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Relative Velocity in One Dimension · Lesson 2 of 5

Relative Velocity on a Line

15 min read3 ideas4 checks1 figure to playBy Asheshanand Acharya
What you will learn 0 / 3
Relative velocity on a line
Maths you need

From relative position to relative velocity

The relative position \(x_{AB}\) tells where A is, seen from B. As the two bodies move, \(x_{AB}\) changes. The rate of change of \(x_{AB}\) is the relative velocity of A with respect to B. It is the velocity that an observer riding on B measures for A.

Deriving \(v_{AB} = v_A - v_B\) step 1 of 3
1

\[ x_{AB} = x_A - x_B \]

Why. the relative position, both positions measured in the ground frame at the same instant.

The result is one subtraction. The velocity of A seen from B is the velocity of A minus the velocity of B, both measured in the same frame. Seen from A, the velocity of B is \(v_{BA} = v_B - v_A = -v_{AB}\). If A moves away from B at \(5\,\text{m/s}\), then B moves away from A at \(5\,\text{m/s}\) in the opposite direction.

RememberRelative velocity on a line
\[ v_{AB} = v_A - v_B, \qquad v_{BA} = -v_{AB} \]
Holds whenboth velocities measured in the same frame, along the same line, with one sign convention.
Symbols\(v_A, v_B\) velocities of A and B, \(v_{AB}\) velocity of A relative to B, \(v_{BA}\) velocity of B relative to A
Unitsm/s

Signs do the work

On a line a velocity is a signed number. Choose one direction as positive and give every velocity its sign before you subtract. Then the one formula covers every case. The relative speed is the magnitude \(|v_{AB}|\), the rate at which the distance between the two bodies changes.

v_AB = v_A − v_B drawn as arrowsVelocity arrows on a line, in metres per second. The first row is v_A, the second v_B. The third row adds minus v_B to the tip of v_A. The last row is the result, v_AB = v_A minus v_B, drawn from zero. Chips choose the same direction, opposite directions, and B faster than A.-15-10-5O51015202530v (m/s) vA vB −vB vAB vAB = +5 m/s
Fig 2.1The subtraction drawn as arrows. Add \(-v_B\) (grey) to the tip of \(v_A\); the purple arrow from zero is \(v_{AB}\). Same direction: \(15 - 10 = +5\,\text{m/s}\). Opposite directions: \(15 - (-10) = +25\,\text{m/s}\). B faster: \(10 - 15 = -5\,\text{m/s}\).

Fig 2.1 has three cases. When both move the same way, the relative velocity is the difference of the two speeds. When they move in opposite directions, \(v_B\) is negative, so subtracting it adds the speeds. When B is faster than A in the same direction, \(v_{AB}\) is negative: seen from B, the body A falls behind.

ExampleTwo trains on parallel tracks

Train A moves east at \(72\,\text{km/h}\). Train B moves west at \(54\,\text{km/h}\) on the next track. Find the velocity of A relative to B and of B relative to A, in m/s.

Reveal one step at a time and finish the rest yourself.
Checkpoint · numericUnanswered

A car moves at \(20\,\text{m/s}\) and a bus moves at \(12\,\text{m/s}\), both towards the east on a straight road. Find the velocity of the car relative to the bus, in m/s, with east positive.

m/s± 1 %
Seeing the motion from a moving body

Sit in one body and watch the others

The formula \(v_{AB} = v_A - v_B\) gives a rule for changing frames. To see everything from B, subtract \(v_B\) from every velocity in the ground frame. B itself gets \(v_B - v_B = 0\), so B is at rest. The ground, whose velocity is zero, gets \(-v_B\), so the ground moves backwards under B.

Predict firstLocked

Train A moves at \(+15\,\text{m/s}\) and train B at \(-10\,\text{m/s}\). Ravi stands on the platform. In the frame of train A, what are the velocities of train B and of Ravi?

Two trains and Ravi, seen from three framesTrain A moves right at 15 metres per second, train B moves left at 10 metres per second, and Ravi stands on the platform. Chips switch the frame between the ground, train A and train B. In each frame the chosen body is at rest, every other velocity arrow is redrawn as its velocity minus the velocity of the frame, and the ground slides past at minus the frame velocity.train — two coaches pulled by a locomotive, side viewperson standing, one arm out Ravi A: +15 m/s B: −10 m/s

In Fig 2.2 the first chip shows the ground frame. The other two chips show the same four seconds from train A and from train B. Look at what changes and what does not. The velocities change: every arrow loses the velocity of the chosen train. The order of events does not change. The trains cross each other at the same moment in all three frames.

principle 4.2
Changing to the frame of a moving body

Let a body B move along a line at a steady velocity \(v_B\). To describe motion as seen from B, subtract \(v_B\) from the velocity of every body, including the ground. In that frame B is at rest.

CheckpointUnanswered

In the same scene, train A moves at \(+15\,\text{m/s}\), train B at \(-10\,\text{m/s}\), and Ravi stands on the platform. In the frame of train B, what is the velocity of Ravi?

Relative acceleration
Maths you need

Relative acceleration

The same step works once more. The rate of change of \(v_{AB}\) is the relative acceleration of A with respect to B.

\[ a_{AB} = a_A - a_B \]

The three equations of motion from the first chapter hold for relative motion too, when \(a_{AB}\) is constant. Replace every quantity by its relative version: \(u_{AB}\), \(a_{AB}\) and the change in \(x_{AB}\).

RememberRelative acceleration and relative motion
\[ a_{AB} = a_A - a_B, \qquad v_{AB} = u_{AB} + a_{AB}\,t, \qquad \Delta x_{AB} = u_{AB}\,t + \tfrac{1}{2} a_{AB}\,t^2 \]
Holds whenall quantities measured in one frame, along one line; the last two need a constant relative acceleration.
Symbols\(a_A, a_B\) accelerations of A and B, \(u_{AB}\) initial relative velocity, \(v_{AB}\) relative velocity at time \(t\), \(\Delta x_{AB}\) change in relative position
Units\(a\) in m/s², \(u, v\) in m/s, \(x\) in m, \(t\) in s

The most useful case is \(a_A = a_B\). Then \(a_{AB} = 0\), so \(v_{AB}\) never changes. Each body may speed up or slow down, but the gap between them changes at a steady rate. Lesson 5 uses this for two balls in the air.

Checkpoint · select all that applyUnanswered

Two cars start side by side. Car A starts at \(10\,\text{m/s}\) and car B at \(4\,\text{m/s}\), and both speed up at \(2\,\text{m/s}^2\) in the same direction. Which statements are correct?

In the world
The run-up of a fast bowler

A fast bowler runs in at about \(7\,\text{m/s}\) before letting the ball go. The arm throws the ball forward relative to the bowler. The speed that the speed gun shows is measured in the ground frame. So the ball's ground speed is its speed relative to the bowler plus the bowler's own speed at release. That is why a bowler who stops dead at the crease bowls slower. It is also why coaches work on the run-up, not only on the arm.

Exam traps
  1. Adding the speeds when the bodies move in the same direction.
    Give each velocity its sign and subtract. Same direction gives a difference of speeds; opposite directions give a sum.
  2. Mixing frames in one subtraction, such as a velocity measured from a train minus a velocity measured from the ground.
    Both velocities in \(v_A - v_B\) must be measured in the same frame, usually the ground frame.
  3. Forgetting that the ground moves in the frame of a moving body.
    In the frame of B, the ground moves at \(-v_B\).