(A) just plug in t=3
(B) v(t) = -(9t^2-6t+8)/(3t^3 - 3t^2 +8t)^2 so t(0) < 0
(C) the denominator is never negative, so where is the numerator negative?
hint: the discriminant is never zero
(D) s = ∫[0,4] √(1+(x')^2) dt
= ∫[0,4] √(1+1/(9t^2-6t+8)^2) dt = 4.01
though actually, since x(0) = ∞, I'd say the distance traveled must also be infinite.
1. A particle moves along the x-axis in such a way that it's position in time t for t is greater or equal to 0 is given by x= 1/3t^3 - 3t^2 +8t
A. Find the position of the particle at time t = 3.
B. Show that at time t = 0, the particle is moving to the right.
C. Find all values of t for which the particle is moving to the left.
D. What is the total distance the particle travels from t = 0 to t = 4?
1 answer
2 answers