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In this Physics tutorial, you will learn:

- The meaning of relative motion
- How does the movement of reference frame affects the motion of an object?
- How do the kinematic quantities change during the relative motion?
- How to calculate the position of objects in different situations during a relative motion?

1. Suppose you are lying on your bed. Do you think you are completely at rest or you are moving in any way? Explain your opinion.

2. You are driving along a straight road. The sideways trees look moving in the opposite direction of you. Are the trees really moving? Explain your opinion again.

3. When you look the Sun, it does not seem to be at the same position it was one or two hours ago. Does the Sun really move or maybe something else happens instead? Explain your opinion.

So far, we have discussed situations involving motions whose kinematic quantities were measured starting from a fixed location known as "reference point (frame)" or "origin." It means the origin was presumed as unmoveable.

But happens if the origin is not stationary? It is obvious the values of kinematic quantities will not be the same as if origin was stationary. In fact, everything in the universe is moveable. Hence, the approach with moveable origin is more realistic than that with stationary one.

In this tutorial, we will discuss about the relationship between the kinematic quantities when the reference point is not stationary. The motion involved in such cases is known as "relative motion."

To help readers understand this point as easier as possible, let's answer to the questions posed in the "Introduction" paragraph.

- In the first question, we can say that if the actual position of our body is taken as reference frame, then we are at rest. However, for somebody who is watching us from outside the Earth, for example from a satellite, we are moving (rotating) together with the Earth. Therefore, it depends on the reference frame chosen saying whether we are moving or not.
- It is obvious for a person who is sitting under a tree, we are moving together with our car past the trees. However, if we chose the reference frame any object inside our car, it seems the trees are moving in the opposite direction of ours.
- If we take our actual position as a reference point (frame), we say the Sun moves from East to West. However, we know this does not occur due to the Sun movement (which takes place in any case), but because of the Earth's rotation around its axis from West to East. Therefore, if we take the Sun as a reference point (frame), we say it the Earth that is moving, not the Sun.

As seen in the above examples, the motion of an object does not depend only on the values of its kinematic quantities but also on the reference frame chosen to study its motion. Therefore, we say "the motion is relative; its parameters may be different in two different reference frames although the rhythm of motion may be the same."

For simplicity, in this tutorial we will focus purely on one-dimensional motions as the same approach can also be used for two and three-dimensional motion as well; only the mathematical apparatus becomes more complex.

This includes the situations described in Uniform Motion explained in the previous tutorials. Thus, if the initial position of the object is x*⃗*_{0} units to the reference point and the object is moving at constant velocity v*⃗*, the equation of motion which gives the position x*⃗* of the object at any instant t, is

For example, if a ball is initially 2m on the right of a tree and it rolls at 3m/s due right, its position after 4 seconds in respect to the tree will be

x*⃗*(4) = 2m + 3 *m**/**s* × 4s

= 2m + 12 m

= 14 m

= 2m + 12 m

= 14 m

on the right of the tree (here, the direction from left to right is considered as positive). It is obvious the displacement of the ball is x*⃗*(4) - x*⃗*_{0} = 14 m-2 m = 12 m.

Look at the figure below:

This situation is similar to the one described in the **Introduction** (the moving car and the sideways trees). In this case, the object's velocity is v*⃗* = 0 but the reference frame's velocity is v*⃗*^{'}. If the reference frame is moving let's say due right (in the positive direction), the object seems moving due left (in the negative direction), although it is at rest.

If the object is initially at x*⃗*_{0}, after t seconds, it will be at

Since the object is at rest (v*⃗* = 0), the second term of the above equation is cancelled, so we obtain

x*⃗*(t) = x*⃗*_{0} - v*⃗*^{'} × t

where v*⃗*^{'} is the velocity by which the reference frame is moving. Let's illustrate this point through a numerical example.

A car is moving at 20 m/s due right as shown in the figure.

What will the position of the tree in respect to the car be after 1.5 seconds?

