Explore the relationship between the acceleration of an object, given by the second derivative of its position function, s(t) = 4t^3 - 2t^2 + t. What is the acceleration at t = 3?
user
Absolutely, let's see. We have the position function s(t) = 4t^3 - 2t^2 + t.

The acceleration, a(t), is given by the second derivative of s(t), which is:

a(t) = ∂²s(t)/∂t² = 12t - 4t + 1

So, when t = 3, the acceleration is:

a(3) = 12(3) - 4(3) + 1 = 29

Therefore, the acceleration at t = 3 is approximately 29 m/s^2. This means that the object is accelerating at a rate of approximately 29 meters per second squared.
assistant
Explore the relationship between the acceleration of an object, given by the second derivative of its position function, s(t) = 4t^3 - 2t^2 + t. What is the acceleration at t = 3?
user
To find the acceleration of an object at a given time, we first need to find the second derivative of its position function, s(t).

Given s(t) = 4t^3 - 2t^2 + t, let's find the first derivative with respect to time, which will give us the velocity function, v(t).

v(t) = ds/dt = d(4t^3 - 2t^2 + t)/dt = 12t^2 - 4t + 1

Now, to find the acceleration function, a(t), we need to find the derivative of the velocity function with respect to time:

a(t) = dv/dt = d(12t^2 - 4t + 1)/dt = 24t - 4

Now that we have the acceleration function, we can find the acceleration at t = 3:

a(3) = 24(3) - 4 = 72 - 4 = 68

So, the acceleration of the object at t = 3 is 68 units per second squared.
assistant