Anita A. answered • 6d

Texas State Certified Secondary Mathematics Instructor

To answer all parts, both the first and the second derivatives of the function will be needed.

s'(t) = 6t^{2} - 42t + 60 s"(t) = 12t - 42

Where s'(t) = 0, the velocity = 0; so the particle changes direction at the t-values where s'(t) = 0.

6(t^{2} - 7t + 10 ) = 0 t - 5 = 0 or t - 2 = 0

The particle changes direction at t = 5 and at t = 0.

s(2) = 28 ft; s(5) = 55 ft

Accelerations at the instants that direction changes are as follow:

s"(2) = 18 ft/sec^{2} and s"(5) = -18 ft/sec^{2}.

Anita A.