You sit by the window of a bus that moves north along a wide road. A scooter crosses the road ahead, moving east. To you, the scooter does not move east. It seems to slide towards the back of the bus and sideways at the same time, along a slanting line. A person standing on the pavement sees the same scooter move straight across the road.
Relative position
Let two bodies A and B move in a plane. Measured from a fixed origin O on the ground, their position vectors are \(\vec r_A\) and \(\vec r_B\). The position of A as seen from B is the vector drawn from B to A. This vector is the relative position of A with respect to B.
\[ \vec r_{AB} = \vec r_A - \vec r_B \]
The order of the subscripts matters. \(\vec r_{AB}\) points from B to A. The vector \(\vec r_{BA}\) points from A to B, so \(\vec r_{BA} = -\vec r_{AB}\).
Relative velocity
The velocity of A relative to B is the rate of change of the relative position. An observer who moves with B measures this velocity for A. It is the relative velocity of A with respect to B. The observer's own moving platform is a frame of reference that moves with B.
\[ \vec r_{AB} = \vec r_A - \vec r_B \]
Why. the position of A seen from B is the vector from B to A, true at every instant.
\[ \dfrac{d\vec r_{AB}}{dt} = \dfrac{d\vec r_A}{dt} - \dfrac{d\vec r_B}{dt} \]
Why. differentiate both sides with respect to time. The derivative of a difference is the difference of the derivatives.
\[ \vec v_{AB} = \vec v_A - \vec v_B \]
Read it. each derivative of a position vector is a velocity. Both velocities on the right are measured from the ground.
The velocity of a body A relative to a body B is the velocity of A as measured by an observer who moves with B. It equals the velocity of A minus the velocity of B, both measured in the same frame, usually the ground: \(\vec v_{AB} = \vec v_A - \vec v_B\).
The two relative velocities \(\vec v_{AB}\) and \(\vec v_{BA}\) have the same magnitude and opposite directions. If you see a friend move north-east away from you, your friend sees you move south-west away from them, at the same speed.
Subtraction is addition of the opposite vector
A difference of two vectors is a sum in disguise. \(\vec v_A - \vec v_B\) is \(\vec v_A + (-\vec v_B)\), where \(-\vec v_B\) has the size of \(\vec v_B\) and points the other way. So draw \(\vec v_A\), then draw \(-\vec v_B\) with its tail at the tip of \(\vec v_A\). This is the tip to tail rule of vector addition. The arrow from the tail of \(\vec v_A\) to the tip of \(-\vec v_B\) is the resultant, \(\vec v_{AB}\).
In Fig 1.1 the two velocities are perpendicular, so the triangle has a right angle. The size of \(\vec v_{AB}\) comes from Pythagoras, \(\sqrt{3^2 + 4^2} = 5\,\text{m/s}\). Its direction is \(\tan^{-1}(4/3) = 53°\) below the east line. In components the same work is one line: \(\vec v_{AB} = (3\,\hat i) - (4\,\hat j) = 3\,\hat i - 4\,\hat j\).
When the two velocities make any angle \(\theta\) with each other, the triangle is no longer right-angled. The size of the difference then follows from a dot product.
\[ |\vec v_{AB}|^2 = (\vec v_A - \vec v_B)\cdot(\vec v_A - \vec v_B) \]
Why. the square of the size of any vector is its dot product with itself.
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Your turn. open the brackets. Which line is it?
\[ |\vec v_{AB}| = \sqrt{v_A^2 + v_B^2 - 2 v_A v_B \cos\theta} \]
Read it. \(\vec v_A\cdot\vec v_B = v_A v_B\cos\theta\), where \(\theta\) is the angle between the two velocities.
Three special cases are worth knowing. If the bodies move in the same direction, \(\theta = 0\) and \(v_{AB} = |v_A - v_B|\). If they move in opposite directions, \(\theta = 180°\) and \(v_{AB} = v_A + v_B\). If they move at right angles, \(\theta = 90°\) and \(v_{AB} = \sqrt{v_A^2 + v_B^2}\).
