Mathematics 2009 JAMB Past Questions - Question 39: The distance traveled by a particle from a fixed point is given as s=(t³ - t² - t + 5)cm.find the minimum distance that the particle can cover from the fixed point?

Correct Answer

B

Explanation

s = t³ - t² -t + 5 ds/dt = 3t² -2t -1(3t² - 3t) + (t -1) =0 3t(t -1) +1( t – 1) = 0(3t X 1) (t-1) =0 T = -1/3 or 1Min = -1/27 - 1/9 + 1/3 + 5/1 = -1-3+9+135/27 = 140/27 = 5.2cmTo find the minimum distance that the particle can cover from the fixed point, we need to find the minimum value of the function s(t) = t³ - t² - t + 5.One way to find the minimum value is by finding the critical points of the function. We can do this by finding the derivative of s(t) and setting it equal to zero:s'(t) = 3t² - 2t - 1Setting s'(t) = 0 and solving for t:3t² - 2t - 1 = 0Using the quadratic formula, we find:t = (-(-2) ± √((-2)² - 4(3)(-1))) / (2(3))Simplifying further:t = (2 ± √(4 + 12)) / 6t = (2 ± √16) / 6t = (2 ± 4) / 6This gives us two possible values for t: t = 1 and t = -1/3.To determine which value gives the minimum distance, we can evaluate s(t) at these points:s(1) = 1³ - 1² - 1 + 5 = 4s(-1/3) = (-1/3)³ - (-1/3)² - (-1/3) + 5 ≈ 5.037Comparing these values, we see that the minimum distance occurs at t = 1, where the distance is 4 cm.Therefore, the minimum distance that the particle can cover from the fixed point is 4 cm.