second derivative question s(t) = t3 + 2t2 + 4t + 1 Find a(t)

Question

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

s(t) = t^3 + 2t^2 + 4t + 1

Find a(t)

Answer 1

One thing to keep in mind here: for added components of a function, the derivative of the whole function is equal to that of the sum of the derivatives of the individual components. For this one, we'll use the power rule: if f(x) = Ax^b, f^'(x) = (A*b)x^b-1.

So, for our case:

displacement: s(t) = t³ + 2t² + 4t + 1

velocity: v(t) = s'(t) = (t³)^' + (2t²)^' + (4t)^' + (1)^' = 3t² + 4t + 4 + 0

acceleration: a(t) = s''(t) = (3t²)^' + (4t)^' + (4)^' = 6t + 4 + 0 = 6t + 4

Answer 2

s(t) = t³ + 2t² + 4t + 1 = position function

v(t) = velocity function = s'(t) = 3t² + 4t + 4

a(t) = acceleration function = s"(t) = 6t + 4