Answer:
(a) 34.47 cm
(b) [tex]24.09^\circ[/tex] south of west
Explanation:
Let us draw a figure representing the individual displacement vectors in the four jumps as shown in the figure attached with this solution.
Now, let us try to write the four displacement vectors in in terms of unit vectors along the horizontal and the vertical axis.
[tex]\vec{d}_1= 31\ cm\ west = -31\ cm\ \hat{i}\\\vec{d}_2= 26\ cm\ south\ of\ west = -26\cos 44^\circ\ \hat{i} -26 \sin 44^\circ\ \hat{j}=(-18.72\ \hat{i}-18.06\ \hat{i})\ cm\\\vec{d}_3= 22\ cm\ south\ of\ east = 22\cos 56^\circ\ \hat{i} -22 \sin 56^\circ\ \hat{j}=(12.30\ \hat{i}-18.23\ \hat{i})\ cm\\\vec{d}_4= 23\ cm\ north\ of\ east = 23\cos 75^\circ\ \hat{i} +23\sin \sin 75^\circ\ \hat{j}=(5.95\ \hat{i}+22.22\ \hat{i})\ cm\\[/tex]
Now, the vector sum of all these vector will give the resultant displacement vector.
[tex]\vec{D} = \vec{d}_1+\vec{d}_2+\vec{d}_3+\vec{d}_4\\\Rightarrow \vec{D} = -31\ cm\ \hat{i}+(-18.72\ \hat{i}-18.06\ \hat{i})\ cm+(12.30\ \hat{i}-18.23\ \hat{i})\ cm+(5.95\ \hat{i}+22.22\ \hat{i})\ cm\\\Rightarrow \vec{D} =(-31.47\ \hat{i}-14.07\ \hat{i})\ cm[/tex]
Part (a):
The magnitude of the resultant displacement vector is given by:
[tex]D=\sqrt{(-31.47)^2+(-14.07)^2}\ m = 34.47\ m[/tex]
Part (b):
Since the resultant displacement vector indicates that the final position of the vector lies in the third quadrant, the vector will make some positive angle in the direction south of west given by:
[tex]\theta = \tan^{-1}(\dfrac{14.07}{31.47})= 24.09^\circ[/tex]
