The question is incomplete. Here is the complete question.
The map (in the attachment) shows Olivia's trip to the coffee shop. She gets on her bike at Loomis and then rides south 0.9mi to Broadway. She turns east onto Broadway, rides 0.8mi to where Broadway turns, and then continues another 1.4mi to the shop.
What is the magnitude of the total displacement of her trip?
Whta is the direction of the total displacement of her trip?
Answer: Magnitude = 2.6mi
Direction: 54.65° east
Explanation: Displacement is the change in postition of a moving object.
There are a few ways to determine total displacement. For this case, the Perpendicular Components of a Vector method will be used.
For this method, total displacement is given by:
[tex]\Delta d_{t}=\sqrt{(\Delta d_{x})^{2}+(\Delta d_{y})^{2}}[/tex]
[tex]\theta=tan^{-1}(\frac{\Delta d_{y}}{\Delta d_{x}})[/tex]
[tex]\Delta d_{x}[/tex] is the x-component of total displacement and it is the sum of each individual x-components;
[tex]\Delta d_{y}[/tex] is the y-component of total displacement and it is the sum of each individual y-components;
θ is the angle the resulting displacement;
For Olivia's trip, there are no x-component of the first part and for the third part, the path she bikes is a hypotenuse of a right triangle. So, that right triangle's x-component is:
[tex]sin30=\frac{x}{1.4}[/tex]
[tex]\frac{1}{2} =\frac{x}{1.4}[/tex]
x = 0.7
Then,
[tex]\Delta d_{x}[/tex] = 0 + 0.8 + 0.7
[tex]\Delta d_{x}[/tex] = 1.5
Related to y, there are no y-component in the second part of Olivia's trip and for the third part:
[tex]cos30=\frac{y}{1.4}[/tex]
[tex]\frac{\sqrt{3} }{2} =\frac{y}{1.4}[/tex]
y = 1.21
Then,
[tex]\Delta d_{y}[/tex] = 0.9 + 0 + 1.21
[tex]\Delta d_{y}[/tex] = 2.11
Total displacement is
[tex]\Delta d_{t}=\sqrt{(1.5)^{2}+(2.11)^{2}}[/tex]
[tex]\Delta d_{t}=\sqrt{6.7021}[/tex]
[tex]\Delta d_{t}=[/tex] 2.6
Magnitude of Olivia's total displacement is 2.6mi
On the map, joining the initial and final points gives a vector pointing towards east at angle:
[tex]\theta=tan^{-1}(\frac{2.11}{1.5})[/tex]
[tex]\theta=tan^{-1}(1.41)[/tex]
θ = 54.65°
Direction of total displacement is 54.65° East.
