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.

Answer: The closest approach is 59 m.