Loading...

In this article, examples of displacement and distance in physics are presented along with their definitions which is helpful for high school physics.

Displacementis a vector quantity describes a change in position of an object or how far an object is displaced from its initial position and is given by below formula

\[ \Delta \vec x = x_f - x_i \]

Where the starting and ending positions are denoted by $x_i$ and $x_f$, respectively.

Distanceis a scalar quantity indicating the total path traveled by a moving object.

Since in both cases, the interval between two points is measured- one as the shortest line and the other as the total traveled path- so the SI unit of displacement and distance is the meter.

The illustration below shows the difference between them clearly.

Now with this brief definition, we can go further and explain those concepts more concisely with numerous examples.

**Example (1): A boy is playing around a rectangle. He starts his game from one corner ($i$) and ends at the same one. **

**Solution:** In this path, since his initial and final points are the same, by definition, __his displacement is zero__.

On the other hand, the __total distance traveled by him is the perimeter__ of that rectangular shape.

In this way, distance traveled (in physics) is consistent with the concept of displacement in everyday language.

**Example: A car travels a circle of the circumference of $100\,\rm m$. In each of the following, find the quantity wanted:
(a) If the car moves once around the track, what is the distance traveled by it? What if it moves twice?
(b) If the car moves once around the track, what is its displacement?**

**Solution**: Another example of displacement that clarifies its difference with the distance traveled.

Distance is defined as the whole path covered by a moving object, but displacement is the difference between the initial and final positions of the moving body.

(a) The whole path taken by the car around the circle is its circumference. Thus, when the car travels once around the circle, its distance traveled is $100\,\rm m$.

Once the car moves twice around the circle, its distance traveled is $200\,\rm m$.

(b) When the car travels a complete circle, its initial and final positions coincide with each other. Thus, according to the definition of displacement, $\Delta x=x_f-x_i$, its displacement becomes zero.

This example explicitly shows us that the distance traveled cannot never get zero.

**Example (2): A biker ride into a horizontal loop of a radius of 10 meters and covers three fourth of it as follows. What are the displacement and distance traveled by him?**

**Solution:** As mentioned earlier, displacement is a quantity that depends only on the position of initial and final points.

Thus, the straightest line between those points, which gives the magnitude of the displacement vector, are computed by Pythagorean theorem as

\begin{align*} D^2 &= r^2 + r^2 \\ &= 2 r^2 \\ \Rightarrow D &= r\sqrt{2} \end{align*} where the magnitude of displacement denoted by $D$. Putting the values gives, $D=10\sqrt{2}\,\rm m$.

Distance is simply the circumference of three fourth of the circle, so \begin{align*} \text{distance}&= \frac{3}{4} \, {\text{perimeter}} \\\\ &=\frac{3}{4} (2\pi\,r) \\\\ &= \frac{3 \times 2\times \pi \times 10}{4} \\\\ &=15\,\pi \,\rm m \end{align*}

**Example (3): A person starts at position 0 and walks 4 meters to the right, returns, and walks 6 meters to the left.
(a) What is displacement?
(b) What is the distance traveled by him?**

**Solution**: denote the initial position as $x_i=0$ and final position as $x_f$. As you can see, from the starting position to the turning point, the person walks 4 meters to the right. From that point ($B$) count 6 steps to the left to reach the final point whose position is on the negative side of the axis i.e. $x_f=-2$.

(a) Displacement in physics is a vector and is defined to be the final position minus the initial position. So, \[\Delta x=x_f-x_i=-2-0=-2\,{\rm m}\] Hence, the person displaced 2 meters to the left (green arrow). Here, the minus sign indicates the direction of the displacement.

(b) For the first part of the trip, the person walks 4 meters, and next, 6 meters more additional to the left. Thus, the total distance traveled is calculated to be 10 meters.

**Example (4): If you walk exactly 4 times around a quarter-meter path, what is your displacement?**

**Solution**: let the initial position to be $x_i=0$. By gluing 4 of these quarter tracks, you reach your initial position. So, the initial and final positions are the same and consequently the displacement becomes zero.

