The distance s (in feet) covered by a car t seconds after starting is given by the following function. s = −t^3 + 7t^2 + 18t (0 ≤ t ≤ 6)

Question

Now

Find a general expression for the car's acceleration (in ft/sec³) at any time t (0 ≤ t ≤ 6). to find the

s''(t)=

At what time t (in sec) does the car begin to decelerate? (Round your answer to one decimal place.)

t=( )in sec

Answer 1

I am assuming that there is a typo in the statement of the problem:

acceleration is measured in ft/sec², not ft/sec³.

The first derivative of s(t) is velocity v(t).

And the derivative of that is acceleration a(t).

s(t) = -t³ + 7t² + 18t

v(t) = -3t² + 14t + 18

a(t) = -6t + 14

At the beginning of the interval, t = 0, the car is accelerating (a(0) > 0)

a(0) = -6×0 + 14 = 14 ft/sec²

At the end of the interval, t = 6, the car is decelerating (a(6) < 0)

a(6) = -6×6 + 14 = -22 ft/sec²

At the time t when a(t) = 0, the car changes from acceleration to deceleration:

-6t + 14 = 0

t = 14/6