related rate question. Given s(t)=3t^3-4t+2, find velocity, v(t) and acceleration, a(t), and initial position?

Answer 1

for the initial position all you have to do y evaluate your ecuation in t=0 (initial time = intial position ;) )so you get s(0)=2

for velocity just derivate:

v(t)=s'(t)=9t^2-4

same for acceleration:

a(t)=v'(t)=18t

Answer 2

velocity is the first derivative: v = 9t^2 -4
acceleration is second derivative a = 18t
the initial position is s at t=0
s(0)=3(0^3) -4(0) +2
s(0) = 2

Answer 3

Initial position is found by putting t = 0 into the distance function s.
Velocity function v(t) is found by differentiating the distance function.
Acceleration function a(t) if found by differentiating the velocity function.
I am not going to do them for you. Try to finish it yourself before looking at any detailed answer that someone leaves here.

Answer 4

v(t) = 9t^2-4 a(t) = 18t initial position s(0)=2

Answer 5

okay, this is a little bit of cal because velocity is the second derivitive and acceleration is the thrid.
v(t)= 9t^2-4
a(t) is 18t

Answer 6

s(t)=3t^3-4t+2
Initial position = s(0) = 2
v(t) = ds/dt = 9t-4
a(t) = dv/dt= 9