Final answer:
The velocity at time t is given by differentiating the position function, which is equivalent to the derivative of f(t). The particle is at rest when the velocity is equal to zero, and it is moving in the positive direction when the velocity is positive. The total distance traveled during the first 7 seconds can be found by integrating the velocity function over that interval.
Step-by-step explanation:
(a) To find the velocity at time t, we need to differentiate the position function f(t) with respect to time. Taking the derivative of f(t), we get f'(t) = 3t^2 - 30t + 72. So, the velocity at time t is given by v(t) = f'(t) = 3t^2 - 30t + 72.
(b) To find the velocity after 5 seconds, we substitute t = 5 into the velocity equation: v(5) = 3(5)^2 - 30(5) + 72 = -27 feet per second.
(c) The particle is at rest when the velocity is equal to zero. Solving the equation 3t^2 - 30t + 72 = 0, we find t = 4 and t = 8. Therefore, the particle is at rest at t = 4 seconds and t = 8 seconds.
(d) The particle is moving in the positive direction when the velocity is positive. From the velocity equation v(t) = 3t^2 - 30t + 72, we can see that the particle is moving in the positive direction when t is in the interval (0, 4) and (8, ∞).
(e) To find the total distance traveled during the first 7 seconds, we need to find the area under the velocity curve from t = 0 to t = 7. This can be done by integrating the velocity function over that interval: ∫[0,7] (3t^2 - 30t + 72) dt = 160.
(f) To find the acceleration at time t, we need to differentiate the velocity function v(t) with respect to time. Taking the derivative of v(t), we get v'(t) = 6t - 30. So, the acceleration at time t is given by a(t) = v'(t) = 6t - 30.
(g) To find the acceleration after 5 seconds, we substitute t = 5 into the acceleration equation: a(5) = 6(5) - 30 = 0 feet per second squared.
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