀/b)[1 − exp(−bt)], which is 5.1 m at 4.0 s and never exceeds v₀/b = 8.0 m: in this model the canoe never quite stops, yet its drift is finite. Separation works because a depends on v alone. A damped spring, with a = −ω²x − bv, depends on x and v together, so neither dv/a nor a dx can be integrated by itself. Its x(t) has to come from solving d²x/dt² = −ω²x − bv as a differential equation, exactly or by stepping it numerically: over a small Δt, update v by aΔt, then x by vΔt, and repeat.