what does 1N equal? It happens to be one of the questions on my "Newton's Laws of Motion Exercises" paper.
1 answer:
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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.
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.