When position is expressed as a function of time $\vec{r}(t)$, differentiation yields velocity and acceleration:

• Velocity: $\vec{v}(t) = \frac{d\vec{r}}{dt}$
• Acceleration: $\vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d^2\vec{r}}{dt^2}$

Example: The position vector of a particle is $\vec{r}(t) = 3t^2\underline{i} + (t^2 - 2t)\underline{j} + t^3\underline{k}$. Find its velocity and acceleration at $t = 2$ seconds.

Solution:

Velocity: $\vec{v} = \frac{d\vec{r}}{dt} = 6t\underline{i} + (2t - 2)\underline{j} + 3t^2\underline{k}$ At $t = 2$: $\vec{v} = 12\underline{i} + 2\underline{j} + 12\underline{k}$

Acceleration: $\vec{a} = \frac{d\vec{v}}{dt} = 6\underline{i} + 2\underline{j} + 6t\underline{k}$ At $t = 2$: $\vec{a} = 6\underline{i} + 2\underline{j} + 12\underline{k}$

Given two vectors $\vec{a}$ and $\vec{b}$, the projection of $\vec{a}$ onto $\vec{b}$ tells us how much of $\vec{a}$ lies in the direction of $\vec{b}$.

Scalar Projection

$\text{Proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$

This gives a scalar (magnitude only).

Vector Projection

$\text{Proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\vec{b}$

This gives a vector in the direction of $\vec{b}$.

Example: Find the vector projection of $\vec{a} = (1, 0, -2)$ onto $\vec{b} = (2, 1, -1)$.

Solution: $\vec{a} \cdot \vec{b} = (1)(2) + (0)(1) + (-2)(-1) = 4$ $|\vec{b}|^2 = 2^2 + 1^2 + (-1)^2 = 6$ $\text{Proj}_{\vec{b}}\vec{a} = \frac{4}{6}(2, 1, -1) = \left(\frac{4}{3}, \frac{2}{3}, -\frac{2}{3}\right)$

When a force $\vec{F}$ acts on a particle causing displacement $\vec{d}$, the work done is the dot product:

$W = \vec{F} \cdot \vec{d} = |\vec{F}||\vec{d}|\cos\theta$

where $\theta$ is the angle between the force and displacement vectors.

Example: Find the work done by $\vec{F} = 2\underline{i} + \underline{j} + 3\underline{k}$ when a particle moves from point $A(2, 2, 3)$ to $B(3, 4, 5)$.

Solution: First find the displacement: $\vec{d} = \vec{AB} = \underline{i} + 2\underline{j} + 2\underline{k}$

Then compute work: $W = \vec{F} \cdot \vec{d} = (2)(1) + (1)(2) + (3)(2) = 10 \text{ units}$

Many physics problems combine these concepts. Consider a particle under multiple forces.

Example: Forces $\vec{F}_1 = 2\underline{i} - \underline{j} + 3\underline{k}$, $\vec{F}_2 = -\underline{i} + 2\underline{j} - \underline{k}$, and $\vec{F}_3 = 2\underline{i} - 5\underline{k}$ act on a particle displaced from $\vec{r}_1 = \underline{i} - 3\underline{j} - 2\underline{k}$ to $\vec{r}_2 = 5\underline{i} + \underline{j} - 3\underline{k}$. Find total work done.

Solution:

Displacement: $\vec{d} = \vec{r}_2 - \vec{r}_1 = (5-1)\underline{i} + (1+3)\underline{j} + (-3+2)\underline{k} = 4\underline{i} + 4\underline{j} - \underline{k}$

Resultant force: $\vec{F} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = (2-1+2)\underline{i} + (-1+2+0)\underline{j} + (3-1-5)\underline{k} = 3\underline{i} + \underline{j} - 3\underline{k}$

Work done: $W = \vec{F} \cdot \vec{d} = (3)(4) + (1)(4) + (-3)(-1) = 19 \text{ units}$

In Tanzania, engineers and surveyors use vector projections when designing roads and bridges. For example, when constructing the new runway at Julius Nyerere International Airport, engineers calculate the projection of forces onto different directions to determine how much load each support structure must bear. Similarly, a Tanzanian taxi driver navigating Dar es Salaam's traffic uses vector reasoning implicitly — calculating the resultant direction when combining the northward flow of traffic with an eastward detour to avoid congestion, with work done representing the energy expended in the journey.