A particle moves with position function

$s=t^4-4t^3-36t^2+15t,t0$

$(a)At$ what time does the particle have a velocity $of15\frac{m}{s}?(Enter$ your answers $as$ a comma$-$separated list.$)$

$t=,s$

$(b)At$ what time $is$ the acceleration $0?(Round$ your answer $to$ two decimal places.$)$

$t=,s$

What $is$ the significance $of$ this?

$At$ this time, the acceleration changes from negative $to$ positive and the velocity attains its maximum value.

$At$ this time, the acceleration changes from negative $to$ positive and the velocity attains its minimum value.

$At$ this time, the acceleration changes from positive $to$ negative and the velocity attains its maximum value.

$At$ this time, the acceleration changes from positive $to$ negative and the velocity attains its minimum value.

$At$ this time, the acceleration changes from positive $to$ negative and the velocity attains a value $of$ zero.