Help with calculus problem please

Question

The position function of a particle in rectilinear motion is given by s(t) = 2t³ - 21t² + 60t + 3 for t ≥ 0 with t measured in seconds and s(t) measured in feet. Find the position and acceleration of the particle at the instant the when the particle reverses direction. Include units in your answer.

Answer 1

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

s'(t) = 6t² - 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(t² - 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/sec² and s"(5) = -18 ft/sec².

Anita A.

Answer 2

The initial function is a third order, or cube function, so there are actually two reversals. For a time vs distance equation, the first derivative is the velocity. Velocity is 0 at the instant the direction changes. The velocity is given by the first differential, V= 6t^2 - 42t +60. V= 0 at t=2 and t=5. The positions are s= 55 and later s=28 (substitute t=2 and t=5 into the original equation).

Acceleration is given by the second derivative of the time vs distance equation, here a=12t - 42. Plug in t=2 and t=5 to give the instantaneous accelerations.

Velocity here is feet per second (change in y over change in x) and acceleration is in feet/sec^2