Consider the following where s is in feet and t is in seconds.
s(t) = t^3 + 4t^2 + 6t + 4
(a) Find v(t).
v(t) =
(b) Find a(t).
a(t) =
(c) Find v(3).
(d) Find a(3).
v(t) is the first derivative of s(t).
a(t) is the second derivative of s(t).
I will be happy to critique your work or thinking.
thanks!
To find v(t), we need to take the derivative of s(t) with respect to t. Let's calculate it step by step:
(a) Find v(t):
First, we differentiate each term of s(t) using the power rule for derivatives:
d/dt(t^3) = 3t^2
d/dt(4t^2) = 8t
d/dt(6t) = 6
d/dt(4) = 0
Then, we can write v(t) as the derivative of s(t):
v(t) = 3t^2 + 8t + 6
(b) Find a(t):
To find a(t), we will differentiate v(t) with respect to t using the power rule:
d/dt(3t^2) = 6t
d/dt(8t) = 8
d/dt(6) = 0
So, a(t) = 6t + 8
(c) Find v(3):
To find v(3), we substitute t = 3 into equation (a):
v(3) = 3(3)^2 + 8(3) + 6
v(3) = 27 + 24 + 6
v(3) = 57
Therefore, v(3) = 57.
(d) Find a(3):
To find a(3), we substitute t = 3 into equation (b):
a(3) = 6(3) + 8
a(3) = 18 + 8
a(3) = 26
Therefore, a(3) = 26.
I hope this helps! Let me know if you have any further questions or if you need clarification on any step.