Question

The x-coordinates of two objects moving along the x-axis are given as a function of time t. x1 = (4 m/s)t and x2 = −(159 m) + (24 m/s)t − (1 m/s2)t2 . Calculate the magnitude of the distance of closest approach of the two objects. x1 and x2 never have the same value.

Answer 1

The x-coordinate of the first object is
x₁(t) = (4 m/s)t

The x-coordinate of the second object is
x₂(t) = -(159 m) + (24 m/s)t - (1 m/s²)t²

The distance between the two objects is
x(t) = x₂ - x₁ 
      =  - 159 + 24t - t² - 4t
      = -t² + 20t  - 159

Write this equation in the standard form for a parabola.
x = -[t² - 20t] - 159
   = -[(t - 10)² - 100] - 159
   = -(t-10)² - 59

This parabola has a vertex at (-10,  -59), and it is downward.
Because the maximum value of x is negative, the two objects never touch
The closest distance between the objects is 59 m.

The two graphs confirm that the analysis is correct.

Answer: The closest approach is 59 m.