The position function of a particle in rectilinear motion

Question

The position function of a particle in rectilinear motion is given by s(t) = 2t^3 + 21t^2 + 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 when the particle reverses direction. Include units in your answer.

Answer 1

The second derivative is 12t + 42 and is zero when t = -42/12 = -21/6 = -7/2

 

However, the domain is restricted to positive values of time t>=0.

 

Moreover, in order to find where the particle changes direction,

we must first find the roots where the polynomial function is zero.

 

Using the rational root theorem, the only possible rational roots

are { +- 1, +-3,  +- 1/2,  +- 3/2} and none of them work.

So it must be solved numerically. Examining the graph, the

function is zero around -1/2, which also is not in the domain.

 

Please check the signs. There has got to be one or two minus signs

in the function somewhere, per Descartes Law of Signs.