Particle travels distance s(t) = 128t^2 -t^4 over interval (0,8).Find time when particle is at maximum speed

velocity will be a maximum when dv/dt=0

we find an expression for v by differentiating s(t)

v = ds/dt = 256 t - 4t^3

then, dv/dt = 256 - 12 t^2

dv/dt is a max when dv/dt =0

256 - 12t^2 = 0 => t = +/- 4.62 s

in the interval in question, the max occurs at t=+4.62 s

let's do the second derivative test to ensure this extremum is a max:

d^2v/dt^2 = -24t

for t>0, the second derivative is <0 therefore this is a max