Question

Acceleration of a Car The distance s (in feet) covered by a car t seconds after starting is given by the following function. -t³ + 15t² + 28t (0 ≤ t ≤ 6) Find a general expression for the car's acceleration (in ft/sec³) at any time t (0 ≤ t ≤ 6). s"(t) = S = -1

At what time t (in sec) does the car begin to decelerate?
t =__________ sec

Answer 1

The acceleration of the car at any time t is given by a(t) = -6t + 30. The car begins to decelerate at t = 5 seconds when the acceleration function transitions from positive to negative.

Step-by-step explanation:

To find the car's acceleration, we need to differentiate the given position function s(t) = -t³ + 15t² + 28t twice with respect to time (t). The first derivative of s(t) with respect to t gives us the velocity function v(t), and the second derivative gives us the acceleration function a(t).

The first derivative of s(t) is:

v(t) = s'(t) = -3t² + 30t + 28

The second derivative, which is the acceleration, is:

a(t) = v'(t) = s''(t) = -6t + 30

To find when the car begins to decelerate, we look for the time t when the acceleration a(t) switches from positive to negative, since positive acceleration indicates speeding up and negative acceleration indicates slowing down.

So, we set a(t) = 0 and solve for t:

-6t + 30 = 0

t = 30 / 6

t = 5 seconds

The car begins to decelerate at t = 5 seconds.