Respuesta :
Answer:
The displacement from t = 0 to t = 10 s, is -880 m
Distance is 912 m
Explanation:
[tex]v = (12 - 3t^2) m/s = ds/dt [/tex]. . . . . . . . . . A
integrate above equation we get
[tex]s = 12t - t^3 + C[/tex]
from information given in the question we have
t = 1 s, s = -10 m
so distance s will be
-10 = 12 - 1 + C,
C = -21
[tex]s(t) = 12t - t^3 - 21[/tex]
we know that acceleration is given as
[tex]a(t) = dv/dt = -6t[/tex]
[FROM EQUATION A]
Acceleration at t = 4 s, a(4) = -24 m/s^2
for the displacement from t = 0 to t = 10 s,
[tex]s(10) - s(0) = (12*10 - 10^3 - 21) - (-21) = -880 m[/tex]
the distance the particle travels during this time period:
let v = 0,
[tex]3t^2 = 12[/tex]
t = 2 s
Distance [tex]= [s(2) - s(0)] + [s(2) - s(10)] = [1\times 2 - 2^3] + [(12\times 2 - 2^3) - (12\times 10 - 10^3)] = 912 m[/tex]