Here the car acts as a reference frame. As stated before, the tree is at rest, so v*⃗* = 0. Also, the car (which acts as a reference frame) is moving past the tree (it leaves the tree behind after a while). Therefore, the car's velocity is denoted as v*⃗*^{'} = 20 m/s (it is positive as the car moves due right as well). Furthermore, the initial position of the tree is x*⃗*_{0} = 10 m as the tree initially is 10 m in front of the car. Therefore, we write

x*⃗*(t) = x*⃗*_{0} + v*⃗* × t - v*⃗*^{'} × t

At t = 1.5 s we can write for the position of the tree x*⃗*:

x*⃗*(1.5) = 10m + 0 *m**/**s* × 1.5 s - 20 *m**/**s* × 1.5 s

= 10 m - 30 m

= -20 m

= 10 m - 30 m

= -20 m

This result means the tree is 20 m behind the car after 1.5 s of motion.

This is the case when two cars for example are moving either in the same or in the opposite direction. One of the cars is taken as a reference frame (for example the car in which we are inside) and the other car is considered as an object moving in respect to the first car. Once again, we use the equation (2) we discussed earlier

x*⃗*(t) = x*⃗*_{0} + v*⃗* × t - v*⃗*^{'} × t

to find the position of the second object in respect to the first object (the reference frame).

Again, let's illustrate this point with a numerical example for a better understanding.

A car is moving due East at 15 m/s. A pedestrian who is initially 100 m in front of the car is walking at 2 m/s due West. What is the position of the pedestrian in respect to the car after 8s?

East is taken as positive direction and West as negative. Given this, we have the following clues:

x*⃗*_{0} = 100 m

v*⃗*^{'} = 15 *m**/**s*

v*⃗* = -2 *m**/**s*

t = 8s

x*⃗* = ?

v

v

t = 8s

x

Using the equation (2),

x*⃗*(t) = x*⃗*_{0} + v*⃗* × t - v*⃗*^{'} × t

we obtain after the substitutions

x*⃗*(8) = 100m + (-2) *m**/**s* × 8s - 15 *m**/**s* × 8s

= 100m - 16m - 120m

= -36m

= 100m - 16m - 120m

= -36m

This result means the man will be 36 m on the West (behind) the car after 8 seconds. Look at the figure:

This case represents the situations described earlier in tutorials involving Physics tutorial motion with constant acceleration. Thus, since v*⃗*^{'} = 0, we obtain for the position of an object at any instant t (if the object which initially was at x*⃗*_{0} and it is moving at constant acceleration a*⃗*):

or

x*⃗*(t) = x*⃗*_{0} + v*⃗*_{0} × t + *a**⃗* × t^{2}*/**2*

since v*⃗*^{'} = 0 and as a result, the last term of equation (3) is cancelled.

In this case, the last term of the equation (3) is not cancelled but it is an active term of the equation. Thus, we have:

x*⃗*(t) = x*⃗*_{0} + v*⃗*_{0} × t + *a**⃗* × t^{2}*/**2* - v*⃗*^{'} × t

Factorizing, we obtain

Let's use the equation (3) or (4) (you can use whichever equation you wish as they are identical; it is up to your choice) to solve a numerical exercise.

A pedestrian initially at rest, starts walking at a*⃗* = 0.5 m/s^{2} in the direction of a car when the car is 150 m away from him as shown in the figure.

The car itself is moving at 20 m/s in the direction of the pedestrian. What will the position of the pedestrian in respect to the car be after 10 seconds?

In this problem, we have the following clues:

x*⃗*_{0} = 150 m

v*⃗*^{'} = 20 m/s

v*⃗*_{0} = 0 m/s

a*⃗* = -0.5 *m**/**s*^{2}

t = 10s

x*⃗* = ?

v

v

a

t = 10s

x

**Note 1.** -0.5 *m**/**s*^{2} - the direction of motion is due left, i.e. a negative number

Using the equation (4), we obtain

x*⃗*(t) = x*⃗*_{0} + (v*⃗*_{0} - v*⃗*^{'}) × t + *a**⃗* × t^{2}*/**2*

Substituting the known values, we obtain

x*⃗*(10) = 150 + (0-20) × 10 + *(-0.5) × 10*^{2}*/**2*

= 150m - 200m - 25m

= -75 m

= 150m - 200m - 25m

= -75 m

This means the pedestrian is 75 m behind the car after 10 s.

Depending on the direction of motion of the object and reference frame, we have

where

- x
*⃗*_{0}is the position of the object in respect to the reference frame at the initial instant, - v
*⃗*_{0}and v*⃗*_{0'}are the initial velocities of the object and the reference frame respectively, - a
*⃗*and a*⃗*^{'}are the accelerations of the object and the reference frame respectively, and - t is the time of motion which is the same for both the object and the reference frame.

