Answer:
a)
is the velocity of the particle at any time t.
b) ![v_5=87\ ft.s^{-1}](https://tex.z-dn.net/?f=v_5%3D87%5C%20ft.s%5E%7B-1%7D)
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) ![s_8=32\ ft](https://tex.z-dn.net/?f=s_8%3D32%5C%20ft)
Explanation:
Given:
The function of displacement dependent on time,
.......(1)
a)
<u>Now as we know that velocity is the time derivative of the displacement:</u>
![v=\frac{d}{dt} s](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bd%7D%7Bdt%7D%20s)
![v=\frac{d}{dt} (t^3-12t^2+36t)](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bd%7D%7Bdt%7D%20%28t%5E3-12t%5E2%2B36t%29)
..........................(2)
b)
Now the velocity after 5 seconds:
put t=5 in eq. (2)
![v_5=3\times 5^2-24\times 5+36](https://tex.z-dn.net/?f=v_5%3D3%5Ctimes%205%5E2-24%5Ctimes%205%2B36)
![v_5=87\ ft.s^{-1}](https://tex.z-dn.net/?f=v_5%3D87%5C%20ft.s%5E%7B-1%7D)
c)
When the particle is at rest it has zero velocity.
Now velocity at time t=6 s:
![v_6=3\times 6^2-24\times 6+36](https://tex.z-dn.net/?f=v_6%3D3%5Ctimes%206%5E2-24%5Ctimes%206%2B36)
![v_6=120\ ft.s^{-1}](https://tex.z-dn.net/?f=v_6%3D120%5C%20ft.s%5E%7B-1%7D)
Now velocity at time t=2 s:
![v_2=3\times 2^2-24\times 2+36](https://tex.z-dn.net/?f=v_2%3D3%5Ctimes%202%5E2-24%5Ctimes%202%2B36)
The particle is at rest at time t=2 seconds.
d)
Put the value t=2 sec. in eq. (1):
![s_2=2^3-12\times 2^2+36\times 2](https://tex.z-dn.net/?f=s_2%3D2%5E3-12%5Ctimes%202%5E2%2B36%5Ctimes%202)
![s_2=32\ ft](https://tex.z-dn.net/?f=s_2%3D32%5C%20ft)
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)
![s_8=8^3-12\times 8^2+36\times 8](https://tex.z-dn.net/?f=s_8%3D8%5E3-12%5Ctimes%208%5E2%2B36%5Ctimes%208)
![s_8=32\ ft](https://tex.z-dn.net/?f=s_8%3D32%5C%20ft)