**Example (5): Consider a car to be a point particle and move in one dimension. To specify the location of the particle in one dimension we need only one axis that we call $x$ and lies along the straight-line path.**

**Solution:** First, we must define an important quantity that the other kinematics quantities are made from it, displacement.

To describe the motion of the car, we must know its position and how that position changes with time.

The change in the car's position from initial position $x_i$ to final position $x_f$ is called displacement, $\Delta \vec x= x_f-x_i$ (in physics we use the Greek letter $\Delta$ to indicate the change in a quantity).

This quantity is a **vector** point from $A$ to $B$ and in $1$-D denoted by $\Delta \vec x=x_B-x_A$.

In the figure below, the car moves from point $A$ at $x=2\, {\rm m}$ and after reaching to $x=9\,{\rm m}$ returns and stops at position $x=6\,{\rm m}$ at point $B$.

Therefore, the car's displacement is $\Delta x=6-2=+4\,{\rm m}$.

Another quantity which sometimes confused with displacement is the **distance traveled** (or simply **distance**) which is defined as the overall distance covered by the particle.

In the above example, the distance from the initial position is computed as follows:

First, calculate the distance to the return point $d_1=x_C-x_A=9-2=7\,{\rm m}$ then from that point ($x_C$) to the final point $x_B$ i.e. $d_2=x_B-x_C=6-9=-3$.

But we should pick the absolute value of it since the distance is a **scalar** quantity and for them, a negative value is non-sense.

Therefore, the total distance covered by our car is $d_{tot}=d_1+|d_2|=7+|-3|=7+3=10\,{\rm m}$ .

In case of several turning points along the straight path or once the motion's path is on a plane or even three-dimensional cases, one should divide the overall path (one, two, or three dimensional) into straight lines (without any turning point), calculate the difference of those initial and final points and then add their *absolute values* of each path to reach to the distance traveled by that particle on that specific path (see examples below).

In more than one dimension, the computations are a bit involved and we need to be armed with additional concepts.

In this section, we can learn how by using vectors one can describe the position of an object, and by manipulating them to characterize the displacement and other related kinematical quantities (like velocity and acceleration).

In a coordinate system, the position of an object is described by a so-called **position vector** that extends from reference origin $O$ to the location of the object $P$ and is denoted by $\vec{r}=\overrightarrow{OP}$.

These vectors, in the Cartesian coordinate system (or other related coordinates), can be expressed as a linear combination of unit vectors, $\hat{i},\hat{j},\hat{k}$, (the ones with unit length) as

\[ \textbf{r}=\sum_1^{n} r_x \hat{i}+r_y \hat{j} +r_z \hat{k} \]

where $n$ denotes the dimension of the problem, i.e., in two and three dimensions $n=2,3$, respectively. $r_x , r_y$ and $r_z$ are called the components of the vector $\vec{r}$.

Now the only thing that remains is adding or subtracting these vectors, known as vector algebra, to provide the kinematical quantities.

To do this, simply add or subtract the terms (components) along a specific axis with each other (as below).

Consider adding of two vector $\vec{a}$ and $\vec{b}$ in two dimension, \begin{array}{cc} \textbf{a}+\textbf{b}&=&\left(a_x \hat{i}+a_y \hat{j} \right)+\left(b_x \hat{i}+b_y \hat{j}\right)\\ &=&\left(a_x+b_x\right)\hat{i}+\left(a_y+b_y\right)\hat{j}\\ &=&c_x\, \hat{i}+c_y\, \hat{j} \end{array}In the last line, the components of the final vector (or resultant vector) are denoted by $c_x$ and $c_y$.

As we learned from vector practice problems that the magnitude and direction of the obtained vector are represented by the following relations \begin{array}{cc} |\textbf{c}| &=& \sqrt{\left(a_x+b_x\right)^2 +\left(a_y+b_y\right)^2 } \qquad \text{magnitude}\\ \theta &=& \tan^{-1} \left(\frac{a_y+b_y}{a_x+b_x}\right) \qquad \text{direction} \end{array} where $\theta$ is the angle with respect to the $x$ axis.

