Question

A person walks in the following pattern:$\mathbf{3}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{km}$north, then$\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{km}$west, and finallyrole="math" localid="1656154713208" $\mathbf{5}\mathbf{.}\mathbf{2}\mathbf{}\mathit{k}\mathit{m}$south. (a) Sketch the vector diagram that represents this motion. (b) How far and (c) in what direction would a bird fly in a straight line from the same starting point to the same final point?

Short answer

a) Resultant displacement of the flying bird from starting point to the final point is$3.2\mathrm{km}$

b) The direction of the flying bird is$221°$

Step-by-step solution

To understand the concept and draw the vector diagram.

The problem involves vector algebra. In vector algebra, various algebraic operations can be performed on vectors. To understand this concept, a vector algebraic diagram can be drawn in which the motion is shown. In this diagram, all the vectors in positive x-direction are added and subtracted the vectors which are in negative x-direction. Similarly, all the vectors in positive y-direction are added and subtracted the vectors which are in negative y-direction. Thus, we would get two components of the resultant vectors, one along x and other along y (it may be positive or negative). Using the Pythagoras theorem, the resultant vectors can be found. Using trigonometry, the angle made by the resultant vector with the x-axis cab be calculated.

Given,

$$\begin{gathered} A=\left(3.1\mathrm{km}\right)\stackrel{\rightharpoonup }{\mathrm{j}} \\ \mathrm{B}=-\left(2.4\mathrm{km}\right)\stackrel{\rightharpoonup }{\mathrm{i}} \\ \mathrm{C}=-\left(5.2\mathrm{km}\right)\stackrel{\rightharpoonup }{\mathrm{j}} \end{gathered}$$

The resultant vector can be written as

$R=\sqrt{A^2+B^2}$

Vector diagram can be drawn as follows.

In the above figure, as the person moves first 3.1 km in the north direction, it is shown in the positive y- direction. Then, the person moves in the west 2.4 km, so it is shown in the negative x- direction, and finally, he is moving in the south 5.2 km, so it is shown in the negative y- direction as shown in the figure.

To find the resultant displacement of the flying bird

As A and C are parallel but opposite in direction, resultant of those two, D can be found as follows:

$$\begin{gathered} D=A+C \\ D=\left(3.0\mathrm{km}\right)\stackrel{\rightharpoonup }{\mathrm{j}}-\left(5.2\mathrm{km}\right)\stackrel{\rightharpoonup }{\mathrm{j}} \\ \mathrm{d}=-\left(2.1\mathrm{km}\right)\stackrel{\rightharpoonup }{\mathrm{j}} \end{gathered}$$

The negative sign implies that the resultant of A and C is along the negative y-direction along the south.

Vector B is along the negative x- direction.

Now, the resultant of D and B is given by,

$R=\sqrt{D^2+B^2}$

$$\begin{gathered} R=\sqrt{{\left(-2.1\right)}^2+{\left(-2.4\right)}^2} \\ R=3.2\mathrm{km} \end{gathered}$$

To find the direction of the flying bird

Direction of the resultant is given by,

$$\begin{gathered} \tan \theta =\frac{D}{B} \\ \tan \theta =\frac{-2.1}{-2.4} \end{gathered}$$

So,

$\theta =41°\mathrm{or}221°$

Value of$\theta$can be, $41°\mathrm{or}{221}^{°}$but as we can see in the vector diagram, the resultant is in the third quadrant, $41°$so cannot be correct. Hence, the angle is$221°$ with the positive x axis or with the east direction.