Respuesta :
Answer:
(a): [tex]\rm meter/ second^2.[/tex]
(b): [tex]\rm meter/ second^3.[/tex]
(c): [tex]\rm 2ct-3bt^2.[/tex]
(d): [tex]\rm 2c-6bt.[/tex]
(e): [tex]\rm t=\dfrac{2c}{3b}.[/tex]
Explanation:
Given, the position of the particle along the x axis is
[tex]\rm x=ct^2-bt^3.[/tex]
The units of terms [tex]\rm ct^2[/tex] and [tex]\rm bt^3[/tex] should also be same as that of x, i.e., meters.
The unit of t is seconds.
(a):
Unit of [tex]\rm ct^2=meter[/tex]
Therefore, unit of [tex]\rm c= meter/ second^2.[/tex]
(b):
Unit of [tex]\rm bt^3=meter[/tex]
Therefore, unit of [tex]\rm b= meter/ second^3.[/tex]
(c):
The velocity v and the position x of a particle are related as
[tex]\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.[/tex]
(d):
The acceleration a and the velocity v of the particle is related as
[tex]\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.[/tex]
(e):
The particle attains maximum x at, let's say, [tex]\rm t_o[/tex], when the following two conditions are fulfilled:
- [tex]\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.[/tex]
- [tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}<0.[/tex]
Applying both these conditions,
[tex]\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}.[/tex]
For [tex]\rm t_o = 0[/tex],
[tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c[/tex]
Since, c is a positive constant therefore, for [tex]\rm t_o = 0[/tex],
[tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0[/tex]
Thus, particle does not reach its maximum value at [tex]\rm t = 0\ s.[/tex]
For [tex]\rm t_o = \dfrac{2c}{3b}[/tex],
[tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.[/tex]
Here,
[tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}<0.[/tex]
Thus, the particle reach its maximum x value at time [tex]\rm t_o = \dfrac{2c}{3b}.[/tex]