Two athletes are facing each other at 80 m initial distance between them as shown in the figure.

Both athletes are initially at rest, then they start moving towards each other at a*⃗*_{1} = 0.6 m/s^{2} and a*⃗*_{2} = 0.4 m/s^{2} respectively. What is the position of athlete 2 in respect to athlete 1 after 14 seconds?

Let's solve this problem in 2 ways and then, the reader will decide which is more suitable to use.

Using equation (5),

x*⃗*(t) = x*⃗*_{0} + v*⃗*_{0} × t + *a**⃗* × t^{2}*/**2*-v*⃗*^{'}_{0} × t + *a**⃗*^{'} × t^{2}*/**2*

and taking the first athlete (the one on the left) as reference frame, we have

x*⃗*(t) = x*⃗*_{0} + v*⃗*_{0}×t+*a**⃗* × t^{2}*/**2*-v*⃗*^{'}_{0}×t+*a**⃗*^{'} × t^{2}*/**2*

= 80 -0×14+*0.4 × 14*^{2}*/**2*-0×14+*0.6 × 14*^{2}*/**2*

= 80m - 39.2m - 58.8m

= -18m

= 80 -

= 80m - 39.2m - 58.8m

= -18m

This result means the second athlete is 18 m on the left of the first one after 14 seconds.

**Remark!** The sign minus before the first pair of brackets shows that the second athlete (not the one taken as reference frame) is moving towards negative. On the other hand, the sign minus before the second pair of brackets shows that the object (here, the first athlete) is taken as a reference frame, although it is moving towards positive (the minus here is contained in the formula itself, not because the direction of motion).

This method is shorter. It consists on taking the relative kinematic quantities between the two athletes, i.e. their approaching velocity v*⃗*_{rel} = v*⃗*_{1} - v*⃗*_{2} or the approaching acceleration a*⃗*_{rel} = a*⃗*_{1} - a*⃗*_{2} and then applying the equation of motion based only on a set of kinematic quantities, i.e. not using two pair of brackets but the following equation instead,

In our example, since the athletes are moving towards each other, we take as relative acceleration the value

a*⃗*_{rel} = a*⃗*_{1} - a*⃗*_{2} = 0.4 *m**/**s*^{2} - (-0.6 *m**/**s*^{2} ) = 1 *m**/**s*^{2}

(One of the accelerations is taken as negative as the athletes are moving in opposite directions).

Therefore, applying the equation (6), we obtain

x*⃗*(t) = x*⃗*_{0} + v*⃗*_{0(rel)} × t + *a**⃗*_{rel} × t^{2}*/**2*

x*⃗*(14) = 80 + 0 × 14 + *1 × 14*^{2}*/**2*

= 80 m + 0m - 98m

= -18m

x

= 80 m + 0m - 98m

= -18m

In this way, we obtained the same results with both methods.

The same approach, i.e. using the concept of relative kinematic quantities can be used in the situations described earlier as well. Hence, we can write v*⃗*_{rel} instead of v*⃗* - v*⃗*^{'}, v*⃗*_{0(rel)} instead of v*⃗*_{0} - v*⃗*_{0}^{'} and a*⃗*_{rel} instead of a*⃗* - a*⃗*^{'} in all exercises involving the relative motion. In this way, the calculations become shorter.

Relative quantities can be used in other equations of motion with constant acceleration as well. Therefore, we can write

v*⃗*_{rel} = v*⃗*_{0(rel)} + a*⃗*_{rel} × t

∆x*⃗*_{rel} = *( v**⃗*_{rel} + v*⃗*_{0(rel)} ) × t*/**2*

v*⃗*^{2}_{rel} - v^{2}_{0(rel)}= 2 × a*⃗*_{rel} × ∆x*⃗*_{rel}

∆x

v

The motion of an object does not depend only on the values of its kinematic quantities but also on the reference frame chosen to study its motion. Therefore, we say "the motion is relative; its parameters may be different in two different reference frames although the rhythm of motion may be the same."

When two objects are moving towards to or away from each other, one of the objects is taken as a reference frame and the other as a normal moving object. Said this, there are six possible cases regarding the objects involved in a relative motion. They are:

This includes the situations described in Uniform Motion explained in the previous tutorials. Thus, if the initial position of the object is x*⃗*_{0} units to the reference point and the object is moving at constant velocity v*⃗*, the equation of motion which gives the position x*⃗* of the object at any instant t, is

In this case, the object's velocity is v*⃗* = 0 but the reference frame's velocity is v*⃗*^{'}. If the reference frame is moving let's say due right (in the positive direction), the object seems moving due left (in the negative direction), although it is at rest.

