Help needed with math question

Question

1.Given s = t⁴ − 4t³ − 20t² + 15t, where t ≥ 0, at what time is the acceleration 0?

2.Given: A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = t³ − 9t² + 24t

1. When is the particle speeding up?
2. When is the particle slowing down?

Any help is appreciated.

Answer 1

1. We are given the position function, as a function of time, which means its first derivative, s^', is velocity, and its second derivative, s^", is acceleration. So we'll find s^" using power rule, set it = 0, and solve:

v(t) = s^'(t) = 4t³ - 12t² - 40t + 15

a(t) = v^'(t) = s^"(t) = 12t² - 24t - 40 = 0

3t² - 8t - 10 = 0

By quadratic formula: t = [8 ± √184] / 6 = 4/3 ± √46/3

2. An object is speeding up whenever its velocity and acceleration have the same sign, and slowing down when velocity and acceleration have opposite signs. We can find both functions by differentiating the position function using power rule, and create sign charts for each as follows:

f^'(t) = v(t) = 3t² - 18t + 24 = 0 ; t² - 6t + 8 = 0 ; (t - 2)(t - 4) = 0 ; t = 2 or 4 ; v(t) < 0 for 2 < t < 4

f^"(t) = v^'(t) = 6t - 18 = 0 ; t = 3 ; a(t) < 0 for t < 3

∴ object slowing down for t < 2 or 3 < t < 4 and object speeding up for 2 < t < 3 or t > 4.