Question

The x-coordinates of two objects moving along the x-axis are given as a function of time (t). x1= (4m/s)t x2= -(161m) + (48m/s)t - (4 m/s^2)t^2 Calculate the magnitude of the distance of closest approach of the two objects. x1 and x2 never have the same value.

Answer 1

The two displacement functions are
x₁ = 4t
x₂ = -161 + 48t - 4t²
where
x₁, x₂ are in meters
t is time, s

The distance between the two objects is
x = x₁ - x₂
   =  4t + 161 - 48t + 4t²
x = 4t² - 44t + 161

Write this equation in the standard form for a parabola.
x = 4[t² - 11t] + 161
  = 4[ (t - 5.5)² - 5.5² ] + 161
 x = 4(t-5)² + 40

Ths is a parabola that faces up and has its vertex (lowest point) at (5, 40).
Therefore the closest approach of the two objects is 40 m.
The graph of x versus t confirms the result.

Answer: The distance of the closest approach is 40 m.