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You are hungry and decide to go to your favorite neighborhood fast-food restaurant. You leave your apartment and take the elevator 10 flights down (each flight is 3.0 m) and then go 15 m south to the apartment exit. You then proceed 0.2 km east, turn north, and go 0.1 km to the entrance of the restaurant.Part ADetermine the displacement from your apartment to the restaurant. Use unit vector notation for your answer, being sure to make clear your choice of coordinates.Take the beginning of the journey as the origin, with north being the y-direction, east the x-direction, and the z-axis vertical.Part BHow far did you travel along the path you took from your apartment to the restaurant?Part CWhat is the magnitude of the displacement you calculated in part A?

Respuesta :

Answer:

A. (200i + 85j - 30k)m

B.  345m

C. 219.37 m

Explanation:

UP: +VE Z

NORTH: +VE Y

EAST: +VE X

A.

DISPLACEMENT : (0.2*1000 i + (-15+0.1*1000)j + (-3*10)k) m

                            = (200i + 85j - 30k)m

B. total distance = 10*3+15+200+100 = 345m

C.

Magnitude = [tex]\sqrt{{200^{2} } +85^{2} +(-30)^{2}[/tex] = 219.37 m

unit vector = [tex]\frac{200i+85j-30k}{219.4}[/tex]

pristina

The displacement from the apartment to the restaurant is given by [tex]\Delta x =200\hat{i}+85\hat{j}-30\hat{k}[/tex].

The distance from the apartment to the restaurant is 345 m

The magnitude of the displacement from the apartment to the restaurant is 219.37 m.

Given that;

  • East is chosen as the X direction,
  • North is chosen as the Y direction,
  • Vertical is chosen as Z direction.

Part A

  • The initial position can be identified as the origin, ie; [tex]x_i= 0 \hat{i}+0 \hat{j}+0 \hat{k}[/tex]

To find the final position, let us find the position in each direction;

  • In X direction, [tex]x_1 = 0.2 \,km= 0.2 m = 200 \hat {i}[/tex]
  • In Y direction, [tex]y_1 = -15\,m +0.1\,km=-15\,m+100\,m = 85 \,m = 85\hat{j}[/tex]
  • In Z direction, [tex]z_1 = 3\times -10\,m = -30\,m = -30\hat{k}[/tex]
  • Therefore, the final position is [tex]x_f = x_1+y_1+z_1=200\hat{i}+85\hat{j}-30\hat{k}[/tex]
  • We know that displacement is the change in position.
  • ie; displacement [tex]\Delta x = x_f - x_i =200\hat{i}+85\hat{j}-30\hat{k}[/tex]

Part B

Here, we are asked to find the total distance covered.

For that, we add all the distances irrespective of the direction.

  • Distance (d) = vertical distance + south distance + east distance + north distance.
  • [tex]d=(3\times 10 m)+ 15\,m + 200\,m + 100\,m = 345\,m[/tex]

Part C

To find the magnitude of the displacement,

  • we know that the magnitude of a vector [tex]a\hat{i}+b\hat{j}+c\hat{k}[/tex] is [tex]\sqrt{(a^2 +b^2 + c^2)}[/tex]
  • Therefore the magnitude of displacement,
  • [tex]|\Delta x|= \sqrt{(200^2+85^2+30^2)} =219.37\,m[/tex]

Learn more about distance and displacement here:

https://brainly.com/question/20115051?referrer=searchResults

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