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A particle moves according to a law of motion s = f(t), t2 0, where t is measured in seconds and s in feet. f(t) = 0.01t4- 0.03t3 (a) Find the velocity at time t (in ft/s). V(t) = 0.043- 0.09/? (b) What is the velocity after 1 second(s)? v(1) = 0.05 X ft/s (c) When is the particle at rest? smaller value t = 0 larger value t = 2.25

A particle moves according to a law of motion s = f(t), t2 0, where t is measured in seconds and s in feet. f(t) = 0.01t4- 0.03t3 (a) Find the velocity at time t (in ft/s). V(t) = 0.043- 0.09/? (b) What is the velocity after 1 second(s)? v(1) = 0.05 X ft/s (c) When is the particle at rest? smaller value t = 0 larger value t = 2.25

College Algebra
College Algebra
10th Edition
ISBN:9781337282291
Author:Ron Larson
Publisher:Ron Larson
Chapter3: Polynomial Functions
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A particle moves according to a law of motion s = f(t), t 2 0, where t is measured in seconds and s in feet. ags=autosave#question4768276_3 f(t) = 0.01t4 - 0.03t3 (a) Find the velocity at time t (in ft/s). v(t) = 0.04 - 0.09/? %3D (b) What is the velocity after 1 second(s)? v(1) = 0.05 X ft/s (c) When is the particle at rest? smaller value larger value t = 2.25 S (d) When is the particle moving in the positive direction? (Enter your answer using interval notation.) t> 2.25 (e) Find the total distance traveled during the first 9 seconds. (Round your answer to two decimal places.) 43.74 (f) Find the acceleration at time t (in ft/s2). a(t) = 0.06 Find the acceleration after 1 second(s). a(1) = ft/s2
Transcribed Image Text:A particle moves according to a law of motion s = f(t), t 2 0, where t is measured in seconds and s in feet. ags=autosave#question4768276_3 f(t) = 0.01t4 - 0.03t3 (a) Find the velocity at time t (in ft/s). v(t) = 0.04 - 0.09/? %3D (b) What is the velocity after 1 second(s)? v(1) = 0.05 X ft/s (c) When is the particle at rest? smaller value larger value t = 2.25 S (d) When is the particle moving in the positive direction? (Enter your answer using interval notation.) t> 2.25 (e) Find the total distance traveled during the first 9 seconds. (Round your answer to two decimal places.) 43.74 (f) Find the acceleration at time t (in ft/s2). a(t) = 0.06 Find the acceleration after 1 second(s). a(1) = ft/s2
Sodium chlorate crystals are easy to grow in the shape of cubes by allowing a solution of water and sodium chlorate to evaporate slowly. If V is the volume of such a cube with side length x, find (in mm /mm) when x 4 mm. xp v(4)31 X mm /mm Explain the meaning of V'(4) in the context of this problem. O v(4) represents the volume as the side length reaches 4 mm. O v represents the rate at which the volume is increasing with respect to the side length as V reaches 12 mm3. O v(4) represents the rate at which the volume is increasing as x reaches 12 mm. O v(4) represents the rate at which the volume is increasing with respect to the side length as x reaches 4 mm. O V(4) represents the rate at which the side length is increasing with respect to the volume as x reaches 4 mm. Need Help? Read It 3:52 PM P Type here to search 日 a 10/2/2021 ..
Transcribed Image Text:Sodium chlorate crystals are easy to grow in the shape of cubes by allowing a solution of water and sodium chlorate to evaporate slowly. If V is the volume of such a cube with side length x, find (in mm /mm) when x 4 mm. xp v(4)31 X mm /mm Explain the meaning of V'(4) in the context of this problem. O v(4) represents the volume as the side length reaches 4 mm. O v represents the rate at which the volume is increasing with respect to the side length as V reaches 12 mm3. O v(4) represents the rate at which the volume is increasing as x reaches 12 mm. O v(4) represents the rate at which the volume is increasing with respect to the side length as x reaches 4 mm. O V(4) represents the rate at which the side length is increasing with respect to the volume as x reaches 4 mm. Need Help? Read It 3:52 PM P Type here to search 日 a 10/2/2021 ..
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