An object moves along a horizontal coordinate line in such a way that its position at time t is specified by s equals t cubed minus 3 t squared minus 24 t plus 8. Here s is measured in centimeters and t in seconds. When is the object slowing down; that is, when is its speed decreasing?

a)

The object slowing down  S = -72 centimetres after t = 4 seconds

b)

The speed is decreasing at t = -2 seconds

The objective function  S = 36 centimetres

Step-by-step explanation:

Step(i):-

Given  S = t³ - 3 t² - 24 t + 8 ...(i)

   Differentiating equation (i) with respective to 'x'

     Equating Zero

      3 t ² - 6 t - 24 = 0

 ⇒   t² - 2 t - 8   = 0

 ⇒  t² - 4 t + 2 t - 8 = 0

⇒ t (t-4) + 2 (t -4) =0

⇒  ( t + 2) ( t -4) =0

⇒ t = -2 and t = 4

 Again differentiating with respective to 'x'

Step(ii):-

Case(i):-

Put t= -2

The  maximum object

S = t³ - 3 t² - 24 t + 8

S = ( -2)³ - 3 (-2)² -24(-2) +8

S = -8-3(4) +48 +8

S = - 8 - 12 + 56

S = - 20 +56

S = 36

Case(ii):-

   put  t = 4

The object slowing down at t =4 seconds

The minimum objective function

S = t³ - 3 t² - 24 t + 8

S = ( 4)³ - 3 (4)² -24(4) +8

S =  64 -48 - 96 +8

S = - 72

The object slowing down  S = -72 centimetres after t = 4 seconds

Final -

The object slowing down  S = -72 centimetres after t = 4 seconds

The speed is decreasing at t = -2 seconds

The objective function  S = 36 centimetres

d≈7.07in

step-by-step explanation: