Could someone help me with a Calc problem? :3

The position function for an object is given by s = 2t^3 - 3 (tan(pi/12 * t)) for t _ 0 , where s is measured in feet and t is measured in seconds.
At time t = 2, is the object speeding up or slowing down? Explain your reasoning.

Question

Could someone help me with a Calc problem? :3

The position function for an object is given by s = 2t^3 - 3 (tan(pi/12 * t))  for t >_ 0 , where s is measured in feet and t is measured in seconds.
At time t = 2, is the object speeding up or slowing down? Explain your reasoning.

jigglypuff314 · Answer

$$s = 2t ^{3}-3\tan(\frac{ \pi }{ 12 } t ) ~~~~ for ~~t \ge 0$$

anonymous · Answer

ummmm

agent0smith · Answer

Differentiate the position function to find velocity, then again to find acceleration. 

Find the acceleration when t=2.

Find the velocity when t=2.

If a and v both have the same sign (ie both are positive or both negative), the object is speeding up. If they have different signs, it's slowing down.

jigglypuff314 · Answer

okay, thank you! :)

jigglypuff314 · Answer

@agent0smith for this problem I got that at t = 2
v = 22.953   and    a = 23.683
so because a and v are both positive, the object is speeding up
would that be correct?

agent0smith · Answer

Yep, if a and v have the same sign (ie same direction) the object is speeding up.

jigglypuff314 · Answer

I was just not sure if I got the correct a and v values :/

agent0smith · Answer

You could check on wolfram alpha.

jigglypuff314 · Answer

lol k, thanks