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Displacement and Distance in 1 Dimension

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In this Physics tutorial, you will learn:

The meaning of Displacement as a change in position
The definition of Distance and its unit
How Distance differs from the Displacement?
How to calculate the Displacement and Distance in one dimension?

Kinematics Learning Material
Tutorial ID	Title	Tutorial	Video Tutorial	Revision Notes	Revision Questions
3.3	Displacement and Distance in 1 Dimension

Introduction

In the previous tutorial "Position. Reference Point" we considered the objects examined as stationary (unmoveable). The only thing we could do was finding the object's coordinates (position) and its distance from the origin (reference point).

But what happens if the object changes its location? Does the position changes or it remains the same?

This and other questions are discussed here in this article. So, let's take a look at them.

Displacement as a Change in Position (in one dimension)

Look at the object shown in the figure.

Physics Tutorials: This image shows an object placed ont a horizontal grid labeled zero on the left increasing in steps of one units through to 13 on the right. A square is positioned as unit 7 and labelled Instant 1

The object initially is at the position x⃗₁ = 7m. After a while, (after losing it from sight) we notice that the object has moved to the position x⃗₂ = 12m as shown below.

Physics Tutorials: This image shows the same horizontal measurement and object as described in the previous image, however, in this image the initial object is not dotted to indicate a previous position. The same object is then shown at positon 12 and labelled Instant 2

Since we have lost the object from sight during its motion, we are not sure in what direction it has moved before stopping at x⃗₂ = 12m. However, we are sure for one thing: at the end of process, the object is 5m on the right of its original position. This means its position has changed by +5m.

This change in position in Kinematics is known as "Displacement". It is denoted by ∆x⃗ and is obtained by subtracting the final and initial position.

Mathematically, we can write:

∆x⃗ = x⃗_final - x⃗_initial
= x⃗₂ - x⃗₁

In the specific case, we have

∆x⃗ = x⃗_final - x⃗_initial
= x⃗₂ - x⃗₁
= 12m - 7m
= 5m

It is obvious that Displacement is obtained by subtracting two vectors: the final and initial position. Therefore, Displacement is a vector quantity as well.

In the figure below, we can show the three vectors involved in our example.

Physics Tutorials: This image shows the previous objects described in image two, in addition this has arrows which run from each start point to object start / end to illustrate visually the values for eachelelemtn therein as described below.

What is Distance? How does it differ from the Displacement?

In the previous example, it was supposed that we had lost the object from sight from a while and when we noticed it again, the object was displaced by +5m due right from its original position. But what if the object after starting its motion from x⃗₁ = 7m had first moved for some metres (for example 6m) due left and then it has moved due right until it went to the final position x⃗₂ = 12m?

It is obvious in such a scenario the object has moved more than 5m, which was the result obtained when we calculated the Displacement. Therefore, we must consider another quantity (measured in metres as well) which shows the entire length of the object's path during its motion. This quantity is known as Distance (in formula, s).

Hence, by definition "Distance (s) is the length of the entire path followed by an object during its motion." Distance is a scalar quantity, as it does not involve any direction. When dealing with distance, we are interested only in the magnitude of movement (in units of length), not in the moving direction. In this regard, we can say that the distance 25m due North is the same as 25m due East as in both cases the object has moved for 25m regardless the direction.

In the previous example, if the object has moved first by 6m due left, it went at the position 7m - 6m = 1m. Then, it moved for other 11m due right as 12m - 1m = 11m (the final position is 12m). So, the object has moved 6m + 11m = 17m in total. Therefore, we obtain two different results for the same event:

Displacement ∆x⃗ = 5m

and

Distance s = 17m

But, what if the object moves only in one direction? Is the Displacement in this case equal to the Distance?

The answer is NO. Although the numerical result (magnitude) is the same in both cases, Displacement has an additional info included - the direction. Thus, if an object has moved 5m due right only, we say, "Distance is 5m" and "Displacement is 5m due right." Therefore, Displacement and Distance are never identical. You must be careful to avoid confusing them with each other.

Remark! Unlike Displacement, the Distance cannot be negative as it represents the length of the object's path. If the object is at rest, this length is zero but if it starts moving, the path's length becomes a positive number. Consider the mileage screen in a car. It show how many km or miles has the car moved since it has been purchased. This number is always positive as it counts the total distance the car has travelled regardless the moving direction.

Physics Tutorials: This image shows a car display with the trip journey showing 201 km and the total distance travelled by the car as 077030

In the figure above, we can learn that the car has travelled 77,030 km in total during its lifetime but we cannot know in which direction it has moved. Every km the car moves is recorded on the screen as a positive number, regardless the moving direction.

Example 1

Find the Displacement and Distance travelled by the object shown in the figure. Physics Tutorials: This image shows the same horizonal grid as in image 1, 2 and 3. This grid however starts on the left at -4 and increases, left to right by a single unit upto and including position 13. Three objects are positioned on the grid. Instant 1 is at -3, instant 2 is at 5 and instant 0 is at position 10

Solution 1

The object's position at the "instant 0" represents the initial position, x⃗_initial = +10m while the position shown at the "instant 2" represents the final position, x⃗_final = +5m. The position shown at the "instant 1" is not considered when dealing with Displacement, as it is a transitory position. When calculating the Displacement, we are interested only at the initial and final position, not at the in-between positions during the transitory stages.

Hence, for the Displacement ∆x⃗ we have:

∆x⃗ = x⃗_final - x⃗_initial
= (+5)m - (+10)m
= (-5)m

This means the final position of the object is 5m on the left of its initial position. You can check the veracity of this statement by looking at the figure, in which it is evident that the final position of the object is 5m on the left of the initial one.

As for the Distance, we must simply count the metres travelled by the object during its motion, regardless the direction. Hence, if we denote as s1 and s2 the two distances travelled by the object, we have

s₁ = 13m

(the positions +10m and-3m are 13 m away from each other)

and

s₂ = 8m

(the positions -3m and+5m are 8 m away from each other)

Therefore, we obtain for the total Distance:

s_total = s₁ + s₂
= 13m + 8m
= 21m

To summarize, we say "the object was displaced by 5m due left but it has moved by 21m in total."

Whats next?

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Continuing learning kinematics - read our next physics tutorial: Displacement and Distance in 2 and 3 Dimensions

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