Respuesta :
Answer:
a) [tex]v=3t^2-24t+36[/tex] is the velocity of the particle at any time t.
b) [tex]v_5=87\ ft.s^{-1}[/tex]
c) The particle is at rest at time t=2 seconds.
d) At time t=2 seconds the particle posses positive velocity being at positive direction.
e) [tex]s_8=32\ ft[/tex]
Explanation:
Given:
The function of displacement dependent on time,[tex]s=t^3-12t^2+36t[/tex] .......(1)
a)
Now as we know that velocity is the time derivative of the displacement:
[tex]v=\frac{d}{dt} s[/tex]
[tex]v=\frac{d}{dt} (t^3-12t^2+36t)[/tex]
[tex]v=3t^2-24t+36[/tex] ..........................(2)
b)
Now the velocity after 5 seconds:
put t=5 in eq. (2)
[tex]v_5=3\times 5^2-24\times 5+36[/tex]
[tex]v_5=87\ ft.s^{-1}[/tex]
c)
When the particle is at rest it has zero velocity.
Now velocity at time t=6 s:
[tex]v_6=3\times 6^2-24\times 6+36[/tex]
[tex]v_6=120\ ft.s^{-1}[/tex]
Now velocity at time t=2 s:
[tex]v_2=3\times 2^2-24\times 2+36[/tex]
[tex]v_2=0\ ft.s^{-1}[/tex]
The particle is at rest at time t=2 seconds.
d)
Put the value t=2 sec. in eq. (1):
[tex]s_2=2^3-12\times 2^2+36\times 2[/tex]
[tex]s_2=32\ ft[/tex]
At time t=2 seconds the particle posses positive velocity being at positive direction.
e)
Distance travelled during the first 8 seconds:
put t=8 in eq. (1)
[tex]s_8=8^3-12\times 8^2+36\times 8[/tex]
[tex]s_8=32\ ft[/tex]
