Distance and Displacement: Differences, Formula and Usage

The total length of the path travelled by an object in a given period of time is called distance. It is a scalar quantity. It is always positive. On the other hand, the shortest distance between the final position of the object and the initial position is called displacement. Displacement is a vector quantity, it has both magnitude and direction. The value of displacement can be anything positive, negative or zero.

Distance and Displacement are those physical quantities that are often used by us in our day to day conversations and most of us use them interchangeably. Are the two same or different, and if they are different, then what are their differences? What are the use cases of each of them? To answer all these questions, let’s understand physical quantities and get a clear picture of both the physical quantities in this article.

What is Distance?

The length of the path travelled by an object between two points is known as distance. It is a scalar quantity that is always non-negative. For a moving object, the distance travelled by it, always increases. The symbol for distance is \(d\).

Example:
A cyclist leaves his home and travels \(4\,{\text{km}}\) towards the west and reaches a hotel. From there, he turns north and travels another \(3\,{\text{km}}\) to reach a stadium.
The distance from his home to the hotel, \(AB\) is \(4\,{\text{km}}\).
The distance from the hotel to the stadium, \(BC\) is \(3\,{\text{km}}\).
The total distance travelled by the cyclist is,
\(d=AB+BC\)
\( \Rightarrow d = 4 + 3 = 7\,{\text{km}}\)

What is Displacement?

The change in position of a moving object is known as displacement. It is a vector quantity that is directed from the initial position of the object to its final position. The magnitude of displacement is equal to the straight line distance between the initial and final position of the object. The symbol for the displacement vector is \(S\). The magnitude of displacement is represented by \(S\).

Example:
Taking the same example of the cyclist travelling west and then north, we found the total distance travelled was \(7\,{\text{km}}\). The first path is \(4\,{\text{km}}\) and the second is \(3\,{\text{km}}\). We see that he makes a \(90^\circ \) turn after the hotel. The straight-line distance between his home and the stadium is the hypotenuse of a right-angled triangle.
Using Pythagoras theorem,
\(A{B^2} + B{C^2} = A{C^2}\)
\({4^2} + {3^2} = 25\)
\(\Rightarrow AC = 5\,{\text{km}}\)

We see that displacement from his home to the stadium is \(5\,{\text{km}}\) in the north-westerly direction.

What is the Difference Between Distance and Displacement?

Distance and Motion

If an object changes its position with respect to an observer, it is said to be in motion. If a body is at point \(A\) at the initial instant, and its position is at point \(B\) at the final instant of time, then we say that the body has displaced from \(A\) to \(B\). The length of the path from \(A\) to \(B\) is the distance between these two points. The rate at which this distance is covered is known as speed.

\({\text{speed=}}\frac{{{\text{distance}}}}{{{\text{time}}}}\)
Since, distance is a scalar quantity, its rate of change – speed, is also a scalar quantity. We can rearrange the above equation to calculate distance travelled as shown below.
\({\text{distance}} = {\text{speed}} \times {\text{time}}\)

Distance Formula (from Coordinate Geometry)

It is the formula that gives the distance between any two points in a horizontal plane. To derive the distance formula, we consider two points on the \(XY\) plane,
\(A\left( {{x_1},\,{y_1}} \right)\) and \(B\left({{x_2},\,{y_2}}  \right)\)
The line joining the two points gives the distance between them.

We can see that the line \(AB\) forms the hypotenuse of a right-angled triangle.
The base of the triangle is \(\left({{x_2} – {x_1}} \right)\).
The height of the triangle is \(\left({{y_2} – {y_1}}\right)\).

According to the Pythagoras theorem,
\({\text{bas}}{{\text{e}}^2} + {\text{heigh}}{{\text{t}}^2} = {\text{hypotenus}}{{\text{e}}^2}\)
Therefore,
\({\text{hypotenus}}{{\text{e}}^2} = {\left( {{x_2} – {x_1}} \right)^2} + {\left( {{y_2} – {y_1}} \right)^2}\)
\(AB = \sqrt {{{\left({{x_2} – {x_1}} \right)}^2} + {{\left({{y_2} – {y_1}} \right)}^2}} \)

This is the length between two points and is called the Distance Formula.

Where is Displacement Used?

Distance is a common everyday usage word. We talk about distances between two places, cities, countries, etc. Then why is displacement needed? Displacement is needed in the study of motion of objects, to know the length travelled and the direction of travel. To give an example, the earth orbits the sun. But the sun too moves in an orbit around the centre of the Milky Way galaxy. So, the position of the earth in one year is not the same but is different as the sun has moved the entire solar system along with it. The shortest length between this year’s position and the previous year’s position of the earth is displacement.

Distance and Displacement – Sample Problems

Q.1. A ball rolls for \(5\,{\text{m}}\) on a floor and returns after hitting a wall. It stops after travelling \(3\,{\text{m}}\). What is the distance travelled and the displacement?
Ans: Distance travelled is the sum of individual paths taken by the ball.
Starting point to the wall: \(5\,{\text{m}}\)
Wall to the stopping point: \(3\,{\text{m}}\)
Total distance travelled by the ball,
\(5 + 3 = 8\,{\text{m}}\)
To find displacement, let us consider right as positive and left as negative. If the ball rolls right for \(5\,{\text{m}}\), then rolls back \(3\,{\text{m}}\) in the left direction, Displacement is,
\(5 + \left({ – 3} \right) = + 2\,{\text{m}}\)
The ball is at \(2\,{\text{m}}\) right from its starting point.

Q.2. An ant is at point \((3, -4)\) and moves in a straight line to point \((-6, 4)\). What is the distance travelled by the ant and its displacement?
Ans:

Distance between two points is given by the distance formula,
\({\text{distance}} = \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2}} \)
\(A\left(x_{1}, y_{1}\right)=A(3,-4)\)
\(B\left(x_{2}, y_{2}\right)=B(-6,4)\)
\({\text{distance}},\,AB = \sqrt {{{( – 6 – 3)}^2} + {{(4 – ( – 4))}^2}} \)
\(A B=\sqrt{9^{2}+8^{2}}\)
\({\text{distance}} = 12.04{\text{ units}}\)
Distance travelled by the ant is \(12.04{\text{ units}}\) units. Since it is a straight line, displacement magnitude is also \(12.04{\text{ units}}\).

FAQs

Q.1. Assuming the orbit of the earth to be circular, what will be the distance travelled by the earth and the displacement after exactly one year?
Ans: To answer this, we make two assumptions: Earth’s orbit is circular, and the sun is fixed in space. After one year, the earth has made one complete circle around the sun. Distance travelled will be equal to the circumference of this circle. Since the earth is in the same place from where it has started one year back, displacement is zero.

Q.2. When are distance and displacement equal in length?
Ans: When the path taken by the object from starting point to its ending point is a straight line, we can see that displacement and distance values are equal in magnitude.

Q.3. Why do we say that displacement can be zero or negative?
Ans: Displacement is a vector and depends on the direction. Depending on the direction, the displacement can be even zero. An object after travelling all over and coming back to its original position has a displacement of zero.

Q.4. Can we call the distance formula as the displacement formula also?
Ans: No. Displacement can be positive, negative or zero. The distance formula is always positive and never negative or zero. This formula does not tell the direction of the path.

Q.5. Can displacement be higher in value than distance?
Ans: No. Displacement can be equal or less than the distance in numerical value, but never more than distance.

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