Answer:
(a): ![\rm meter/ second^2.](https://tex.z-dn.net/?f=%5Crm%20meter%2F%20second%5E2.)
(b): ![\rm meter/ second^3.](https://tex.z-dn.net/?f=%5Crm%20meter%2F%20second%5E3.)
(c): ![\rm 2ct-3bt^2.](https://tex.z-dn.net/?f=%5Crm%202ct-3bt%5E2.)
(d): ![\rm 2c-6bt.](https://tex.z-dn.net/?f=%5Crm%202c-6bt.)
(e): ![\rm t=\dfrac{2c}{3b}.](https://tex.z-dn.net/?f=%5Crm%20t%3D%5Cdfrac%7B2c%7D%7B3b%7D.)
Explanation:
Given, the position of the particle along the x axis is
![\rm x=ct^2-bt^3.](https://tex.z-dn.net/?f=%5Crm%20x%3Dct%5E2-bt%5E3.)
The units of terms
and
should also be same as that of x, i.e., meters.
The unit of t is seconds.
(a):
Unit of ![\rm ct^2=meter](https://tex.z-dn.net/?f=%5Crm%20ct%5E2%3Dmeter)
Therefore, unit of ![\rm c= meter/ second^2.](https://tex.z-dn.net/?f=%5Crm%20c%3D%20meter%2F%20second%5E2.)
(b):
Unit of ![\rm bt^3=meter](https://tex.z-dn.net/?f=%5Crm%20bt%5E3%3Dmeter)
Therefore, unit of ![\rm b= meter/ second^3.](https://tex.z-dn.net/?f=%5Crm%20b%3D%20meter%2F%20second%5E3.)
(c):
The velocity v and the position x of a particle are related as
![\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.](https://tex.z-dn.net/?f=%5Crm%20v%3D%5Cdfrac%7Bdx%7D%7Bdt%7D%5C%5C%3D%5Cdfrac%7Bd%7D%7Bdx%7D%28ct%5E2-bt%5E3%29%5C%5C%3D2ct-3bt%5E2.)
(d):
The acceleration a and the velocity v of the particle is related as
![\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.](https://tex.z-dn.net/?f=%5Crm%20a%20%3D%20%5Cdfrac%7Bdv%7D%7Bdt%7D%5C%5C%3D%5Cdfrac%7Bd%7D%7Bdt%7D%282ct-3bt%5E2%29%5C%5C%3D2c-6bt.)
(e):
The particle attains maximum x at, let's say,
, when the following two conditions are fulfilled:
![\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.](https://tex.z-dn.net/?f=%5Crm%20%5Cleft%20%28%5Cdfrac%7Bdx%7D%7Bdt%7D%5Cright%20%29_%7Bt%3Dt_o%7D%3D0.)
![\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}](https://tex.z-dn.net/?f=%5Crm%20%5Cleft%20%28%20%5Cdfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%5Cright%20%29_%7Bt%3Dt_o%7D%3C0.)
Applying both these conditions,
![\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.](https://tex.z-dn.net/?f=%5Crm%20%5Cleft%20%28%20%5Cdfrac%7Bdx%7D%7Bdt%7D%5Cright%20%29_%7Bt%3Dt_o%7D%3D0%5C%5C2ct_o-3bt_o%5E2%3D0%5C%5Ct_o%282c-3bt_o%29%3D0%5C%5Ct_o%3D0%5C%20%5C%20%5C%20%5C%20%5C%20or%5C%20%5C%20%5C%20%5C%20%5C%202c%3D3bt_o%5CRightarrow%20t_o%20%3D%20%5Cdfrac%7B2c%7D%7B3b%7D.)
For
,
![\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c](https://tex.z-dn.net/?f=%5Crm%20%5Cleft%20%28%20%5Cdfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%5Cright%20%29_%7Bt%3Dt_o%7D%3D2c-6bt_o%20%3D%202c-6%5Ccdot%200%3D2c)
Since, c is a positive constant therefore, for
,
![\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0](https://tex.z-dn.net/?f=%5Crm%20%5Cleft%20%28%20%5Cdfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%5Cright%20%29_%7Bt%3Dt_o%7D%3E0)
Thus, particle does not reach its maximum value at ![\rm t = 0\ s.](https://tex.z-dn.net/?f=%5Crm%20t%20%3D%200%5C%20s.)
For
,
![\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.](https://tex.z-dn.net/?f=%5Crm%20%5Cleft%20%28%20%5Cdfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%5Cright%20%29_%7Bt%3Dt_o%7D%3D2c-6bt_o%20%3D%202c-6b%5Ccdot%20%5Cdfrac%7B2c%7D%7B3b%7D%3D2c-4c%3D-2c.)
Here,
![\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}](https://tex.z-dn.net/?f=%5Crm%20%5Cleft%20%28%20%5Cdfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%5Cright%20%29_%7Bt%3Dt_o%7D%3C0.)
Thus, the particle reach its maximum x value at time