# Physics Tutorial: Relative Motion

[ 2 Votes ]

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?
Kinematics Learning Material
Tutorial IDTitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
3.13Relative Motion

## Introduction

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.

## What is Relative Motion?

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.

1. 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.
2. 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.
3. 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.

### Case 1 - Reference frame is at rest and the object is moving at constant velocity

This includes the situations described in Uniform Motion explained in the previous tutorials. Thus, if the initial position of the object is x0 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

### Equation 1

x(t) = x0 + v × t

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

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) - x0 = 14 m-2 m = 12 m.

Look at the figure below:

### Case 2 - Reference frame is moving at constant velocity v⃗' and the object is at rest

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 x0, after t seconds, it will be at

### Equation 2

x(t) = x0 + v × t - v' × t (2)

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

x(t) = x0 - v' × t

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

#### Example 1

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?

#### Solution 1

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 x0 = 10 m as the tree initially is 10 m in front of the car. Therefore, we write

x(t) = x0 + 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

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

### Case 3 - Reference frame is moving at constant velocity v⃗' and the object is moving at constant velocity v⃗

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) = x0 + 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.

#### Example 2

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?

#### Solution 2

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

x0 = 100 m
v' = 15 m/s
v = -2 m/s
t = 8s
x = ?

Using the equation (2),

x(t) = x0 + v × t - v' × t

we obtain after the substitutions

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

### Case 4 - Reference point is at rest and the object is moving at constant acceleration a⃗

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 x0 and it is moving at constant acceleration a):

### Equation 3

x(t) = x0 + v0 × t + a × t2/2 - v' × t

or

x(t) = x0 + v0 × t + a × t2/2

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

### Case 5 - Reference point is moving at constant velocity v⃗ and the object is moving at constant acceleration a⃗

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) = x0 + v0 × t + a × t2/2 - v' × t

Factorizing, we obtain

### Equation 4

x(t) = x0 + (v0 - v') × t + a × t2/2

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.

### Example 3

A pedestrian initially at rest, starts walking at a = 0.5 m/s2 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?

### Solution 3

In this problem, we have the following clues:

x0 = 150 m
v' = 20 m/s
v0 = 0 m/s
a = -0.5 m/s2
t = 10s
x = ?

Note 1. -0.5 m/s2 - the direction of motion is due left, i.e. a negative number

Using the equation (4), we obtain

x(t) = x0 + (v0 - v') × t + a × t2/2

Substituting the known values, we obtain

x(10) = 150 + (0-20) × 10 + (-0.5) × 102/2
= 150m - 200m - 25m
= -75 m

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

### Case 6 - Both the object and the reference frame are moving at constant acceleration.

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

### Equation 5

x(t) = x0 + v0 × t + a × t2/2-v'0 × t + a' × t2/2

where

• x0 is the position of the object in respect to the reference frame at the initial instant,
• v0 and v0' 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.

#### Example 4

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 a1 = 0.6 m/s2 and a2 = 0.4 m/s2 respectively. What is the position of athlete 2 in respect to athlete 1 after 14 seconds?

#### Solution 4

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

##### Method 1

Using equation (5),

x(t) = x0 + v0 × t + a × t2/2-v'0 × t + a' × t2/2

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

x(t) = x0 + v0×t+a × t2/2-v'0×t+a' × t2/2
= 80 - 0×14+0.4 × 142/2-0×14+0.6 × 142/2
= 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).

##### Method 2

This method is shorter. It consists on taking the relative kinematic quantities between the two athletes, i.e. their approaching velocity vrel = v1 - v2 or the approaching acceleration arel = a1 - a2 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,

### Equation 6

x(t) = x0 + v0(rel) × t + arel × t2/2

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

arel = a1 - a2 = 0.4 m/s2 - (-0.6 m/s2 ) = 1 m/s2

(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) = x0 + v0(rel) × t + arel × t2/2
x(14) = 80 + 0 × 14 + 1 × 142/2
= 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 vrel instead of v - v', v0(rel) instead of v0 - v0' and arel 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

vrel = v0(rel) + arel × t
∆xrel = ( vrel + v0(rel) ) × t/2
v2rel - v20(rel)= 2 × arel × ∆xrel

## Whats next?

Enjoy the "Relative Motion" physics tutorial? People who liked the "Relative Motion" tutorial found the following resources useful:

1. Physics tutorial Feedback. Helps other - Leave a rating for this tutorial (see below)
2. Kinematics Revision Notes: Relative Motion. Print the notes so you can revise the key points covered in the physics tutorial for Relative Motion
3. Kinematics Practice Questions: Relative Motion. Test and improve your knowledge of Relative Motion with example questins and answers
4. Check your calculations for Kinematics questions with our excellent Kinematics calculators which contain full equations and calculations clearly displayed line by line. See the Kinematics Calculators by iCalculator™ below.
5. Continuing learning kinematics - read our next physics tutorial: Motion. Types of Motion

### Kinematics Calculators

The following Physics Calculators are provided in support of the Kinematics tutorials.

## Physics Calculators

You may also find the following Physics calculators useful.