Lesson Explainer: Relative Velocity | Nagwa Lesson Explainer: Relative Velocity | Nagwa

Lesson Explainer: Relative Velocity Mathematics • Second Year of Secondary School

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.

When a particle moves in a straight line from one point, 𝐴, to another, 𝐡, we can describe its displacement using the vector 𝐴𝐡=⃑𝑠.

If the particle moves from 𝐴 to 𝐡 in the time interval Δ𝑑, then the velocity of this particle is given by ⃑𝑣=⃑𝑠Δ𝑑.

This velocity is called the average velocity of the particle as it moves from 𝐴 to 𝐡. If the velocity of the particle is constant, then this average velocity is equal to the velocity of the particle at any point during its motion. There are some situations where we may want to consider a different kind of velocity, which we call the relative velocity.

Suppose you are sat on a train, looking out the window at a car driving in the opposite direction. Suppose also that the car is moving with a constant speed of 50 km/h and the train is moving with a constant speed of 80 km/h. Using this, we can say that the velocity of the train is 80 km/h, while the velocity of the car is βˆ’50 km/h. However, since you are sat on the train, which is moving at 80 km/h, while observing the car move in the opposite direction, it will appear as though the car is moving faster than its speed of 50 km/h. This is due to relative velocities.

Relative velocities tell us that movement is a relative concept, which means that it differs depending on the observer. In every case, while the viewer observes the movement of the other object, they consider themself to be at rest even if they are not. The velocity that the viewer observes the object moving at is not the actual velocity of the object, it is the relative velocity.

Let us now consider how we can calculate relative velocities. Suppose that we have two bodies, 𝐴 and 𝐡, such that the velocity of 𝐴 is οƒŸπ‘£οŒ  and the velocity of 𝐡 is οƒŸπ‘£οŒ‘. A stationary observer would see the two bodies moving with their velocities οƒŸπ‘£οŒ  and οƒŸπ‘£οŒ‘.

Now, let us suppose the observer is sat on body 𝐴, moving with velocity οƒŸπ‘£οŒ . The observer assumes that they are stationary, so when we are finding the relative velocity of 𝐡, we will need to take this into consideration.

First, suppose that οƒŸπ‘£=0, so 𝐡 is stationary. The observer is moving away from 𝐡 with velocity οƒŸπ‘£οŒ , so they will see 𝐡 move away from them with velocity βˆ’οƒŸπ‘£οŒ . This is the relative velocity of 𝐡 with respect to 𝐴, which we can call οƒ π‘£οŒ‘οŒ . In the case where οƒŸπ‘£=0, 𝑣=βˆ’οƒŸπ‘£.

Next, we will consider when οƒŸπ‘£β‰ 0. Since the observer is still moving away with velocity οƒŸπ‘£οŒ , we will again need to take this into consideration. This time, we also have οƒŸπ‘£οŒ‘. To find the relative velocity of 𝐡 with respect to 𝐴, we need to simply add οƒŸπ‘£οŒ‘ to the βˆ’οƒŸπ‘£οŒ  we had before. This gives us that the relative velocity is 𝑣=οƒŸπ‘£βˆ’οƒŸπ‘£.

Definition: Relative Velocity

If bodies 𝐴 and 𝐡 are moving with velocities οƒŸπ‘£οŒ  and οƒŸπ‘£οŒ‘, respectively, then we can say that the relative velocity of 𝐡 with respect to 𝐴, οƒ π‘£οŒ‘οŒ , is given by 𝑣=οƒŸπ‘£βˆ’οƒŸπ‘£οŒ‘οŒ οŒ‘οŒ  and the relative velocity of 𝐴 with respect to 𝐡, οƒ π‘£οŒ οŒ‘, is given by 𝑣=οƒŸπ‘£βˆ’οƒŸπ‘£.

These formulas are given in terms of vectors, but they still hold for scalar quantities too.

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

Fill in the blank: If ⃑𝑣=25βƒ‘π‘–οŒΊοŒ» and ⃑𝑣=40βƒ‘π‘–οŒΊ, then ⃑𝑣=βƒ‘π‘–οŒ».

Answer

The difference between βƒ‘π‘£οŒΊ and βƒ‘π‘£οŒ» is given by ⃑𝑣=βƒ‘π‘£βˆ’βƒ‘π‘£.

Rearranging to find βƒ‘π‘£οŒ» gives ⃑𝑣=βƒ‘π‘£βˆ’βƒ‘π‘£.

Substituting known values gives ⃑𝑣=40βƒ‘π‘–βˆ’25⃑𝑖=15⃑𝑖.

Now, let us look at another such example.

Example 2: Finding Relative Velocity Using Unit Vectors

If ⃑𝑣=50βƒ‘π‘’οŒ  and ⃑𝑣=βˆ’15βƒ‘π‘’οŒ‘, then the relative velocity ⃑𝑣=βƒ‘π‘’οŒ οŒ‘.

Answer

⃑𝑒 is a unit vector in some fixed direction.

The difference between βƒ‘π‘£οŒ  and βƒ‘π‘£οŒ‘ is given by ⃑𝑣=βƒ‘π‘£βˆ’βƒ‘π‘£.

Substituting known values gives ⃑𝑣=50βƒ‘π‘’βˆ’ο€Ήβˆ’15⃑𝑒=65⃑𝑒.

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 π‘£οŒΊ be the velocity of the car and π‘£οŒ» be the velocity of the motorbike. The direction of the car is positive; hence, 𝑣=84/kmh and 𝑣=βˆ’45/.kmh

The velocity of the motorbike relative to the car is given by 𝑣=π‘£βˆ’π‘£π‘£=βˆ’45βˆ’84=βˆ’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=402βˆ’366=36/.kmh

The time required to travel 144 km at a speed of 36 km/h is given by 𝑑=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 π‘£οŒΊ and π‘£οŒ», respectively, the speed of either body relative to the other, 𝑣, is given by 𝑣=|π‘£βˆ’π‘£|.

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 𝑣Δ𝑑=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=234Γ—10003600=65/.helicopterms

For the first time interval, 21(65βˆ’π‘£)=Δ𝑑.train

For the second time interval, 14ο€Όπ‘£βˆ’652=Δ𝑑.train

The length of the train remains constant, so the two Δ𝑑 terms can be equated to give 21(65βˆ’π‘£)=14ο€Όπ‘£βˆ’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(65βˆ’π‘£)=π‘£βˆ’652.traintrain

The bracket can be expanded to give 32(65)βˆ’32(𝑣)=π‘£βˆ’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(65βˆ’52)=Δ𝑑21(13)=Δ𝑑=273.m

If the second of the time intervals is used, this gives us 14ο€Ό52βˆ’652=Δ𝑑14ο€Ό392=Δ𝑑=273.m

The length of the train is 273 metres.

Key Points

  • If bodies 𝐴 and 𝐡 are moving with velocities οƒŸπ‘£οŒ  and οƒŸπ‘£οŒ‘, respectively, then we can say that the relative velocity of 𝐡 with respect to 𝐴, οƒ π‘£οŒ‘οŒ , is given by 𝑣=οƒŸπ‘£βˆ’οƒŸπ‘£οŒ‘οŒ οŒ‘οŒ  and the relative velocity of 𝐴 with respect to 𝐡, οƒ π‘£οŒ οŒ‘, is given by 𝑣=οƒŸπ‘£βˆ’οƒŸπ‘£.
  • When finding relative velocities, it is useful to establish a sign convention for solving problems.

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