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A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. (If an answer does not exist, enter DNE.)
A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. (If an answer does not exist, enter DNE.) f(t) = t^3 − 8t^2 + 28t
a. Find the velocity at time t.
b. What is the velocity after 1 second?
c. When is the particle at rest?
d. When is the particle moving in the positive direction?
e. Find the total distance traveled during the first 6 seconds.
Answer:a)v = 3 t² - 16 t + 28 , b) v = 15 ft / s , c) the speed is never zero ,d) the particle always moves in the positive direction , e) s = 96 ftExplanation:a) For this exercise how should we use the definition of speedv = ds / dtlet's make the derivativev = 3 t² - 16 t + 28b) we calculate with t = 1 sv = 3 1² - 16 1 +28v = 15 ft / sc) the particle is at rest when the velocity is zero0 = 3 t² - 16 t + 283t² - 16t + 28 = 0we solve the quadratic equationt = [16 ±√ (16² - 4 3 28)] / (2 3)t = [5.33 ±√ (256 - 336)] / 6the square root of a negative number is not defined by which the speed is never zerod) the particle moves in the positive directional when y> 0, therefored (t)> 0t³ - 8 t² + 28 t> 0t (t² - 8t + 28)> 0the expression is true fort> 0t² - 8t +28> 0we solve the quadratic equationt = [8 ±√ (8² - 4 28)] / 2t = [8 ±√ (64 -112)] / 2the square root of a negative number is not defined, therefore the expression never becomes zero, it is always positiveIn short the particle always moves in the positive directione) we calculate the distance even t = 6 ss = 6³ - 8 6² +28 6s = 96 ft...