Final answer:
The velocity function can be found by differentiating the position function. The velocity after 1 second is -0.02 ft/s. The particle is at rest at t = 0 and t = 1.5 seconds.
Step-by-step explanation:
(a) To find the velocity at time t, we can differentiate the position function. The derivative of f(t) = 0.01t^4 - 0.02t^3 is v(t) = 0.04t^3 - 0.06t^2. So, the velocity function is v(t) = 0.04t^3 - 0.06t^2 ft/s.
(b) To find the velocity after 1 second, we substitute t = 1 into the velocity function. v(1) = 0.04(1)^3 - 0.06(1)^2 = -0.02 ft/s.
(c) The particle is at rest when its velocity is equal to zero. To find the time(s) the particle is at rest, we set v(t) = 0 and solve for t. So, the particle is at rest at t = 0 and t = 1.5 seconds.
(d) The particle is moving in the positive direction when its velocity is positive. To find the time(s) the particle is moving in the positive direction, we look for the intervals where v(t) > 0. The particle is moving in the positive direction when 0 < t < 1.5 seconds.
(e) The total distance traveled during the first 12 seconds can be found by finding the area under the velocity-time graph. We integrate v(t) from 0 to 12 seconds. The total distance traveled during the first 12 seconds is approximately 96 ft.
(f) To find the acceleration at time t, we differentiate the velocity function. The derivative of v(t) = 0.04t^3 - 0.06t^2 is a(t) = 0.12t^2 - 0.12t ft/s^2.
(g) To find the acceleration after 1 second, we substitute t = 1 into the acceleration function. a(1) = 0.12(1)^2 - 0.12(1) = 0 ft/s^2.