Given s (t) equals 3 t squared plus 6 ts (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 tequals=22 sec

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.