Kinematics describes motion without reference to forces. Vectors allow motion in 2D and 3D to be handled compactly.
| Quantity | Symbol | Vector or Scalar? | Units |
|---|---|---|---|
| Position vector | $\mathbf{r}(t)$ | Vector | m |
| Displacement | $\Delta\mathbf{r} = \mathbf{r}(t_2)-\mathbf{r}(t_1)$ | Vector | m |
| Distance travelled | $\displaystyle\int_{t_1}^{t_2} | \mathbf{v} | \,dt$ |
$$\mathbf{v}(t) = \dot{\mathbf{r}}(t) = \frac{d\mathbf{r}}{dt}$$
$$\mathbf{a}(t) = \dot{\mathbf{v}}(t) = \frac{d^2\mathbf{r}}{dt^2}$$
Differentiate and integrate componentwise:
$$\mathbf{r}(t) = x(t)\mathbf{i}+y(t)\mathbf{j}+z(t)\mathbf{k} \Rightarrow \mathbf{v}(t) = \dot x\mathbf{i}+\dot y\mathbf{j}+\dot z\mathbf{k}$$
Speed (scalar): $v = |\mathbf{v}| = \sqrt{\dot x^2+\dot y^2+\dot z^2}$.
Direction of motion: unit tangent $\hat{\mathbf{v}} = \mathbf{v}/|\mathbf{v}|$.
Given $\mathbf{a}(t)$ and initial conditions $\mathbf{r}(0)$ and $\mathbf{v}(0)$:
$$\mathbf{v}(t) = \mathbf{v}(0) + \int_0^t \mathbf{a}(\tau)\,d\tau$$
$$\mathbf{r}(t) = \mathbf{r}(0) + \int_0^t \mathbf{v}(\tau)\,d\tau$$
A particle has $\mathbf{a}(t) = 2t\,\mathbf{i} - \mathbf{j}$, $\mathbf{v}(0) = \mathbf{i}+3\mathbf{j}$, $\mathbf{r}(0) = \mathbf{0}$.
$$\mathbf{v}(t) = \int (2t\,\mathbf{i}-\mathbf{j})\,dt = t^2\mathbf{i} - t\mathbf{j} + \mathbf{C}$$
IC: $\mathbf{v}(0) = \mathbf{C} = \mathbf{i}+3\mathbf{j}$. So $\mathbf{v}(t) = (t^2+1)\mathbf{i}+(3-t)\mathbf{j}$.
$$\mathbf{r}(t) = \int\mathbf{v}\,dt = \left(\frac{t^3}{3}+t\right)\mathbf{i}+\left(3t-\frac{t^2}{2}\right)\mathbf{j} + \mathbf{D}$$
IC: $\mathbf{r}(0) = \mathbf{D} = \mathbf{0}$. So $\mathbf{r}(t) = \left(\dfrac{t^3}{3}+t\right)\mathbf{i}+\left(3t-\dfrac{t^2}{2}\right)\mathbf{j}$.
Speed at $t=2$: $\mathbf{v}(2) = 5\mathbf{i}+\mathbf{j}$, so $|\mathbf{v}(2)| = \sqrt{26}$ m/s.
Velocity of $B$ relative to $A$: $\mathbf{v}_{B/A} = \mathbf{v}_B - \mathbf{v}_A$.
Example: A boat travels at $\mathbf{v}_B = 4\mathbf{i}+3\mathbf{j}$ km/h. Wind blows at $\mathbf{v}_W = -\mathbf{i}+2\mathbf{j}$ km/h. Velocity of boat relative to wind: $(4\mathbf{i}+3\mathbf{j}) - (-\mathbf{i}+2\mathbf{j}) = 5\mathbf{i}+\mathbf{j}$ km/h.
Two particles at $\mathbf{r}_1(t)$ and $\mathbf{r}_2(t)$. Distance$^2 = |\mathbf{r}_1-\mathbf{r}_2|^2$. Minimise by differentiating w.r.t. $t$ and setting to zero.
KEY TAKEAWAY: Differentiation gives velocity from position, and again gives acceleration. Integration reverses this — always include the constant vector and apply initial conditions componentwise.
EXAM TIP: Distance travelled requires integrating the speed (magnitude of velocity), not the magnitude of displacement. These are different unless motion is in a straight line without reversing.
COMMON MISTAKE: Confusing displacement ($\Delta\mathbf{r}$, a vector) with distance (a scalar integral of speed).
