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.