Car A is \(200\,\text{m}\) west of a crossing on a highway near Nagpur, driving east at \(15\,\text{m/s}\). Car B is \(100\,\text{m}\) south of the same crossing, driving north at \(20\,\text{m/s}\). Neither driver slows down. The cars do not collide, because they reach the crossing at different times. But how close do they come, and when?
The position vector
Fix an origin O and a set of axes. The position vector of a point P is the arrow from O to P, \(\vec r = x\hat i + y\hat j + z\hat k\). Its components are the coordinates of P. Its magnitude \(r = \sqrt{x^2 + y^2 + z^2}\) is the distance of P from O. A different origin gives a different position vector for the same point. So every position vector belongs to a chosen origin.
A body that moves from \(P_1\) to \(P_2\) has the displacement \(\Delta\vec r = \vec r_2 - \vec r_1\). By the triangle law, \(\vec r_1 + \Delta\vec r = \vec r_2\), so the displacement is the arrow from \(P_1\) to \(P_2\). The origin cancels in the subtraction. Two observers with different origins disagree about the position vectors, but they agree about the displacement.
Where one body is, seen from another
Let two bodies A and B have position vectors \(\vec r_A\) and \(\vec r_B\) from one origin. The relative position of A with respect to B is \(\vec r_{AB} = \vec r_A - \vec r_B\). It is the arrow from B to A. An observer riding on B sees A along exactly that arrow. The origin cancels again, so every observer agrees on it.
The relative position changes as the bodies move. Its rate of change is the velocity of A relative to B. Differentiating each term gives \(\vec v_{AB} = \vec v_A - \vec v_B\), and once more gives the relative acceleration \(\vec a_{AB} = \vec a_A - \vec a_B\). Every relative velocity of the chapters on relative motion is this one subtraction.
Car A is at \(\vec r_A = -80\hat i\,\text{m}\) and car B at \(\vec r_B = 60\hat j\,\text{m}\) from a crossing. Which statement about \(\vec r_{AB} = \vec r_A - \vec r_B\) is correct?
Equations of motion as vectors
The equations of motion hold for each component while the acceleration is constant. Written with vectors, they hold for all components at once: \(\vec v = \vec u + \vec a\,t\) and \(\vec r = \vec r_0 + \vec u\,t + \tfrac12\vec a\,t^2\). Take \(\vec r_0 = 2\hat i + \hat j\) m, \(\vec u = 4\hat i + 6\hat j\) m/s and \(\vec a = \hat i - 2\hat j\) m/s². At \(t = 2\,\text{s}\), \(\vec v = 6\hat i + 2\hat j\) m/s and \(\vec r = 12\hat i + 9\hat j\) m, which is \(15\,\text{m}\) from the origin.
The closest approach of two bodies
Two bodies that move at constant velocities have a constant relative velocity \(\vec v_{AB}\). So their relative position is \(\vec r_{AB}(t) = \vec r_0 + \vec v_{AB}\,t\). The tip of that arrow moves along a straight line, as the figure showed. The shortest arrow from a point to a straight line meets the line at right angles. So the bodies are closest when \(\vec r_{AB}\) is perpendicular to \(\vec v_{AB}\), and the dot product turns that into a formula.
\[ \vec r(t) = \vec r_0 + \vec v\,t \]
Why. write \(\vec r\) for \(\vec r_{AB}\) and \(\vec v\) for \(\vec v_{AB}\). The relative velocity is constant, so the relative position changes by \(\vec v\,t\).
\[ \vec r(t_{min})\cdot\vec v = 0 \]
Why. at the closest approach the relative position is perpendicular to the line along which its tip moves, the line of \(\vec v\).
\[ \vec r_0\cdot\vec v + v^2\,t_{min} = 0 \]
Why. substitute step 1 and use \(\vec v\cdot\vec v = v^2\).
\[ t_{min} = -\dfrac{\vec r_0\cdot\vec v}{v^2} \]
Why. solve for the time. It is positive only when \(\vec r_0\cdot\vec v\) is negative, that is, while the bodies are still approaching.
\[ d_{min} = |\vec r_0 + \vec v\,t_{min}| \]
Read it. the least distance is the size of the relative position at that time.
