Lesson Explainer: Relative Velocity Mathematics

In this explainer, we will learn how to calculate the relative velocity of a particle with respect to another and how to calculate a relative velocity vector.

If a body travels from a start point 𝐴 to an endpoint 𝐵, it has a displacement from 𝐴 that can be represented by the vector 𝐴𝐵=⃑𝑠.

The velocity vector for a body that has this displacement in a time interval Δ𝑡 is given by ⃑𝑣=⃑𝑠Δ𝑡.

The displacement of the body is displacement from 𝐴, and the velocity of the body is velocity relative to 𝐴.

A coordinate system 𝑆 can be defined so that the origin of the coordinate system is at 𝐴. This is, however, an arbitrarily chosen coordinate system. If a different position for the origin of the coordinate system is chosen for a different coordinate system 𝑆′, then the displacement of 𝐵 from 𝐴 can be defined in 𝑆 and in 𝑆′.

In the following figure, the blue 𝑦-axis is the 𝑦-axis for the coordinate system 𝑆 and the red 𝑦-axis is the 𝑦-axis for the coordinate system 𝑆′. Both coordinate systems have the same 𝑥-axis.

A particle, 𝑝, is shown.

In 𝑆, the coordinates of 𝑝 are (𝑑,0), but in 𝑆′ the coordinates of 𝑝 are (𝐷+𝑑,0).

Suppose that a particle moves from a point 𝐴 to a point 𝐵, where the straight-line distance between 𝐴 and 𝐵 is 𝑑, as shown in the following figure.

We can see that 𝐴𝐵=𝑑⃑𝑖 and 𝐵𝐴=−𝑑⃑𝑖, where ⃑𝑖 is a unit vector in the 𝑥-direction.

The magnitude of the displacement is the same in both 𝑆 and 𝑆′: ‖‖⃑𝑠‖‖=‖‖⃑𝑠‖‖.

The time interval for the motion is the same in both 𝑆 and 𝑆′, and so ‖‖⃑𝑣‖‖=‖‖⃑𝑣‖‖.

Suppose though that, in the time interval Δ𝑡, the particle is displaced from 𝐴 to 𝐵. The coordinate system 𝑆 is displaced from 𝐴 to 𝐵, as shown in the following figure.

In 𝑆′, the velocity of the particle is given by ⃑𝑣=𝑑⃑𝑖Δ𝑡.

In the coordinate system 𝑆, both 𝐴 and 𝐵 are at the origin of 𝑆. The displacement of the particle from the origin of 𝑆 is zero, and so the velocity of the particle in 𝑆 is zero.

Suppose that a point, 𝑄, is at the origin of 𝑆′, as shown by the green dot in the following figure.

Let us define a reference point 𝑃 that is always at the origin of 𝑆, so it moves with 𝑆.

In the time interval Δ𝑡, the displacement of 𝑄 from the point 𝑃 increases by 𝑑.

The displacement of 𝑃 from 𝑄 when 𝑃 has moved from 𝐴 to 𝐵 is given by ⃑𝑠′=𝑑⃑𝑖.

The displacement of 𝑄 from 𝑃 when 𝑃 has moved from 𝐴 to 𝐵 is given by ⃑𝑠=−𝑑⃑𝑖.

The velocity of 𝑄 relative to 𝑃 is, therefore, given by ⃑𝑣=−𝑑⃑𝑖Δ𝑡=−⃑𝑣.

The relative velocity of 𝑄 and 𝑃 has a magnitude of ‖‖⃑𝑣‖‖=‖‖⃑𝑣‖‖=‖‖⃑𝑣‖‖=𝑑Δ𝑡.

In a coordinate system that is at the origin of 𝑆, point 𝑃 is at rest and point 𝑄 moves away from it at speed ‖‖⃑𝑣‖‖. In a coordinate system that is at the origin of 𝑆′, point 𝑄 is at rest and point 𝑃 moves away from it at speed ‖‖⃑𝑣‖‖.

The signs of ⃑𝑣 and ⃑𝑣 are opposite. The velocity that is positive is determined by the initial choice made for the positive value of the 𝑥-axis direction.

Suppose that two particles move along the 𝑥-axis of a coordinate system, each at some speed.

The velocities of the particles can be represented as one-dimensional vectors in the coordinate system. The magnitudes of the velocities of the particles are equal to the speeds of the particles, and the velocities have the same sign if the particles move in the same direction and opposite signs if the particles move in opposite directions.

