The position s of a point (in feet) is given as a function of time t (in seconds).

s = −10 + t − 15t2;  t = 8

(a) Find the point's acceleration as a function of t.

s''(t)=

I know this answer, it is -30

(b) Find the point's acceleration at the specified time.

s''(8) = I don't know this one

s = -10 + t - 15t^2

s' = v = 1 - 30t
s'' = a = -30 , which is a constant

so no matter what value of t you take, e.g. t = 30
there is no variable to sub into, so

a = -30

Well, let me help you with that. To find the acceleration at a specific time, you need to differentiate the function twice.

Let's start with finding the first derivative of the function:
s'(t) = d/dt (-10 + t - 15t^2)
s'(t) = 1 - 30t

Now, let's take the second derivative by differentiating s'(t):
s''(t) = d/dt (1 - 30t)
s''(t) = -30

So, as you correctly mentioned, the acceleration function is -30.

Now, to find the acceleration at t = 8, we substitute t = 8 into the acceleration function:
s''(8) = -30

So, the answer is -30.

To find the point's acceleration at the specified time t = 8, we need to plug in the value of t into the expression for the acceleration function s''(t).

Given that s''(t) = -30, we can substitute t = 8 into this expression to find the acceleration at t = 8:

s''(8) = -30

To find the point's acceleration as a function of time, we need to differentiate the position function twice with respect to time. Let's break down the process step by step:

Step 1: Differentiate the position function once to find the velocity function.

Given: s = -10 + t - 15t^2

Differentiating with respect to t, we get:
v(t) = ds/dt = d/dt (-10 + t - 15t^2)

The derivative of -10 with respect to t is 0 since it's a constant.
The derivative of t with respect to t is 1.
The derivative of -15t^2 with respect to t is -30t.

So the velocity function is:
v(t) = 1 - 30t

Step 2: Differentiate the velocity function to find the acceleration function.

Differentiating v(t) with respect to t, we get:
a(t) = dv/dt = d/dt (1 - 30t)

The derivative of 1 with respect to t is 0 since it's a constant.
The derivative of -30t with respect to t is -30.

So the acceleration function is:
a(t) = -30

Therefore, the point's acceleration as a function of t is given by a(t) = -30.

(Solution to part (a): s''(t) = -30)

To find the point's acceleration at the specified time t = 8 (part b), we substitute t = 8 into the acceleration function:

a(8) = -30

Therefore, the point's acceleration at t = 8 is -30.