Please help me
A person starts walking from home and walks: 6 miles East 6 miles Southeast 3 miles South 5 miles Southwest 2 miles East This person has walked a total of 22Correct miles Find the total displacement vector for this walk: If this person walked straight home, they'd have to walk miles

Respuesta :

Answer:

1) The total displacement vector is ((16 + √2)/2, -(6+11·√2)/2)

2) The number of miles they'd have to walk is approximately 13.856 miles

Step-by-step explanation:

1) The distance, direction, and location of the path of the walk the person takes,  are listed as follows;

Start location, (0, 0)

6 miles East walk to location, (6, 0)

6 miles Southeast to location, (6 + 3·√2, -3·√2)

3 miles South to location, (6 + 3·√2, -3·√2 - 3)

5 miles Southwest to location, (6 + 3·√2 - 2.5·√2, -3·√2 - 3 - 2.5·√2)

2 miles East to location, (6 + 3·√2 - 2.5·√2 + 2, -3·√2 - 3 - 2.5·√2)

(6 + 3·√2 - 2.5·√2 + 2, -3·√2 - 3 - 2.5·√2) = ((16 + √2)/2, -(6+11·√2)/2)

Therefore the destination coordinates is ((16 + √2)/2, -(6+11·√2)/2)

The total displacement vector, [tex]\underset{d}{\rightarrow}[/tex] = ((16 + √2)/2, -(6+11·√2)/2)

d =  (16 + √2)/2)·i - (6+11·√2)/2)·j

2) If the person walked straight home, the number of miles they'd have to walk, [tex]\left | \underset{d}{\rightarrow} \right |[/tex], is given as follows;

[tex]\left | \underset{d}{\rightarrow} \right | = \sqrt{\left(\dfrac{16 +\sqrt{2} }{2} \right)^2 + \left(-\dfrac{6 + 11 \cdot \sqrt{2} }{2} \right)^2 } = \sqrt{134 + 41 \cdot \sqrt{2} }[/tex]

Therefore;

If the person walked straight home, the number of miles they'd have to walk [tex]\left | \underset{d}{\rightarrow} \right | \approx 13.856 \ miles[/tex]