Taking the signs of the velocities into account, the difference between the velocities of two particles gives the relative velocity of the particles.

The relative velocity of two particles can be taken from the position of either particle, where the velocity is positive if taken from one of the particles and negative if taken from the other particle.

Consider two velocity vectors ⃑𝑣 and ⃑𝑣 that have the same magnitude but opposite directions, as shown in the following figure.

We denote with 𝑣 and 𝑣 the components of ⃑𝑣 and ⃑𝑣 along the motion axis.

The difference between these components can be expressed as 𝑣−𝑣=𝑣=𝑣−(−𝑣)=2𝑣.

Subtraction of a component with a negative sign is equivalent to addition of a component with a positive sign; hence, the difference between components with opposite signs is the sum of the components.

The result of subtracting ⃑𝑣 from ⃑𝑣 is equivalent to the result of adding −⃑𝑣 to ⃑𝑣, as is shown in the following figure.

The difference between these components can also be expressed as 𝑣−𝑣=𝑣=−𝑣−𝑣=−2𝑣.

The result of subtracting ⃑𝑣 from ⃑𝑣 is equivalent to the result of adding −⃑𝑣 to ⃑𝑣, as is shown in the following figure.

Consider now two velocity vectors ⃑𝑣 and ⃑𝑣 that have the same magnitude and the same direction. We denote with 𝑣 and 𝑣 the components of ⃑𝑣 and ⃑𝑣 along the motion axis. The difference between these components can be expressed as 𝑣−𝑣=𝑣=𝑣−𝑣=0 or as 𝑣−𝑣=𝑣=𝑣−𝑣=0.

The two expressions of the difference between these components is shown in the following figure.

The relative velocity for two particles in one-dimensional motion can be defined as follows.

Definition: Relative Velocity for Two Particles in One-Dimensional Motion

For two particles 𝑎 and 𝑏 with velocities ⃑𝑣 and ⃑𝑣 along one axis, and denoting with 𝑣 and 𝑣 their respective components along that axis, the component of the velocity of 𝑎 relative to 𝑏 is given by 𝑣=𝑣−𝑣, and the component of the velocity of 𝑏 relative to 𝑎 is given by 𝑣=𝑣−𝑣.

Let us look at an example that considers relative velocity purely in terms of the differences between velocity vectors.

Example 1: Finding Relative Velocity Using Unit Vectors

If ⃑𝑣=20⃑𝑖 and ⃑𝑣=45⃑𝑖, then ⃑𝑣=⃑𝑖.

Answer

The difference between ⃑𝑣 and ⃑𝑣 is given by ⃑𝑣=⃑𝑣−⃑𝑣.

Rearranging to find ⃑𝑣 gives ⃑𝑣=⃑𝑣−⃑𝑣.

Substituting known values gives ⃑𝑣=45⃑𝑖−20⃑𝑖=25⃑𝑖.

Now, let us look at another such example.

Example 2: Finding Relative Velocity Using Unit Vectors

If ⃑𝑣=60⃑𝑒 and ⃑𝑣=−40⃑𝑒, then ⃑𝑣=⃑𝑒.

Answer

⃑𝑒 is a unit vector in some fixed direction.

The difference between ⃑𝑣 and ⃑𝑣 is given by ⃑𝑣=⃑𝑣−⃑𝑣.

Substituting known values gives ⃑𝑣=60⃑𝑒−−40⃑𝑒=100⃑𝑒.

Let us consider an example where there is a context supplied, involving a body that measures the velocity of a second body relative to the first body.

Example 3: Relative Speed of Bodies Moving in Opposite Directions

A car is moving on a straight road at 84 km/h, and in the opposite direction, a motorbike is moving at 45 km/h. Suppose that the direction of the car is positive. Find the velocity of the motorbike relative to the car.

Answer

Let 𝑣 be the velocity of the car and 𝑣 be the velocity of the motorbike. The direction of the car is positive; hence, 𝑣=84/kmh and 𝑣=−45/.kmh

The velocity of the motorbike relative to the car is given by 𝑣=𝑣−𝑣𝑣=−45−84=−129/.kmh

If two bodies move one-dimensionally in opposite directions, their speeds are added to determine the speed of either body relative to the other. Let us look at another example where this occurs.

