The correct question is:
Given s(t) = 3t² + 6t, where s(t) is in feet and t is in seconds, find each of the following.
a) v(t)
b) a(t)
c) The velocity and acceleration when t =22 sec
Solution:
Given s(t) = 3t² + 6t
(a) The velocity, v(t) is the derivative of s(t). That is v(t) = s'(t)
s'(t) = v(t) = 6t + 6
(b) The acceleration, a(t) is the derivative of v(t). That is a(t) = v'(t)
v'(t) = a(t) = 6
(c) When t = 22 sec,
v(t) = v(22) = 6(22) + 6 = 132 + 6
= 138
a(t) = a(22) = 6
At t = 22, the velocity is 138, and acceleration is 6.