Answer:
![\textsf{a)} \quad v=12t^2-6t-18](https://tex.z-dn.net/?f=%5Ctextsf%7Ba%29%7D%20%5Cquad%20v%3D12t%5E2-6t-18)
![\textsf{b)} \quad a=24t-6](https://tex.z-dn.net/?f=%5Ctextsf%7Bb%29%7D%20%5Cquad%20a%3D24t-6)
c) 1.5 s
d) -19.25 m
e) 24.5 m
Step-by-step explanation:
<u>Displacement</u>
![x=4t^3-3t^2-18t+1](https://tex.z-dn.net/?f=x%3D4t%5E3-3t%5E2-18t%2B1)
(where t ≥ 0 and x is in meters)
<h3><u>Part (a)</u></h3>
To find the equation for <u>velocity</u>, differentiate the equation for displacement:
![\implies v=\dfrac{\text{d}x}{\text{d}t}=12t^2-6t-18](https://tex.z-dn.net/?f=%5Cimplies%20v%3D%5Cdfrac%7B%5Ctext%7Bd%7Dx%7D%7B%5Ctext%7Bd%7Dt%7D%3D12t%5E2-6t-18)
(where t ≥ 0 and v is in meters per second)
<h3><u>Part (b)</u></h3>
To find the equation for <u>acceleration</u>, differentiate the equation for velocity:
![\implies a=\dfrac{\text{d}v}{\text{d}t}=24t-6](https://tex.z-dn.net/?f=%5Cimplies%20a%3D%5Cdfrac%7B%5Ctext%7Bd%7Dv%7D%7B%5Ctext%7Bd%7Dt%7D%3D24t-6)
(where t ≥ 0 and a is in meters per second squared)
<h3><u>Part (c)</u></h3>
The particle comes to <u>rest</u> when its velocity is zero:
![\begin{aligned}v & = 0\\\implies 12t^2-6t-18 & = 0\\6(2t^2-t-3) & = 0\\2t^2-t-3 & = 0\\2t^2-3t+2t-3 & = 0\\t(2t-3)+1(2t-3) & = 0\\(t+1)(2t-3) & = 0\\\implies t & =-1, \dfrac{3}{2}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dv%20%26%20%3D%200%5C%5C%5Cimplies%2012t%5E2-6t-18%20%26%20%3D%200%5C%5C6%282t%5E2-t-3%29%20%26%20%3D%200%5C%5C2t%5E2-t-3%20%26%20%3D%200%5C%5C2t%5E2-3t%2B2t-3%20%26%20%3D%200%5C%5Ct%282t-3%29%2B1%282t-3%29%20%26%20%3D%200%5C%5C%28t%2B1%29%282t-3%29%20%26%20%3D%200%5C%5C%5Cimplies%20t%20%26%20%3D-1%2C%20%5Cdfrac%7B3%7D%7B2%7D%5Cend%7Baligned%7D)
As t ≥ 0, the particle comes to rest at 1.5 s.
<h3><u>Part (d)</u></h3>
<u>Substitute</u> the found value of t from part (c) into the equation for displacement to find where the particle comes to rest:
![\implies 4(1.5)^3-3(1.5)^2-18(1.5)+1=-19.25\: \sf m](https://tex.z-dn.net/?f=%5Cimplies%204%281.5%29%5E3-3%281.5%29%5E2-18%281.5%29%2B1%3D-19.25%5C%3A%20%5Csf%20m)
<h3><u>Part (e)</u></h3>
We have determined that the <u>particle is at rest at 1.5 s</u>.
Therefore, to find how far the particle traveled in the first 2 seconds, we need to divide the journey into two parts: before and after it was at rest.
The first leg of the journey is the first 1.5 s and the second leg of the journey is the next 0.5 s.
At the beginning of the journey, t = 0 s.
![\textsf{when }t=0: \quad x=4(0)^3-3(0)^2-18(0)+1=1](https://tex.z-dn.net/?f=%5Ctextsf%7Bwhen%20%7Dt%3D0%3A%20%20%5Cquad%20%20x%3D4%280%29%5E3-3%280%29%5E2-18%280%29%2B1%3D1)
Therefore, when t = 0, x = 1
When t = 1.5 s, x = -19.25 m (from part (d)).
⇒ Total distance traveled in first 1.5 s = 1 + 19.25 = 20.25 m
When t = 2, x = -15 m.
Therefore, the particle has traveled 19.25 - 15 = 4.25 m in the last 0.5 s of its journey.
<u>Total distance traveled</u>:
1 + 19.25 + 4.25 = 24.5 m
<u>Refer to the attached diagram</u>
When the particle is at rest, it <u>changes direction</u>. If we model its journey using the x-axis, for the first leg of its journey (0 - 1.5 s) it travels in the negative direction (to the left). At 1.5 s it stops, then changes direction and travels in the positive direction (to the right), arriving at -15 at 2 seconds.