We have two types of problems on the topic of displacement.

In the first case, the initial and final coordinates (position) of an object are given.

Write position vectors for every point. The vector which extends from the tail of the initial point to the tail of the final point is a displacement vector and computed as the difference of those vectors i.e. $\vec{c}=\vec{b}-\vec{a}$.

In the second case, the overall path of an object, between initial and final points, is given as consecutive vectors as the figure below.

Here, one should decompose each vector with respect to its origin, then add components along $x$ and $y$ axes separately.

The displacement vector is the one that points from the tip of the first vector to the tail of the last vector and its magnitude is the vector addition of those vectors i.e. $\vec{d}=\vec{a}+\vec{b}+\vec{c}$.

**Example 1: A moving object is displaced from A(2,-1) to B(-5,3) in a two-dimensional plane. What is the displacement vector of this object?**

**Solution**: First, this is the first case mentioned above. So construct the position vectors of point $A$ and $B$ as below \begin{gather*} \overrightarrow{OA}=2\,\hat{i}+(-1)\,\hat{j}\\ \overrightarrow{OB}=-5\,\hat{i}+3\,\hat{j} \end{gather*} Now, by definition, the difference of initial and final points or simply position vectors gets the displacement vector $\vec{d}$ as \begin{align*}\vec{d} &=\overrightarrow{OB}-\overrightarrow{OA}\\\\ &= \left(-5\,\hat{i}+3\,\hat{j}\right)-\left(2\,\hat{i}+(-1)\,\hat{j}\right)\\\\ &= -7\,\hat{i}+4\,\hat{j} \end{align*} its magnitude and direction is also obtained as follows \begin{align*} |\vec{d}|&=\sqrt{(-7)^2 +4^2} \\\\ &=\sqrt{49+16} \\\\ &\approx 8.06\,{\rm m} \end{align*} and \[\theta = \tan^{-1} \left(\frac{d_y}{d_x}\right)=\tan^{-1}\left(\frac{4}{-7}\right)\] The angle $\theta$ may be $-29.74^\circ$ or $150.25^\circ$ but since $d_x$ is negative and $d_y$ is positive so the resultant vector lies on the second quarter of coordinate system. Therefore, the desired angle with $x$-axis is $150.25^\circ$.

**Example (2): An airplane flies $276.9\,{\rm km}$ $\left[{\rm W}\, 76.70^\circ\, {\rm S}\right]$ from Edmonton to Calgary and then continues $675.1\,{\rm km}$ $\left[{\rm W}\, 11.45^\circ\,{\rm S}\right]$ from Calgary to Vancouver. Using components, calculate the plane's total displacement. (Nelson 12, p. 27). **

**Solution**: In these problems, there is a new thing that appears in many textbooks which is the compact form of direction as stated in brackets. $\left[{\rm W}\, 76.70^\circ\, {\rm S}\right]$ can be read as "point west, and then turn $76.70^\circ$ toward the south".

To solve such practices, first sketch a diagram of all vectors, decompose them, and next using vector algebra, explained above, compute the desired quantity (here $ \vec{d}$).

