Question

A particle moves along a line so that its position at any time is t>0 or t=0 is given by the function s(t)=(t^3)-(6t^2)+(8t)+2 where s is measured in meters and t is measured in seconds
a) find instantaneous velocity at any time t
b) find acceleration of the particle at any time t
c) when is the particle at rest?
d) describe motion of the particle. at what values of t does the particle change directions

Answer 1

a) velocity is first derivative of position
b) acceleration is first derivative of velocity (second derivative of position)
c) rest is such that velocity = 0.
d) try finding critical points and graphing. changing directions is when value of velocity changes sign +/-

Answer 2

ok so would the answer to a) be 3t^2 -12t +8
b) 6t-12

Answer 3

Looks good so far.

Answer 4

c) when 3t^2 -12t+8 = 0 but since that isnt factorable do you use quadratic equation?

Answer 5

Quadratic formula always works. ;-)

Answer 6

ok thank you very much for all of your help!!!

Answer 7

My pleasure! Good luck.