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.
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.