Two successive paths denoted by vectors $\vec{d_1}$ and $\vec{d_2}$ and in terms of components read as \begin{align*} \vec{d_1} &= |\vec{d_1}|\cos \theta (-\hat{i})+|\vec{d_1}|\sin \theta (-\hat{j}) \\ \vec{d_2} &=|\vec{d_2}|\cos \alpha (-\hat{i})+|\vec{d_2}| \sin \alpha (-\hat{j}) \end{align*} Substituting the numerical values into the above expression, we can find \begin{align*} \vec{d_1} &= 276.9\cos 76.70^\circ (-\hat{i})+276.9\sin 76.7^\circ (-\hat{j}) \\ &= 63.700 (-\hat{i})+269.47 (-\hat{j}) \quad [{\rm km}] \\\\ \vec{d_2} &= 675.1\cos 11.45^\circ (-\hat{i})+675.1\sin 11.45^\circ (-\hat{j}) \\ &= 661.664 (-\hat{i})+134.016(-\hat{j}) \quad [{\rm km}] \end{align*} The total displacement is drawn from the tail of $\vec{d_1}$ to the tip of $\vec{d_2}$. In the language of vector addition $\vec{d}=\vec{d_2}+\vec{d_1}$, so \begin{align*} \vec{d} &= \vec{d_2}+\vec{d_1}\\\\ &= 725.364(-\hat{i})+403.486(-\hat{j}) \quad [{\rm km}] \end{align*} Therefore, the length of total path flied by the airplane from Edmonton to Vancouver and its direction with $x$-axis is \begin{align*} |\vec{d}| &=\sqrt{(725.364)^2 +(403.486)^2}\\\\ &= 830.032\quad {\rm km} \end{align*} and \begin{align*} \gamma &=\tan^{-1} \left(\frac{d_y}{d_x}\right) \\\\ &= \tan^{-1} \left(\frac{403.486}{725.364}\right) \\\\ &= 29.09^\circ \end{align*} As one can see, the resultant vector points to the south west or $\left[{\rm W}\,29.09^{\circ}\,{\rm S}\right]$

Two successive paths denoted by vectors $\vec{d_1}$ and $\vec{d_2}$ and in terms of components read as \begin{align*} \vec{d_1} &= |\vec{d_1}|\cos \theta (-\hat{i})+|\vec{d_1}|\sin \theta (-\hat{j}) \\ \vec{d_2} &=|\vec{d_2}|\cos \alpha (-\hat{i})+|\vec{d_2}| \sin \alpha (-\hat{j}) \end{align*} Substituting the numerical values into the above expression, we can find \begin{align*} \vec{d_1} &= 276.9\cos 76.70^\circ (-\hat{i})+276.9\sin 76.7^\circ (-\hat{j}) \\ &= 63.700 (-\hat{i})+269.47 (-\hat{j}) \quad [{\rm km}] \\\\ \vec{d_2} &= 675.1\cos 11.45^\circ (-\hat{i})+675.1\sin 11.45^\circ (-\hat{j}) \\ &= 661.664 (-\hat{i})+134.016(-\hat{j}) \quad [{\rm km}] \end{align*} The total displacement is drawn from the tail of $\vec{d_1}$ to the tip of $\vec{d_2}$. In the language of vector addition $\vec{d}=\vec{d_2}+\vec{d_1}$, so \begin{align*} \vec{d} &= \vec{d_2}+\vec{d_1}\\\\ &= 725.364(-\hat{i})+403.486(-\hat{j}) \quad [{\rm km}] \end{align*} Therefore, the length of total path flied by the airplane from Edmonton to Vancouver and its direction with $x$-axis is \begin{align*} |\vec{d}| &=\sqrt{(725.364)^2 +(403.486)^2}\\\\ &= 830.032\quad {\rm km} \end{align*} and \begin{align*} \gamma &=\tan^{-1} \left(\frac{d_y}{d_x}\right) \\\\ &= \tan^{-1} \left(\frac{403.486}{725.364}\right) \\\\ &= 29.09^\circ \end{align*} As one can see, the resultant vector points to the south west or $\left[{\rm W}\,29.09^{\circ}\,{\rm S}\right]$

**Example (3): A moving particle moves over the surface of a solid cube in such a way that passes through $A$ to $B$. What is the magnitude of the displacement vector in this change of location of the particle?**

**Solution**: In three-dimensional cases like $2-D$ ones, we should only know the location(coordinates) of the object, then use the following relations to obtain the displacement of a moving particle.Points $A$ and $B$ lies on the $x-z$ plane and $y$ axis, respectively so their coordinates are $(10,0,10)$ and $(0,10,0)$ which parenthesis denote the $(x,y,z)$. This is type one of the problems. \begin{align*} \overrightarrow{OA}&= 10\,\hat{i}+0\,\hat{j}+10\,\hat{k} \\ \overrightarrow{OB} &= 0\,\hat{i}+10\,\hat{j}+0\,\hat{k} \end{align*} and displacement vector is \begin{align*} \vec{d} &=\overrightarrow{OB}-\overrightarrow{OA} \\ &= -10\,\hat{i}+10\,\hat{j}-10\,\hat{k} \end{align*} Therefore, the wanted vector in terms of its components was computed as above. Also, its magnitude is the square root of the sum of the squares of each component ,i.e., \[|\vec{d}|=\sqrt{(-10)^2 +(-10)^2 + (10)^2}=10\sqrt{3}\rm m\]

