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?

(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.

To answer these questions, we need to analyze the position function and understand its properties. The given position function is x = (1/3)t^3 - 3t^2 + 8t.

A. To find the position of the particle at time t = 3, substitute t = 3 into the position function:
x = (1/3)(3^3) - 3(3^2) + 8(3)
x = (1/3)(27) - 3(9) + 24
x = 9 - 27 + 24
x = 6
Therefore, the position of the particle at time t = 3 is x = 6.

B. To determine if the particle is moving to the right or left at time t = 0, we need to find the particle's velocity at that specific time. Velocity is given by the derivative of the position function with respect to time, which is dx/dt.

Let's differentiate the position function:
dx/dt = d/dt((1/3)t^3 - 3t^2 + 8t)
= (1/3)(3t^2) - 3(2t) + 8
= t^2 - 6t + 8

Replace t = 0:
dx/dt at t = 0 = (0)^2 - 6(0) + 8
= 0 + 0 + 8
= 8

Since the velocity at t = 0 is positive (8), the particle is moving to the right.

C. To find the values of t for which the particle is moving to the left, we need to determine when the velocity (dx/dt) is negative. Set the velocity function equal to zero and solve for t:

t^2 - 6t + 8 = 0

To solve this quadratic equation, we can factor it:
(t - 2)(t - 4) = 0

Setting each factor equal to zero gives us:
t - 2 = 0 or t - 4 = 0

Solving for t, we get:
t = 2 or t = 4

So, when t is between 2 and 4 (exclusive), the particle is moving to the left.

D. To find the total distance the particle travels from t = 0 to t = 4, we need to calculate the sum of the absolute values of the displacements. Displacement is the difference between initial position and final position. Recall that distance cannot be negative.

At t = 0, the initial position is x = (1/3)(0)^3 - 3(0)^2 + 8(0) = 0.
At t = 4, the final position is x = (1/3)(4)^3 - 3(4)^2 + 8(4) = -4.

To find the total distance traveled, we take the absolute value of the displacement:
|Displacement| = |-4 - 0| = 4

Therefore, the total distance the particle travels from t = 0 to t = 4 is 4 units.