Example 4: Finding the Time to Complete a Journey Using Relative Velocities

A ship was sailing with a uniform velocity directly toward a port that is 144 km away. A patrol aircraft passed over the ship traveling in the opposite direction at 366 km/h. When the aircraft measured the ship’s speed, it appeared to be traveling at 402 km/h. Determine the time required for the ship to reach the port.

Answer

To determine the time required for the ship to reach the port, it is necessary to know the speed at which the ship approaches the port. The port is assumed to be stationary.

The speed of the ship measured by the aircraft is 402 km/h. As the ship and aircraft travel in opposite directions, 402 km/h is the sum of their speeds. The speed of the aircraft is stated to be 366 km/h, so the speed of the ship is given by 𝑣=402−366=36/.kmh

The time required to travel 144 km at a speed of 36 km/h is given by 𝑡=14436=4.hours

Let us consider an application of relative velocity in a context involving two bodies moving in the same direction. For two bodies moving in the same direction at speeds 𝑣 and 𝑣, respectively, the speed of either body relative to the other, 𝑣, is given by 𝑣=|𝑣−𝑣|.

The trivially obvious case of this is the case corresponding to both bodies having the same speed, and hence the position of one body relative to the other is constant throughout the motion of the bodies.

Example 5: Using Relative Velocity to Find the Length of a Train given the Time Taken by a Moving Object to Pass It

A helicopter flew in a straight line at 234 km/h above a train moving in the same direction. It took the helicopter 21 seconds to travel the length of the train. Following this, the pilot halved the helicopter’s speed. Given that it took the train 14 seconds to pass the helicopter traveling at this speed, find the length of the train in metres.

Answer

The most important thing to appreciate in this question is that because the helicopter and the train move in the same direction throughout, relative to the ground, their velocities have the same sign. The difference in their velocities is thus equal to the difference in their speeds, and the speed of the train relative to the helicopter (and vice versa), 𝑣, is simply the difference between their speeds.

In the first 21 seconds, the helicopter has a greater velocity relative to the ground than the train has, and in the 14 following seconds, the train has a greater velocity relative to the ground than the helicopter has. The change in the velocity of the helicopter between the first and the second time intervals is assumed to occur in negligible time.

In each time interval, it is the case that 𝑣Δ𝑡=Δ𝑑, where Δ𝑑 is the distance that the helicopter moves relative to the train (and vice versa), which is also the length of the train. The difference between the time intervals allows us to determine the velocity of the train.

To simplify finding the length of the train in metres, the speed of the helicopter is converted to a speed in metres per second as follows: 𝑣=234×10003600=65/.helicopterms

For the first time interval, 21(65−𝑣)=Δ𝑑.train

For the second time interval, 14𝑣−652=Δ𝑑.train

The length of the train remains constant, so the two Δ𝑑 terms can be equated to give 21(65−𝑣)=14𝑣−652.traintrain

This can be rearranged to determine the velocity of the train. Both sides of the equation can be divided by 14 to give 32(65−𝑣)=𝑣−652.traintrain

The bracket can be expanded to give 32(65)−32(𝑣)=𝑣−652.traintrain

The rearrangement is then completed as follows: 32(65)+652=𝑣+32(𝑣)32(65)+652=52(𝑣)4(65)=5(𝑣)2605=𝑣=52/.traintraintraintraintrainms

This value for 𝑣train can now be substituted into the equation for Δ𝑑 in either time interval. If the first of the time intervals is used, this gives us 21(65−52)=Δ𝑑21(13)=Δ𝑑=273.m

If the second of the time intervals is used, this gives us 1452−652=Δ𝑑14392=Δ𝑑=273.m

The length of the train is 273 metres.

Key Points

  • For two bodies 𝑎 and 𝑏 with velocities ⃑𝑣 and ⃑𝑣 along one axis, and denoting with 𝑣 and 𝑣 their respective components along that axis, the component of the velocity of 𝑎 relative to 𝑏 is given by 𝑣=𝑣−𝑣, and the component of the velocity of 𝑏 relative to 𝑎 is given by 𝑣=𝑣−𝑣.
  • For two bodies moving in one dimension in opposite directions, the speed of either body relative to the other is the sum of the speeds of the bodies.
  • For two bodies moving in one dimension in the same direction, the speed of either body relative to the other is the difference between the speeds of the bodies.

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