Strategy (draw a picture): Put the starting point on a simple coordinate grid, let east be +x and north be +y, mark each move or each point from that baseline, then use the Pythagorean theorem (distance = sqrt(dx^2 + dy^2)).
Problem 1
W swims: 60 mi north, 30 mi east, 30 mi north, 150 mi west. Draw a path or combine the north/south and east/west moves:
- North/south total: 60 + 30 = 90 miles north.
- East/west total: 30 east then 150 west = 30 - 150 = -120 miles (which means 120 miles west).
So W's final position relative to the start is 90 miles north and 120 miles west. Treat those as the legs of a right triangle:
Distance = sqrt((90)^2 + (120)^2) = sqrt(8100 + 14400) = sqrt(22500) = 150 miles.
Answer to Problem 1: 150 miles from the starting point.
Problem 2
We place B at the origin (0,0). Take east as +x and north as +y. Points given only by east/west are on y = 0; points given north/south off that line share the x of the point they are north/south of.
From the text:
- A is 50 m east of B: A = (50, 0).
- A is 30 m west of C, so C is 30 m east of A: C = (80, 0).
- D is 60 m east of C: D = (140, 0).
- D is 40 m east of E, so E = D - 40 in x: E = (100, 0).
- F is 50 m north of E, so F = (100, 50).
- F is 80 m north of G, so G is 80 m south of F: G = (100, 50 - 80) = (100, -30).
We need the distance between B and G. B = (0,0), G = (100, -30). Compute the horizontal and vertical differences:
dx = 100 - 0 = 100 meters
dy = -30 - 0 = -30 meters (use the magnitude 30 for the triangle leg)
Distance BG = sqrt(100^2 + (-30)^2) = sqrt(10000 + 900) = sqrt(10900) ≈ 104.403064... m.
Round to the nearest tenth: 104.4 m.
Answer to Problem 2: 104.4 meters (to the nearest tenth).
Connection to standards: this uses A-CED.1 (translate directions into coordinates/equations) and A-SSE.1 thinking about structure of the displacement (split into orthogonal components) and then the Pythagorean theorem to compute the magnitude.