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.

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.

The closest approach is 59 m.

The answer is a i think