Respuesta :
Answer:
Explanation:
Given that
initial velocity ,[tex]v= 3 m/s[/tex]
[tex]a=-1.1v^{\dfrac{1}{2}}[/tex]
We know that
[tex]a=v\dfrac{dv}{dx}[/tex]
Lets take x is the distance before coming to the rest.
The final speed of the particle = 0 m/s
[tex]v\dfrac{dv}{dx}=-1.1v^{\dfrac{1}{2}}[/tex]
[tex]\dfrac{dv}{dx}=-1.1v^{-\dfrac{1}{2}}[/tex]
[tex]v^{\dfrac{1}{2}}{dv}=-1.1dx[/tex]
[tex]\int_{3}^{0}v^{\dfrac{1}{2}}{dv}=-\int_{0}^{x}1.1dx[/tex]
[tex]\left [v^{\dfrac{3}{2}}\times \dfrac{2}{3}\right]_3^0=-1.1x[/tex]
[tex]3^{\dfrac{3}{2}}\times \dfrac{2}{3}=1.1x[/tex]
[tex]x=\dfrac{3.46}{1.1}\ m\\x=3.14\ m[/tex]
(b)time taken by it
[tex]a=\frac{\mathrm{d} v}{\mathrm{d} t}=-1.1\sqrt{v}[/tex]
[tex]\int_{3}^{0}\frac{dv}{\sqrt{v}}=-1.1\int_{0}^{t}dt[/tex]
[tex]\int_{0}^{3}\frac{dv}{\sqrt{v}}=1.1\int_{0}^{t}dt[/tex]
[tex]2\times 3\sqrt{3}=1.1t[/tex]
[tex]t=9.44\ s[/tex]
