Final answer:
To determine the direction of a particle's motion, we analyze the sign of its velocity, which is the derivative of its position function. The particle described by x(t)=(t-2)(t-6)^3 moves to the left when t is between 2 and 6 seconds, as during this interval the velocity is negative.
Step-by-step explanation:
To determine when a particle is moving to the left, we need to examine the sign of its velocity function, which is the derivative of its position function. For the given position function x(t) = (t-2)(t-6)^3, the velocity function v(t) is:
v(t) = dx/dt = d/dt[(t-2)(t-6)^3].
Without performing full differentiation, we can use the fact that if a particle's velocity is positive, it is moving to the right, and if the velocity is negative, it is moving to the left. For the particle to be moving to the left, v(t) must be less than 0. We'll find the intervals where the velocity is negative by analyzing the factors of the position function (t-2) and (t-6).^3
1. We know that the cubic term (t-6)^3 will determine the sign of v(t) near t=6, where it changes from positive to negative as t increases through 6.
2. Similarly, the linear term (t-2) will change sign at t=2.
To find the exact intervals of motion to the left, we would differentiate x(t) fully to get v(t) and analyze the sign changes. Nevertheless, without full differentiation, we can still deduce the following:
t < 2, both factors (t-2) and (t-6)^3 are negative, and their product is positive, so the particle is moving right.
Between t = 2 and t = 6, (t-2) is positive, and (t-6)^3 is negative, so the particle is moving left.
t > 6, both factors are positive, the product is positive, and the particle is moving right again.
Therefore, the particle is moving to the left when t is between 2 and 6 seconds.