Respuesta :
Answer:633 m
Explanation:
First we have moved 300 m in North
let say it as point a and its vector is [tex]300\hat{j}[/tex]
after that we have moved 500 m northeast
let say it as point b
therefore position of b with respect to a is
r[tex]_{ba}=500cos(45)\hat{i}+500sin(45)\hat{j}[/tex]
Therefore position of b w.r.t to origin is
[tex]r_b=r_a+r_{ba}[/tex]
[tex]r_b=300\hat{j}+500cos(45)\hat{i}+500sin(45)\hat{j}[/tex]
[tex]r_b=500cos(45)\hat{i}+\left [ 250\sqrt{2}+300\right ]\hat{j}[/tex]
after this we moved 400 m [tex]60^{\circ}[/tex] south of east i.e. [tex]60^{\circ}[/tex] below from positive x axis
let say it as c
[tex]r_{cb}=400cos(60)\hat{i}-400sin(60)\hat{j}[/tex]
[tex]r_c=r_{b}+r_{cb}[/tex]
[tex]r_c=500cos(45)\hat{i}+\left [ 250\sqrt{2}+300\right ]\hat{j}+400cos(60)\hat{i}-400sin(60)\hat{j}[/tex]
[tex]r_c=\left [ 250\sqrt{2}+200\right ]\hat{i}+\left [ 250\sqrt{2}+300-200\sqrt{3}\right ]\hat{j}[/tex]
magnitude is [tex]\sqrt{\left [ 250\sqrt{2}+200\right ]^2+\left [ 250\sqrt{2}+300-200\sqrt{3}\right ]^2}[/tex]
=633.052
for direction[tex]tan\theta =\frac{250\sqrt{2}+300-200\sqrt{3}}{250\sqrt{2}+200}[/tex]
[tex]tan\theta =\frac{307.139}{553.553}[/tex]
[tex]\theta =29.021^{\circ}[/tex] with x -axis
