finding acceleration

Question

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

s = −17 + t − 17t²; t = 8

(a)

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

s″(t) = ____ ft/sec²

(b)

Find the point's acceleration at the specified time.

s″(8) = ____ ft/sec²

Answer 1

s"(t)= -34 ft/sec^2

s"(8) = -34 ft/sec^2

Answer 2

s = -17 + t - 17t² ; t = 8

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

1. [s(t) = -17 + t - 17t²] d/dt
2. ds/dt = 1 - 34t
3. s’(t) = 1 - 34t; This is our velocity function in ft/sec
4. [ds/dt = 1 - 34t] d/dt
5. ds/(dt)² = -34
6. s’’(t) = -34; This is our acceleration function in ft/sec²

(b) Find the point’s acceleration at the specified time; t = 8

s’’(t) = -34; s’’(8) = -34 [Edit: ft/sec²]