Respuesta :
Answer:
a)  t = 2.0 s,  b)  x_f = - 24.56 m,  Δx = 16.56 m
Explanation:
This is an exercise in kinematics, the relationship of position and time is indicated
     x = 5 t³ - 9t² -24 t - 8
a) ask when the velocity is zero
speed is defined by
     v = [tex]\frac{dx}{dt}[/tex]
let's perform the derivative
    v = 15 t² - 18t - 24
    0 = 15 t² - 18t - 24
let's solve the quadratic equation
   [tex]t = \frac{18 \pm \sqrt{18^2 + 4 \ 15 \ 24} }{2 \ 15}[/tex]
    t = [tex]\frac{18 \pm 42}{30}[/tex]
    t1 = -0.8 s
   t2 = 2.0 s
the time has to be positive therefore the correct answer is t = 2.0 s
b) the position and distance traveled for a = 0
acceleration is defined by
    a = dv / dt
    a = 30 t - 18
    a = 0
    30 t = 18
    t = 18/30
    t = 0.6 s
we substitute this time in the expression of the position
   Â
    x = 5 0.6³ - 9 0.6² - 24 0.6 - 8
    x = 1.08 - 3.24 - 14.4 - 8
    x = -24.56 m
we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s
the position for t = 0
    x₀ = -8 m
the position for t = 0.6 s
   x_f = - 24.56 m
the distance
   ΔX = x_f - x₀
   Δx = | -24.56 -(-8) |
   Δx = 16.56 m
