Two cars approach a crossing on two roads at right angles, each at a steady speed. The distance between them shrinks, reaches a least value, and then grows again. The least distance does not come when either car is at the crossing. It comes at some moment in between, when the two cars are on different roads.
One body at rest, the other on a straight line
Let two bodies A and B move with constant velocities. At \(t = 0\) the position of A relative to B is \(\vec r_0\), the vector from B to A. The velocity of A relative to B is \(\vec v_{AB} = \vec v_A - \vec v_B\), also constant. In the frame of B, the relative position at time \(t\) is the start plus the relative velocity times the time.
\[ \vec r_{AB}(t) = \vec r_0 + \vec v_{AB}\, t \]
This is the equation of a straight line traced at a steady speed. So, seen from B, the body A moves along a fixed straight line, as it did in Fig 1.2.
Watch the graph in Fig 5.1. The separation falls from \(224\,\text{m}\), reaches \(100\,\text{m}\) at \(8\,\text{s}\), and rises again. At that moment car B is \(60\,\text{m}\) past the crossing and car A is still \(80\,\text{m}\) short of it. The closest approach happens with neither car at the crossing.
The least distance is a perpendicular
Now sit on car B. Car B is at rest, and car A moves along a straight line with velocity \(\vec v_{AB}\). The distance from a fixed point to the points of a straight line is least at the foot of the perpendicular. So the minimum distance between the two bodies is the perpendicular distance from B to the relative path of A.
Let \(\varphi\) be the angle between \(\vec v_{AB}\) and the line from A to B at the start. The start, the foot F and the point B make a right triangle with hypotenuse \(r_0\). The side opposite \(\varphi\) is the least distance, and the side next to \(\varphi\) is the distance A travels to reach F.
\[ d_{\min} = r_0\sin\varphi, \qquad AF = r_0\cos\varphi \]
The time of closest approach
In the frame of B, the body A moves at the steady speed \(v_{AB}\) along its line. It reaches F after travelling \(r_0\cos\varphi\). So the time of closest approach is that distance divided by the relative speed.
Car A is \(200\,\text{m}\) west of a crossing, moving east at \(15\,\text{m/s}\). Car B is \(100\,\text{m}\) south of it, moving north at \(20\,\text{m/s}\). Find the least distance between the cars and the time at which it occurs.
Origin at the crossing, \(\hat i\) east, \(\hat j\) north. \(\vec r_A = -200\,\hat i\), \(\vec r_B = -100\,\hat j\), \(\vec v_A = 15\,\hat i\), \(\vec v_B = 20\,\hat j\).
\(\vec r_0 = \vec r_A - \vec r_B = -200\,\hat i + 100\,\hat j\), so \(r_0 = 224\,\text{m}\). \(\vec v_{AB} = 15\,\hat i - 20\,\hat j\), so \(v_{AB} = 25\,\text{m/s}\). The line from A to B is \(-\vec r_0 = 200\,\hat i - 100\,\hat j\).
\(\cos\varphi = \dfrac{(200)(15) + (-100)(-20)}{224 \times 25} = \dfrac{5000}{5590} = 0.894\), so \(\sin\varphi = 0.447\). \(d_{\min} = 224 \times 0.447 = 100\,\text{m}\). \(t_{\min} = 224 \times 0.894 / 25 = 8\,\text{s}\).
The cars come within \(100\,\text{m}\) of each other at \(t = 8\,\text{s}\). Check on the ground: A is at \(-80\,\hat i\) and B at \(60\,\hat j\), and \(\sqrt{80^2 + 60^2} = 100\,\text{m}\).
Ship A is \(10\,\text{km}\) due north of ship B. Ship A sails east at \(8\,\text{km/h}\) and ship B sails north at \(6\,\text{km/h}\). Find the least distance between the ships and the time after which it occurs.
Relative velocity: \(\vec v_{AB} = 8\,\hat i - 6\,\hat j\), so \(v_{AB} = 10\,\text{km/h}\). A starts at \(\vec r_0 = 10\,\hat j\) from B.
Angle: the line from A to B is \(-10\,\hat j\). \(\cos\varphi = \dfrac{(0)(8) + (-10)(-6)}{10 \times 10} = 0.6\), so \(\sin\varphi = 0.8\).
Least distance: \(d_{\min} = 10 \times 0.8 = 8\,\text{km}\).
Time: \(t_{\min} = 10 \times 0.6 / 10 = 0.6\,\text{h} = 36\,\text{min}\). Check: A is at \((4.8, 10)\) and B at \((0, 3.6)\), and \(\sqrt{4.8^2 + 6.4^2} = 8\,\text{km}\).
Particle A starts at the origin and moves north at \(3\,\text{m/s}\). At the same moment particle B is \(30\,\text{m}\) north of the origin and moves east at \(4\,\text{m/s}\). Find the least distance between them, in metres.
Calculus gives the same answer with more algebra. The square of the separation is \(s^2(t) = |\vec r_0 + \vec v_{AB} t|^2 = r_0^2 + 2(\vec r_0\cdot\vec v_{AB})t + v_{AB}^2 t^2\). Setting its derivative to zero gives \(t_{\min} = -\vec r_0\cdot\vec v_{AB} / v_{AB}^2\). For the two cars, \(\vec r_0\cdot\vec v_{AB} = -5000\), so \(t_{\min} = 5000/625 = 8\,\text{s}\). The perpendicular in the frame of B is this minimum, seen as a picture.
Why is the least distance between two bodies moving at constant velocities the perpendicular from one body to a line?
In the frame of B, the body B is at rest and A moves with the constant velocity \(v_{AB}\). So A moves along a fixed straight line. The distance between A and B is then the distance from a fixed point to a moving point on a straight line. Of all the points on a line, the one nearest to a given point is the foot of the perpendicular. So the least separation is the perpendicular distance from B to the path of A in the frame of B.
When the closest approach is a collision
The least distance \(r_0\sin\varphi\) is zero only when \(\varphi = 0\). So two bodies collide if and only if their relative velocity points straight from one towards the other. The collision then comes after \(t = r_0 / v_{AB}\).
Seen from A, a body on a collision course stays in the same direction and only grows larger. Its bearing does not change. A body that will pass safely drifts across the view. Sailors and pilots use this rule: a boat that holds a constant bearing while it comes closer is a danger.
Two bodies move with constant velocities. Which condition guarantees that they collide?
On a straight rural road, two cars can approach an open crossing on a collision course. For each driver, the other car keeps the same bearing, so it stays at one spot in the windscreen. If that spot is behind the pillar beside the windscreen, the other car stays hidden until the last moment. A driver who moves the head a little, or changes speed early, breaks the constant bearing and sees the danger. Ships at sea follow the same rule: a constant compass bearing and a shrinking range mean a collision course.
- Assuming the closest approach comes when one body reaches the crossing.Work in the frame of B. The least distance is the perpendicular from B to A's relative path, and it may come before, after, or between the two arrivals.
- Measuring \(\varphi\) from \(\vec r_0\) instead of from the line from A to B.\(\vec r_0\) points from B to A. The angle that gives \(d_{\min} = r_0\sin\varphi\) is between \(\vec v_{AB}\) and \(-\vec r_0\), the direction from A towards B.
- Using ground velocities for the relative path.In the frame of B, A moves with \(\vec v_A - \vec v_B\), never with \(\vec v_A\) alone.