Two boats start from the same point with constant velocities…
Two boats start from the same point with constant velocities $\vec v_1$ and $\vec v_2$ whose directions form an angle $\alpha$. (a) What is the relative velocity of the boats? (b) What is the distance between them a time $t$ after the start?
Explanation of Method
Relative velocity of two boats at an angle.
Step 1: (a) Relative velocity
The velocity of boat 2 as seen from boat 1 is the difference of the two velocities: $\vec v_{21} = \vec v_2 - \vec v_1$.
The vectors $\vec v_1$, $\vec v_2$ and $\vec v_2 - \vec v_1$ form a triangle with angle $\alpha$ between $\vec v_1$ and $\vec v_2$. By the law of cosines: $v_{21} = \sqrt{v_1^2 + v_2^2 - 2 v_1 v_2 \cos\alpha}$.
Step 2: (b) Distance after time $t$
After time $t$ the boats are at $\vec r_1 = \vec v_1 t$ and $\vec r_2 = \vec v_2 t$, so the vector from boat 1 to boat 2 is $\vec r_2 - \vec r_1 = (\vec v_2 - \vec v_1)\,t$. Its length is
\[ d = \sqrt{(v_1 t)^2 + (v_2 t)^2 - 2 (v_1 t)(v_2 t)\cos\alpha} \]
\[ \phantom{d} = t\sqrt{v_1^2 + v_2^2 - 2 v_1 v_2 \cos\alpha}. \]
\[d = v_{21}\, t\]
Special cases
| Angle | Relative speed | Meaning |
|---|---|---|
| $\alpha = 0^\circ$ | $|v_1 - v_2|$ | same direction |
| $\alpha = 90^\circ$ | $\sqrt{v_1^2 + v_2^2}$ | perpendicular (021: 30 and 40 give 50 km/h) |
| $\alpha = 180^\circ$ | $v_1 + v_2$ | opposite directions |
Result
\[\boxed{\,v_{21} = \sqrt{v_1^2 + v_2^2 - 2 v_1 v_2 \cos\alpha}\,}\]
\[\boxed{\,d = v_{21}\, t\,}\]
Remark
Note. Because both boats start from the same point, their distance grows in proportion to time: the triangle in the figure keeps its shape and just gets bigger.