**Example (4): A car moves around a circle of radius of $20\,{\rm m}$ and returns to its starting point. What is the distance and displacement of the car? ($\pi = 3$)**

**Solution**: As mentioned above, displacement depends on the initial and final points of the motion. Because the car returns to its initial position so no displacement is made by the car. But the amount of distance traveled is simply the perimeter of the circle (since this scalar quantity depends on the form of the path). So $d=2 \pi r=2 \times 3 \times 20 =120\,{\rm m}$, where $r$ is radius of the circle.

**Example (5): A moving object is covering a square path -with one end left open- as shown in the figure below. What is the desired displacement and distance traveled between the specified points? point $p$ lies in the middle of $BC$.**

**Solution: **Displacement is the shortest and straightest line between initial and final points.

So using the Pythagorean theorem, we get \begin{align*} D^2 &= (iB)^2 +(Bf)^2 \\\\ &= 5^2+(2.5)^2 \\\\ \Rightarrow D &= \sqrt{25 + 6.25} \\\\ &\approx 5.6\quad (\rm m) \end{align*} The direction of displacement vector is also obtained as \begin{align*} \tan \theta &= \frac{Bp}{iB} \\\\ &= \frac{2.5}{5} \\\\ \Rightarrow \theta &= \arctan \frac{2.5}{5} \\\\ &= 26.5^\circ\, \left[\text{South east}\right] \end{align*} Distance is simply the perimeter of the path traveled, so \begin{align*} \text{distance} &=5 + 2.5 \\ &=7.5\quad \rm m \end{align*}

**Example (6): you walk once around an oval track for 6.0 minutes at an average speed of 1.7 m/s.**

**(a) What is the distance traveled?**

**(b) What is the displacement? **

**Solution**:(a) here, the distance covered by the person is the perimeter of the ellipse. Since the time elapsed and average speed are given, we can use the definition of average speed and solve for the distance traveled to find it as below \begin{align*} \text{average speed}&=\frac{\text{distance}}{\text{time interval}}\\1.7&=\frac{d}{6\times 60\,{\rm s}}\\ &\\ \Rightarrow d&=\big(1.7\,{\rm \frac ms}\big)(360\,{\rm s})\\&=612\quad {\rm m} \end{align*}

**(b)** Displacement is the difference between initial and final positions. In this problem, since the person had returned to his original position so his * displacement is zero*.

**Example (7): The velocity versus time graph of a car is shown below. Find the displacement in 4 seconds.**

**Solution**: The area under a velocity-time graph in a given time interval indicates the displacement in that interval.

In this case, the bounded area is a triangle whose base is 5 units and height is 2 units. So, its area is \begin{align*} \text{area=displacement}&=\frac 12 \times base\times height\\\\&=\frac 12 \times 5\times 2\\\\&=5\quad {\rm m}\end{align*} Thus, this car displaces 5 meters in 5 s.

As you can see in the examples of displacement in above, displacement is a vector that depends only on the initial and final positions of the object, and not on the details of the motion and path.

These vector quantities require both a length and a direction to be identified.

It is also possible to find the displacement vector using a position vs. time graph.

On the contrary, distance is a quantity that is characterized only by a simple value, which is called **scalar,** and is path dependence.

In general, the distance traveled and the magnitude of the displacement vector between two points is not the same.

If the moving object changes its direction in the course of travel, then the total distance traveled is greater than the magnitude of the displacement between those points.

The SI units of both quantities are meters.

Distance and displacement are frequently used in solving velocity and acceleration problems.

**Author**: Ali Nemati

**Date Published:** 26 July 2017

**Last Update**: 8-28-2021

© 2015 All rights reserved. by Physexams.com