If the object is initially at x*⃗*_{0}, after t seconds, it will be at

Since the object is at rest (v*⃗* = 0), the second term of the above equation is cancelled, so we obtain

x*⃗*(t) = x*⃗*_{0} - v*⃗*^{'} × t

where v*⃗*^{'} is the velocity by which the reference frame is moving.

This is the case when two objects are moving either in the same or in the opposite direction. One of the objects is taken as a reference frame (for example the car in which we are inside) and the other object is considered as an object moving in respect to the first. Once again, we use the equation (2)

x*⃗*(t) = x*⃗*_{0} + v*⃗* × t - v*⃗*^{'} × t

to find the position of the second object in respect to the first one (the reference frame).

This case represents the situations described earlier in tutorials involving motion with constant acceleration. Thus, since v*⃗*^{'} = 0, we obtain for the position of an object at any instant t (if the object which initially was at x*⃗*_{0} and it is moving at constant acceleration a*⃗*):

or

x*⃗*(t) = x*⃗*_{0} + v*⃗*_{0} × t + *a**⃗* × t^{2}*/**2*

since v*⃗*^{'} = 0 and as a result, the last term of equation (3) is cancelled.

In this case, the last term of the equation (3) is not cancelled but it is an active term of the equation. Thus, we have:

x*⃗*(t) = x*⃗*_{0} + v*⃗*_{0} × t + *a**⃗* × t^{2}*/**2* - v*⃗*^{'} × t

Factorizing, we obtain

Depending on the direction of motion of the object and reference frame, we have

where,

- x
*⃗*_{0}is the position of the object in respect to the reference frame at the initial instant, - v
*⃗*_{0}and v*⃗*_{0}^{'}are the initial velocities of the object and the reference frame respectively, - a
*⃗*and a*⃗*^{'}are the accelerations of the object and the reference frame respectively, and - t is the time of motion which is the same for both the object and the reference frame.

We can also use the help of relative kinematic quantities to approach the abovementioned situations. Hence, we can write v*⃗*_{rel} instead of v*⃗* - v*⃗*^{'}, v*⃗*_{0}(rel) instead of v*⃗*_{0} - v*⃗*_{0}^{'} and a*⃗*_{rel} instead of a*⃗* - a*⃗*^{'} in all exercises involving the relative motion. In this way, the calculations become easier and shorter.

Relative quantities can be used in other equations of motion with constant acceleration as well. Therefore, we can write

v*⃗*_{rel} = v*⃗*_{0(rel)} + a*⃗*_{rel} × t

∆x*⃗*_{rel} = *( v**⃗*_{rel} + v*⃗*_{0(rel)} ) × t*/**2*

v*⃗*^{2}_{rel} - v^{2}_{0(rel)}= 2 × a*⃗*_{rel} × ∆x*⃗*_{rel}

∆x

v

*1. A car is moving due East at 16 m/s. A cyclist initially at 60 m on the East of the car is riding at 2 m/s due East as shown in the figure. *

What will be the position of the cyclist in respect to the car after 5 s?

- 20 m due West
- 20 m due East
- 10 m due West
- 10 m due East

**Correct Answer: C**

*2. Two pedestrians initially at a distance of 36 m away between them, are moving towards each other. The first pedestrian (on the left) moves due East at constant velocity of 1 m/s while the other pedestrian starts moving from rest and accelerates at 0.2 m/s2 due West as shown in the figure. *

- 14.6 s
- 39.2 s
- 29.2 s
- 30.0 s

**Correct Answer: A**

*3. Two cars are moving at constant velocities in opposite directions as shown in the figure. *

The initial distance between the cars is 200 m. The blue car stops immediately after 5 s and the red car stops immediately as well but after 15 s. What is the position of the blue car (the car 2) in respect to the red car (the car 1) at the end of their motion?

- 20 m due East of the red car
- 20 m due West of the red car
- 40 m due West of the red car
- 100 due East of the red car

**Correct Answer: B**

We hope you found this Physics tutorial "Relative Motion" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we expand your insight and knowledge of kinematics with our Physics tutorial on What Causes the Motion? The Meaning of Force.

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