Answer:
a) magnitude of direction = 387.92m
b)Angle of displacement = 25.54°
c) Total distance=415m
Explanation:
Let +x axis be east
Let +y axis be north
d1=250<(90-48)= 250<42°
d2=165<0°
Remember all angles are measured anti clockwise in +axis
total =(250cos42° + 165cos0°)I + (250sin42°)j
dtotal= 350i (m) + 167.28j(m)
/dtotal/ =sqrt(350^2 + 167.28^2)
/dtotal/= sqrt150482.59
/dtotal/= 387.92m
b) Theta = Tan^-1(167.28/350)
Theta = tan^-1 0.5779
Theta= 25.54°
c)Total distance covered by the woman =250+165= 415m
Answer:
(a) [tex]Magnitude=380.69[/tex]
(b) [tex]Angle=29.21^{o}[/tex]
(c) [tex]Distance=415m[/tex]
Explanation:
Given data
The Woman walks 250m in 48° east of north
Then 165m due to east
Let east be the +x axis and the north is +y direction
So the two displacements is written in polar form as:
d₁=250∠(90°-48°)=250∠42°
d₂=165∠0°
All angles are measured in counterclockwise from +x direction
So the total displacement is given dot product as:
[tex]d_{dot}=d_{1}.d_{2}\\d_{dot}=(250Cos(48^{o} )+165Cos(0_{o} ))i+(250Sin(48^{o} )j) \\d_{dot}=332.28(m)i+185.786(m)j[/tex]
Part (a)
The magnitude of women total displacement
[tex]|d_{dot} |=\sqrt{(332.28)^{2} +(185.786)^{2} } \\|d_{dot} |=380.69[/tex]
Part (b)
The angle of woman displacement from her starting point is
[tex]\alpha =tan^{-1}(\frac{185.786}{332.28} )\\\alpha =29.21^{o}[/tex]
Part (c)
The Total distance that the women walks is given as:
[tex]distance=250m+165m\\distance=415m[/tex]
