A Boat Aimed Straight Across a River Never Lands Straight Opposite

On a straight road, a single number with a sign tells you position and velocity. On a curved flyover ramp or across a river, direction keeps changing, so every quantity becomes a vector with an x-part and a y-part.

The good news: each part obeys the same straight-line equations, and the two parts do not interfere with each other.

This part covers the kinematic vectors, the vector form of the kinematic equations, and splitting motion into independent components.

**Position is $\vec{r}=x\hat{i}+y\hat{j}$, displacement is the change in $\vec{r}$, velocity is $\vec{v}=\frac{d\vec{r}}{dt}=v_x\hat{i}+v_y\hat{j}$, and acceleration is $\vec{a}=\frac{d\vec{v}}{dt}=a_x\hat{i}+a_y\hat{j}$.

The velocity at any instant points along the tangent to the path**, at angle ${\tan }^{-1}\frac{v_y}{v_x}$ to the x-axis. Average velocity is $\frac{\Delta \vec{r}}{\Delta t}$.

Worked example. A particle has $\vec{r}=3t\hat{i}+2t^2\hat{j}$ m.

Velocity and acceleration.

$$\vec{v}=3\hat{i}+4t\hat{j}\text{ m/s}\vec{a}=4\hat{j}{\text{ m/s}}^2$$

**At $t=2$ s.**

$$\vec{v}=3\hat{i}+8\hat{j}|\vec{v}|=\sqrt{9+64}=\sqrt{73}\approx 8.54\text{ m/s}\theta ={\tan }^{-1}\frac{8}{3}\approx 69.4°$$

**Displacement and average velocity over the first $2$ s.**

$$\Delta \vec{r}=6\hat{i}+8\hat{j}\text{ m}(10\text{ m}){\vec{v}}_{avg}=3\hat{i}+4\hat{j}\text{ m/s}(5\text{ m/s})$$

An everyday example. A car on a curved flyover ramp always has its velocity pointing along the road at that spot, even as the direction keeps turning.

The substance. Acceleration need not point along velocity — here $\vec{a}$ is always along $y$ while $\vec{v}$ turns.

**For constant $\vec{a}$, $\vec{v}=\vec{u}+\vec{a}t$ and $\vec{r}={\vec{r}}_0+\vec{u}t+\frac{1}{2}\vec{a}{t}^2$, and each vector equation is really two scalar equations, one for $x$ and one for $y$.**

$$v_x=u_x+a_{x}tx=x_0+u_{x}t+\frac{1}{2}{a}_x{t}^2$$

$$v_y=u_y+a_{y}ty=y_0+u_{y}t+\frac{1}{2}{a}_y{t}^2$$

Worked example. A particle starts from the origin with $\vec{u}=5\hat{i}$ m/s and constant $\vec{a}=3\hat{i}+2\hat{j}$ m/s${}^2$.

**At $t=4$ s.**

$$\vec{v}=(5+12)\hat{i}+8\hat{j}=17\hat{i}+8\hat{j}|\vec{v}|=\sqrt{353}\approx 18.8\text{ m/s}$$

$$\vec{r}=20\hat{i}+\frac{1}{2}(3\hat{i}+2\hat{j})(16)=44\hat{i}+16\hat{j}\text{ m}r=\sqrt{2192}\approx 46.8\text{ m}$$

**When is $x=84$ m?**

$$5t+1.5t^2=84\Rightarrow t=\frac{-5+\sqrt{25+504}}{3}=\frac{-5+23}{3}=6\text{ s}$$

At that moment, $y=\frac{1}{2}(2)(36)=36$ m.

An everyday example. A drone flying forward while a steady crosswind pushes it sideways follows these equations, one along its heading and one across it.

The substance. **$v^2=u^2+2as$ becomes $\vec{v}\cdot \vec{v}=\vec{u}\cdot \vec{u}+2\vec{a}\cdot \Delta \vec{r}$** — the vector version needs the dot product.

**Motion along $x$ and motion along $y$ happen independently, linked only by the shared time $t$, so you solve each direction as a separate straight-line problem and combine the results as vectors.

Worked example 1 — crossing a river.** A boat moves at $4$ m/s relative to the water, pointed straight across a $200$ m wide river flowing at $3$ m/s.

- Across ($y$): $t=\frac{200}{4}=50$ s
- Downstream ($x$): drift $=3\times 50=150$ m
- Resultant speed $=\sqrt{4^2+3^2}=5$ m/s

The current changes where the boat lands but not how long the crossing takes.

Worked example 2 — reaching the opposite point. To land straight across, the boat must head upstream at angle $\theta$ with $4\sin \theta =3$, so $\sin \theta =0.75$ and $\theta \approx 48.6°$ from the straight-across line. Its speed across is $4\cos \theta =\sqrt{16-9}\approx 2.65$ m/s, so the crossing takes $\frac{200}{2.65}\approx 75.6$ s.

Worked example 3 — two coins. One coin is dropped from a $1.25$ m table and another is flicked horizontally off it at the same moment. With $g=10$ m/s${}^2$, both fall $1.25=5t^2$, so **both land after $0.5$ s.

An everyday example. A ferry crossing a river has to point upstream to reach the jetty directly opposite.

The substance. Horizontal speed does not change the time of fall** — only the vertical motion sets it.