Expert: Ahmed Salami - 3/25/2009

Question
Hey, I'm trying to figure out the answer to this question:
If the displacement of a particle P at time t is given by 's=3t3/2 - t1/2 + 3t -12' , find it’s (a) velocity when t=2s, t=5s and t=7s  (b) its acceleration at t=4s and t = 5s.
I have this formula to look them both out, so can you please confirm that they are correct:

s'(t) = 3 Sqrt(t)  - 1/(2Sqrt(t)) + 3  this is the velocity

s''(t) =  (3/(2 Sqrt(t)) + 1/(4 t^(3/2))  this is the acceleration

Plug the values in for t to get the answers.

Thanks

Answer
Hi John,
Did you mean s = 3t^3/2 - t^1/2 + 3t -12 ?
If so, the velocity and acceleration functions are successive derivatives of the displacement function.
s' = (3/2)3t^1/2 - (1/2)t^-1/2 + 3
  = (9/2)t^1/2 - (1/2)t^-1/2 + 3
  = (9/2)sqrt(t) - 1/[2sqrt(t)] + 3

s'' = (1/2)(9/2)t^-1/2 - (-1/2)(1/2)t^-3/2
   = (9/4)t^-1/2 + (1/4)t^-3/2
   = 9/[4sqrt(t)] + 1/[4t.sqrt(t)]

Hope its clear.

Regards