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.

step-by-step explanation:

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