This is the worked example "The two cars of Fig 5.1" of Relative Velocity in Two Dimensions, lesson 5, done with the tool. 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. \(\vec r_A = -200\hat i\), \(\vec r_B = -100\hat j\), \(\vec v_A = 15\hat i\), \(\vec v_B = 20\hat j\).
\(\vec r_0 = \vec r_A - \vec r_B = -200\hat i + 100\hat j\) m and \(\vec v = \vec v_A - \vec v_B = 15\hat i - 20\hat j\) m/s, so \(v^2 = 625\,\text{m}^2/\text{s}^2\).
\(\vec r_0\cdot\vec v = (-200)(15) + (100)(-20) = -5000\), so \(t_{min} = 5000/625 = 8\,\text{s}\). Then \(\vec r = (-200 + 120)\hat i + (100 - 160)\hat j = -80\hat i - 60\hat j\) m.
\(d_{min} = \sqrt{80^2 + 60^2} = 100\,\text{m}\) at \(t = 8\,\text{s}\), the same answer as the physics chapter.
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 \(6\,\text{km/h}\). After how many minutes are they closest?
Two bodies move at constant velocities. At the instant of their closest approach, how are their relative position \(\vec r_{AB}\) and relative velocity \(\vec v_{AB}\) related?
A weighted sum of position vectors
Position vectors can be averaged. Give each point a weight, multiply each position vector by its weight, add them, and divide by the total weight. The result is a point between the given points, drawn towards the heavier weights. With masses as the weights, this point is the centre of mass. In components it is three ordinary weighted averages, one for each coordinate.
The origin of the figure is below the plank, yet C always lands on the plank. Place the origin at the lighter child instead, and \(x_C = (30 \times 0 + 45 \times 3)/75 = 1.8\,\text{m}\), the same point. A change of origin adds one fixed vector to every position vector, and the weighted average adds the same vector to \(\vec r_C\). The point itself does not move.
Point masses of \(1\,\text{kg}\) at \((2, 1)\), \(2\,\text{kg}\) at \((6, 1)\), \(3\,\text{kg}\) at \((6, 4)\) and \(2\,\text{kg}\) at \((2, 4)\), in metres, lie at the corners of a rectangle. Find the \(y\) coordinate of their centre of mass, in m.
Two cars move at constant velocities. Why does the tip of their relative position vector, drawn from a fixed point, move along a straight line?
The relative position is \(\vec r_{AB}(t) = \vec r_0 + \vec v_{AB}\,t\). The relative velocity \(\vec v_{AB} = \vec v_A - \vec v_B\) is constant, because both velocities are constant. So in every second the tip moves by the same vector \(\vec v_{AB}\), always in the same direction. A point that moves by equal steps in one fixed direction moves along a straight line.
Relative vectors need care in three places. Both positions must be measured from one origin, at the same instant. The simple subtraction \(\vec v_A - \vec v_B\) holds for everyday speeds, far below the speed of light. The closest-approach formula assumes constant velocities. If it gives a negative \(t_{min}\), the bodies are already moving apart, and they are closest now, at \(t = 0\).
The two cars of the opening are answered. With one subtraction the problem became one arrow, \(\vec r_0 + \vec v\,t\), and with one dot product the closest moment, \(8\,\text{s}\). At that moment car A is \(80\,\text{m}\) west of the crossing and car B is \(60\,\text{m}\) north of it. They are \(100\,\text{m}\) apart, and after that they separate.
Motion in Two Dimensions writes the equations of motion as vectors. The two chapters on relative velocity subtract positions, velocities and accelerations. They do so for trains that overtake, boats on a river, rain on a moving person and two bodies at their closest approach. Centre of Mass finds the centre of mass as a weighted sum of position vectors. Rolling Motion finds the velocity of any point of a wheel as the velocity of its centre plus a velocity relative to the centre. Rigid Bodies uses relative velocities to show that two points of a rigid body have equal velocity components along the line joining them.
- Taking \(\vec r_B - \vec r_A\) for the position of A relative to B.\(\vec r_{AB} = \vec r_A - \vec r_B\). The body looked at comes first, the observer second.
- Measuring the two position vectors from different origins.Choose one origin and one set of axes for both bodies before subtracting.
- Accepting a negative time of closest approach as a moment in the future.A negative \(t_{min}\) means the bodies are already separating; from now on they are closest at \(t = 0\).