The relative velocity is a difference, not a sum. Students who add the two arrows tip to tail get \(\vec v_A + \vec v_B\), which has no meaning here. Always reverse the velocity of the observer before you add.
A bus moves north at \(12\,\text{m/s}\). A scooter moves east at \(5\,\text{m/s}\). Find the velocity of the scooter relative to the bus.
Take \(\hat i\) east and \(\hat j\) north. \(\vec v_S = 5\,\hat i\) and \(\vec v_B = 12\,\hat j\), both relative to the ground.
\(\vec v_{SB} = \vec v_S - \vec v_B = 5\,\hat i - 12\,\hat j\).
\(v_{SB} = \sqrt{5^2 + 12^2} = 13\,\text{m/s}\). The angle with the south line is \(\tan^{-1}(5/12) = 23°\), towards the east.
From the bus, the scooter moves at \(13\,\text{m/s}\), \(23°\) east of south. It slides towards the back of the bus while it crosses. This is the slanting motion of the hook.
Two cars move at \(10\,\text{m/s}\) each on straight roads that make an angle of \(60°\) with each other. What is the speed of one car relative to the other?
A moves east at \(8\,\text{m/s}\) and B moves south at \(6\,\text{m/s}\). Find the speed of A relative to B, in m/s.
Stand on B
A relative velocity is most useful when you imagine yourself on the second body. From there, the second body does not move at all. Only the first body moves, with the velocity \(\vec v_{AB}\). If both ground velocities are constant, \(\vec v_{AB}\) is constant too. So, in the frame of B, A moves in a straight line at a steady speed.
Ball A moves east and ball B moves north, both at steady speeds. What path does A follow, as seen from B?
Play Fig 1.2 with each chip. From the ground, A runs east and B runs north. From B, the ball B sits still, and A runs along a slanting straight line. Every second, A moves \(1.5\,\text{m}\) east and \(2\,\text{m}\) south, which is \(2.5\,\text{m/s}\) along \(\vec v_A - \vec v_B\). The rest of this chapter uses this one picture again and again. It serves for a boat, for rain, for an aeroplane, and for two bodies that come close.
Relative motion is like sitting in a Metro coach that runs smoothly. You feel at rest, and the platform seems to move backward. Nothing about the platform has changed. You have only chosen the coach as your frame, and in that frame every velocity loses the velocity of the coach.
The velocity of A relative to B is \(3\,\hat i + 4\,\hat j\) m/s. What is the velocity of B relative to A?
Two aeroplanes fly at constant velocities in different directions. Why must each one appear to the other to move in a straight line?
The velocity of one aeroplane relative to the other is the difference of their ground velocities. Both ground velocities are constant, so their difference is constant in size and direction. A body that moves with a constant velocity moves in a straight line at a steady speed. So each aeroplane sees the other move along a straight line, though not along its ground track.
Galileo asked his readers to shut themselves in the cabin of a large ship with flies, fish in a bowl and a dripping bottle. While the ship is at rest, the flies fly equally in all directions and the drops fall straight down. Now let the ship move at a steady speed. Galileo argued that nothing inside changes: the drops still fall straight down into the vessel below. Every motion in the cabin is measured relative to the ship, and the ship's own velocity is subtracted from all of them alike. This is the rule \(\vec v_{AB} = \vec v_A - \vec v_B\), and it is why you cannot feel the steady motion of a train with your eyes closed.
- Adding the two velocities instead of subtracting them.Relative velocity is \(\vec v_A - \vec v_B\). Reverse the observer's velocity, then add tip to tail.
- Using \(\vec v_B - \vec v_A\) when the question asks for the velocity of A relative to B.The first subscript is the body you watch, the second is the observer. \(\vec v_{AB}\) starts with \(\vec v_A\).
- Using the cosine rule with the angle between the arrows after one has been reversed.In \(v_{AB} = \sqrt{v_A^2 + v_B^2 - 2 v_A v_B\cos\theta}\), \(\theta\) is the angle between the original velocities drawn from one point.