Explanation:
Given that,
She walks in east,
Speed = 0.80 m/s
Time = 4.0 min
In north,
Speed = 0.50 m/s
Time = 5.5 min
In west,
Speed = 1.1 m/s
Time = 2.8 min
(a). We need to calculate the unit-vector velocities for each of the legs of her journey.
The velocity of her in east
[tex]\vec{v_{1}}=0.80\ \hat{x}\ m/s[/tex]
[tex]\vec{v_{2}}=0.50\ \hat{y}\ m/s[/tex]
[tex]\vec{v_{3}}=1.1\ \hat{-x}\ m/s[/tex]
(b). We need to calculate the unit-vector displacements for each of the legs of her journey
Using formula of displacement
[tex]\vec{d_{1}}=v_{1}\times t_{1}[/tex]
In east ,
[tex]\vec{d_{1}}=0.80\times4.0\times60[/tex]
[tex]\vec{d_{1}}=192\ \hat{x}\ m[/tex]
In north,
[tex]\vec{d_{2}}=0.50\times5.5\times60[/tex]
[tex]\vec{d_{2}}=165\ \hat{y}\ m[/tex]
In west,
[tex]\vec{d_{3}}=1.1\times2.8\times60[/tex]
[tex]\vec{d_{3}}=184.8\ \hat{-x}\ m[/tex]
(c). We need to calculate the net displacement from the postal truck after her journey is complete
[tex]\vec{d}=\vec{d_{1}}+\vec{d_{2}}+\vec{d_{3}}[/tex]
Put the value in the formula
[tex]\vec{d}=192\hat{x}+165\hat{y}+184.8\hat{-x}[/tex]
[tex]\vec{d}=7.2\hat{x}+165\hat{y}[/tex]
We need to calculate the magnitude of the displacement
[tex]d=\sqrt{(7.2)^2+(165)^2}[/tex]
[tex]d=165.16\ m[/tex]
The magnitude of the displacement is 165.16 m.
Hence, This is the required solution.