Time–Speed–Distance (TSD) – Advanced Study Notes
(MIT‑Level, First Principles & Edge‑Case Analysis)
1. Foundations: Kinematics in One Dimension
| Symbol | Meaning | Units |
|---|---|---|
| (x(t)) | Position of a particle at time (t) | meters (m) |
| (v(t) = \dot{x}(t)) | Instantaneous velocity | m s(^{-1}) |
| (a(t) = \ddot{x}(t)) | Instantaneous acceleration | m s(^{-2}) |
| (t) | Time | seconds (s) |
| (d) | Displacement | m |
| (s) | Distance travelled (arc length) | m |
| (u) | Initial speed | m s(^{-1}) |
| (v) | Final speed | m s(^{-1}) |
The basic time–speed–distance relation is a special case of the kinematic equations:
- ( v = u + a t )
- ( d = ut + \frac12 a t^2 )
- ( v^2 = u^2 + 2 a d )
These follow directly from integrating the definitions of velocity and acceleration.
1.1 Derivation from Calculus
Starting from (v = \dot{x}):
[ x(t) = \int_0^t v(\tau), d\tau + x(0) ]
If (v(t) = u + a t) (constant acceleration), then
[ x(t) = x(0) + \int_0^t (u + a \tau), d\tau = x(0) + ut + \frac12 a t^2 . ]
Subtracting (x(0)) gives the displacement formula above.
2. Generalized Motion: Variable Acceleration
When (a(t)) is not constant, the relation between (d) and (t) requires solving an integral:
[ d = \int_0^t v(\tau), d\tau = \int_0^t \Bigl(u + \int_0^\tau a(s), ds \Bigr) d\tau . ]
If (a(t)) is a known function, this integral can be evaluated analytically or numerically.
Example: If (a(t) = k,t) (linearly increasing acceleration), then
[ v(t) = u + \frac{k}{2} t^2, \qquad d(t) = ut + \frac{k}{6} t^3 . ]
3. Vector Extension – Two and Three Dimensions
Let (\mathbf{r}(t)) be the position vector. Then
[ \mathbf{v}(t) = \dot{\mathbf{r}}(t), \qquad \mathbf{a}(t) = \dot{\mathbf{v}}(t). ]
With constant acceleration (\mathbf{a}),
[ \mathbf{r}(t) = \mathbf{r}_0 + \mathbf{u},t + \frac12 \mathbf{a},t^2 . ]
The scalar distance travelled is
[ s(t) = \int_0^t |\mathbf{v}(\tau)|, d\tau . ]
In uniform circular motion ((|\mathbf{v}| = v) constant, (\mathbf{a}) centripetal), (s = v t) while displacement (d = r \theta = r v t / r = v t) as well because (\theta = v t / r). However, if (v) varies with time, (d \neq s).
4. Edge Cases & Singularities
| Situation | What Happens | How to Handle |
|---|---|---|
| Zero acceleration ((a=0)) | Motion is uniform; (d = u t). | Use the limit (a\to 0) in the general equations; all terms involving (a) vanish. |
| Negative acceleration (deceleration) | (v) decreases; may reach zero or become negative (direction change). | Solve (v = u + a t = 0) for (t = -u/a); after this time reverse sign of velocity. |
| Infinite acceleration (impulsive force) | Instantaneous change in velocity; (t=0) impulse. | Use Dirac delta in acceleration: (a(t)=A\delta(t)). Then (v(t)=u+A). |
| Zero speed ((u=0, v=0)) | Particle remains at rest. | All distances are zero for all (t). |
| Negative time | Physically meaningless; interpret as backward motion in time. | Typically ignore; restrict (t\ge 0). |
| Non‑linear distance–time relation | Requires integral evaluation. | Use symbolic or numerical integration; check for convergence. |
| Relativistic speeds ((v\approx c)) | Classical equations break down; use Lorentz transformation. | Replace (v = u + a t) with (v(t) = \frac{u + a t}{\sqrt{1+(u+at)^2/c^2}}). |
| Periodic motion (oscillations) | Displacement and distance differ due to back‑and‑forth motion. | Compute (s(t)=\int_0^t |
5. Deeper Theory: Differential Geometry of Motion
5.1 Arc Length in Curvilinear Motion
For a trajectory (\mathbf{r}(t)), the infinitesimal arc length element is
[ ds = |\mathbf{v}(t)| dt . ]
Hence
[ s(t) = \int_0^t |\mathbf{v}(\tau)|, d\tau . ]