Below are step-by-step solutions using coordinates and the Pythagorean theorem.
Problem 1
W swims: 60 miles north, 30 miles east, 30 miles north, 150 miles west. Find how far W is from the starting point.
- Choose a coordinate system where the start is at (0,0). Let north be +y and east be +x.
- Compute net displacement in each direction:
- Net north (y): 60 + 30 = 90 miles.
- Net east (x): 30 - 150 = -120 miles (which means 120 miles west).
- Distance from start is the magnitude of the displacement vector (x,y):
Distance = sqrt(x^2 + y^2) = sqrt((-120)^2 + 90^2) = sqrt(14400 + 8100) = sqrt(22500) = 150 miles.
Answer (1): 150 miles.
Problem 2
Given: A is 50 m east of B and 30 m west of C. D is 60 m east of C, and 40 m east of E. F is 50 m north of E and 80 m north of G. Find distance between B and G to the nearest tenth of a meter.
- Place B at the origin: B = (0,0). Let east be +x and north be +y.
- Find x-coordinates step by step:
- 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 = (50 + 30, 0) = (80,0).
- D is 60 m east of C: D = (80 + 60, 0) = (140,0).
- D is 40 m east of E, which means E is 40 m west of D: E = (140 - 40, 0) = (100,0).
- Now use the vertical positions:
- F is 50 m north of E: F = (100, 50).
- F is 80 m north of G, so G is 80 m south of F: G = (100, 50 - 80) = (100, -30).
- So B = (0,0) and G = (100, -30). Distance BG = sqrt((100 - 0)^2 + (-30 - 0)^2) = sqrt(100^2 + 30^2) = sqrt(10000 + 900) = sqrt(10900).
Compute sqrt(10900) ≈ 104.403065... Rounding to the nearest tenth gives 104.4 m.
Answer (2): 104.4 meters.