for cos(3.14t/2) find when the particle is speeding up and slowing down, and the acceleration after 3 seconds, and the distance traveled in 4 seconds
May I be the first to say
"What????"
I presume that your position function for the particle is
$\displaystyle x = cos \left ( \frac{ \pi t}{2} \right )$
So the speed function is the first time derivative and the acceleration is the second time derivative.
Thus
$\displaystyle v = \frac{dx}{dt} = -\frac{\pi}{2}~sin \left ( \frac{ \pi t}{2} \right )$
and
$\displaystyle a = \frac{d^2x}{dt^2} = -\left ( \frac{\pi}{2} \right ) ^2 ~cos \left ( \frac{ \pi t}{2} \right )$
Can you do the rest?
-Dan