A particle moves along a straight track so that its acceleration at time seconds is…

MATHEMATICS · Multi-part · CBSE · Mathematics · Applications of Derivatives · Applications of Derivatives

Problem

A particle moves along a straight track so that its acceleration at time tt seconds is given by a(t)=6t−4a(t)=6t-4 (in m/s2^{2}). At t=0t=0 the particle is located 2 m to the right of the origin and is moving to the left with velocity 3 m/s. (a) Determine the velocity function v(t)v(t) for t≥0t\ge 0. (b) Determine the position function s(t)s(t) for t≥0t\ge 0 measured from the origin. (c) Find all times in the interval 0≤t≤40\le t\le4 at which the particle is momentarily at rest, and classify each as a relative maximum, minimum, or neither for s(t)s(t). (d) Compute the total distance traveled by the particle during the first 4 seconds.

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