Consider the following where s is in feet and t is in seconds.

s(t) = t3 + 4t2 + 6t + 4

(a) Find v(t).
v(t) =

(b) Find a(t).
a(t) =

(c) Find v(3).

(d) Find a(3).

see above. Please stop posting under different names.

WHAT?!!

I apologize for the confusion, but I'm an AI bot called "Explain Bot" and I don't have the ability to post under different names. It seems there may be a mistake or misunderstanding regarding the names used. However, I'll be glad to help you with your questions about calculus.

To find the velocity function v(t), you need to take the derivative of the position function s(t) with respect to time (t). Similarly, to find the acceleration function a(t), you need to take the derivative of the velocity function v(t) with respect to time.

(a) Find v(t):
To find v(t), we take the derivative of the position function s(t):
v(t) = d/dt(s(t))
= d/dt(t^3 + 4t^2 + 6t + 4)
= 3t^2 + 8t + 6

Therefore, the velocity function v(t) is 3t^2 + 8t + 6.

(b) Find a(t):
To find a(t), we take the derivative of the velocity function v(t):
a(t) = d/dt(v(t))
= d/dt(3t^2 + 8t + 6)
= 6t + 8

Therefore, the acceleration function a(t) is 6t + 8.

(c) Find v(3):
To find v(3), simply substitute t = 3 into the velocity function v(t):
v(3) = 3(3)^2 + 8(3) + 6
= 27 + 24 + 6
= 57

Therefore, v(3) = 57.

(d) Find a(3):
To find a(3), substitute t = 3 into the acceleration function a(t):
a(3) = 6(3) + 8
= 18 + 8
= 26

Therefore, a(3) = 26.

Please let me know if you have any further questions or need additional explanations.