# Math Help - rate of change

1. ## rate of change

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

2. Originally Posted by pageraets
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
$x = cos \left ( \frac{ \pi t}{2} \right )$

So the speed function is the first time derivative and the acceleration is the second time derivative.

Thus
$v = \frac{dx}{dt} = -\frac{\pi}{2}~sin \left ( \frac{ \pi t}{2} \right )$
and
$a = \frac{d^2x}{dt^2} = -\left ( \frac{\pi}{2} \right ) ^2 ~cos \left ( \frac{ \pi t}{2} \right )$

Can you do the rest?

-Dan