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Displacement and Distance

Displacement and Distance
Category : Straight Line Motion

The motion of objects in one dimension is described in a problem-solution based approach.

 


Velocity and acceleration
Displacement and Distance

Displacement and Distance

Displacement in one dimension

Let's explain this concept with an example. We want to study a car's motion along a straight line. Let the car be a point particle and moves in one dimension. To specify the location of the particle in one dimension we need only one axis that we called it $x$ and lies along the straight line path. First, we must define an important quantity that the other kinematic 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, $\Delta x= x_f-x_i$ (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 $\Delta x=x_B-x_A$. 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 $\Delta x=6-2=+4\,{\rm m}$.

Displacement along a straight line


Another quantity which sometimes confused with displacement is the distance traveled (or simply distance) and defined as the overall distance covered by the particle. In our 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 distance is a scalar quantity and for them, a negative value is nonsense. Therefore, the total distance covered by our car is $d_{tot}=d_1+|d_2|=7+|-3|=7+3=10\,{\rm m}$ . In fact, if there are several turning points along the straight path or once the motion's path to be 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 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).

Displacement in two and three dimension

In more than one-dimension, the computations are a bit involved and we need to 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 of them to characterize the displacement and other related kinematical quantities (like velocity and acceleration). 

In a coordinate system, the position of an object is describe by a so-called position vector which is extends from reference origin $O$ to the location of object $P$ and denoted by $\vec{r}=\vec{OP}$. This vectors, in 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 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 $\vec{r}$.
Illustration of position vector in two-dimension in terms of components

Now the only thing remains is adding or subtracting of 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{eqnarray*}
 \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{eqnarray*}
in the last line, the components of the final vector (or resultant vector) is denoted by $c_x$ and $c_y$.

The magnitude and direction of the obtained vector is represented by the following relations
\begin{eqnarray*}
|\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{eqnarray*}

where $\theta$ is the angle with respect to the $x$ axis.

We have two types of problems in the topic of displacement. In the first case, initial and final coordinates (position) of an object are given. Write position vectors for every point. The vector which extends from the tail of initial point to the tail of final point is displacement vector and computed as the difference of those vectors i.e. $\vec{c}=\vec{b}-\vec{a}$.

Position vectors of two point particle

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. 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 addition vector of those vectors i.e. $\vec{d}=\vec{a}+\vec{b}+\vec{c}$. 

Vector addition of three consecutive vector

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 which mentioned above. So construct the position vectors of point $A$ and $B$ as below
\begin{eqnarray*}
\overrightarrow{OA}&=&2\,\hat{i}+(-1)\,\hat{j}\\
\overrightarrow{OB}&=&-5\,\hat{i}+3\,\hat{j}
\end{eqnarray*}
Now, by definition, the difference of initial and final points or simply position vectors gets the displacement vector $\Delta\vec{d}$ as
\begin{eqnarray*}
\Delta\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{eqnarray*}
its magnitude and direction is also obtained as follows
\begin{eqnarray*}
|\Delta\vec{d} |&=&\sqrt{\left(-7\right)^2 +\left(4\right)^2 }\\
&=&\sqrt{49+16}\approx 8.06\,{\rm m}
\end{eqnarray*}
\[\theta = \tan^{-1} \left(\frac{d_y}{d_x}\right)=\tan^{-1}\left(\frac{4}{-7}\right)\]
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 which appears in many Textbooks that 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 $70^\circ$ toward the south".

Total displacement of fly of airplane

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 $\Delta \vec{d}$).
Vectors $\vec{d_1}$ and $\vec{d_2}$ in terms of components read as
\begin{eqnarray*}
\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{eqnarray*}
putting numbers in the above, one obtain 
\begin{eqnarray*}
\vec{d_1} &=& 276.9\,\cos 76.70^{\circ} \, (-\hat{i})+276.9\,\sin 76.7^{\circ} \, (-\hat{j})\\
                    &=& 63.700\,\left(-\hat{i}\right)+269.47\,\left(-\hat{j}\right) \qquad [{\rm km}]\\
\vec{d_2} &=& 675.1\,\cos 11.45^{\circ} \, (-\hat{i})+675.1\,\sin 11.45^{\circ} \, (-\hat{j})\\
                    &=& 661.664\,\left(-\hat{i}\right)+134.016\,\left(-\hat{j}\right) \qquad [{\rm km}]
\end{eqnarray*}
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{eqnarray*}
\Delta \vec{d} &=& \vec{d_2}+\vec{d_1}\\
                             &=& \left(661.664+63.700 \right)\,\left(-\hat{i}\right)+\left(134.016+269.47 \right)\,\left(-\hat{j}\right)\\
                             &=& \left(725.364 \right)\,\left(-\hat{i}\right)+\left(403.486 \right)\,\left(-\hat{j}\right) \qquad [{\rm km}]
\end{eqnarray*}
Therefore, the length of total path flied by the airplane from Edmonton to Vancouver and its direction with $x$-axis is 
\begin{eqnarray*}
|\Delta \vec{d}| &=& \sqrt{\left(725.364 \right)^2 +\left(403.486 \right)^2 }\\
                    &=& 830.032\,{\rm km}
\end{eqnarray*}
\begin{eqnarray*}
\gamma &=& \tan^{-1} \left(\frac{d_y}{d_x}\right)\\
&=& \tan^{-1} \left(\frac{403.486}{725.364}\right)\\
&=& 29.09^\circ
\end{eqnarray*}
A 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 passes through $A$ to $B$. What is the magnitude of displacement vector in this change of location of the particle?

 

Displacement on cube

Solution:
In three-dimensional cases as $2-D$ ones, we should only know the location(coordinates) of the object and then use the following relations, one can obtain the displacement of a moving particle on any dimensions.
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{eqnarray*}
\overrightarrow{OA}&=& 10\,\hat{i}+0\,\hat{j}+10\,\hat{k}\\
\overrightarrow{OB}&=& 0\,\hat{i}+10\,\hat{j}+0\,\hat{k}
\end{eqnarray*}
\begin{eqnarray*}
\Delta \vec{d} &=&\overrightarrow{OB}-\overrightarrow{OA}\\
&=& -10\,\hat{i}+10\,\hat{j}-10\,\hat{k}
\end{eqnarray*}
Therefore, the desired vector in terms of its components computed as above. Also its magnitude is the square root of sum of the squares of each components.
\[
|\Delta \vec{d}|=\sqrt{(-10)^2 +(-10)^2 + (10)^2 }=10\sqrt{3}
\].

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. since car returns to its initial position so in fact 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.

In summary, Displacement is a vector that depends only on initial and final positions of the particle and not on the details of the motion and path. These vector quantities, require both a length and a direction to be identified. On the contrary, distance is a quantity which 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.