Respuesta :
Answer:
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'
[tex]\frac{dS}{dt} = 3 t^{2} - 3 (2 t) - 24[/tex]
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'
[tex]\frac{d^{2} S}{dt^{2} } = 6 t - 6[/tex]
Step(ii):-
Case(i):-
Put t= -2
[tex]\frac{d^{2} S}{dt^{2} } = 6 t - 6 = 6 ( -2) -6 = -12 -6 = -18 <0[/tex]
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
[tex]\frac{d^{2} S}{dt^{2} } = 6 t - 6 = 6 ( 4) -6 = 24 -6 = 18 >0[/tex]
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 answer:-
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
