A particle moves along the x-axis so that its position at time t>0 is given by x(t)=(t^2 -9)/(3t^2 +8)

Question

a. Show that the velocity of the particle at time t is given by v(t) = 70t/(3t^2 +8)^2

b. Is the particle moving toward the origin or away from the origin at time t=2? Give a reason for your answer

c. The acceleration of the particle is given by a(t). Write an expression for a(t), and find the value of a(2).

d. What position does the particle approach as t approaches infinity?

Answer 1

a. v(t) = x'(t) = 2t(3t² + 8) - 6t(t² - 9)/(3t² + 8)² = (6t³ + 16t - 6t³ + 54t)/(3t² + 8)² = 70t/(3t² + 8)².

b. v(2) = 140/(3·4 + 8)² = 7/20 > 0, so at t = 2 away from the origin or in positive direction of x-axis.

c. a(t) = v'(t) = 70·[1(3t² + 8)² - t·2(3t² + 8)6t]/(3t² + 8)⁴ = 70(3t² + 8 - 12t²)/(3t² + 8)³ =

70(8 - 9t²)/(3t² + 8)³; a(2) = 70·(8 - 36)/20³ = - 0.24

d. lim x(t) as t →∞ = lim (t² - 9)/(3t² + 8) = 1/3