if the velocity can be expressed as

v(t)=1-∛t what is the acceleration function.
so i started by bringing in acceleration formula a=v1-vo/t1-to
took my velocity function and found the derivative. v=-1/3t^2/3 have i started it correct so far and where do i go from here besides just plugging it into the acceleration form

a = dv/dt

if v = 1 - ∛t, then
a = -1/3 t^-2/3 = -1/(3∛t2)

that is the acceleration function

Yes, you have started correctly. To find the acceleration function, you need to take the derivative of the velocity function with respect to time.

In this case, the velocity function is given by v(t) = 1 - ∛t. Let's find its derivative:

- Differentiate 1 with respect to t: d(1)/dt = 0
- For the second term, use the chain rule: d(∛t)/dt = 1/3 * (1/∛t^(2/3)) * d(t)/dt = 1/3 * (1/∛t^(2/3)) * 1

Combining the derivatives, we get:

v'(t) = 0 - 1/3 * (1/∛t^(2/3))

Simplifying further:

v'(t) = -1/3∛(t^(-2/3))

Now, this expression represents the derivative of velocity, so it corresponds to the acceleration function. Therefore, the acceleration function is:

a(t) = v'(t) = -1/3∛(t^(-2/3))

You can plug in any value of 't' to this equation to find the acceleration at that specific time.