An object is moving linearly according to (x, y, z) = (100-4t, 60+2t, 60-4t). For what values of t is it 120m away from (0,0,0)? What is the closest it comes to (0,0,0)?

Question

t = time in minutes; x, y, and z are measured in meters

Answer 1

A) 120² = (100 - 4t)² + (60 + 2t)² + (60 - 4t)²; 14400 = 10000 - 800t + 16t² + 3600 + 240t + 4t² + 3600 + 2800 -

480t + 16t²; 36t² - 1040t + 2800 = 0; 9t² - 260t + 700 = 0; t = (130 ± √16900 - 6300)/9 = (130 ± 102.96)/9;

t = 3 s or t = 25.88 s

B) Distance square f(t) = 36t² - 1040t + 17200; f'(t) = 72t - 1040 = 0; t = 1040/72 = 130/9;

Closest distance d = √f(130/9) = 98.43 m