Respuesta :
Answer:
We know that the acceleration of the particle is defined as
[tex]a(t)=\frac{dv}{dt}[/tex]
Since it is given that
[tex]v(t)=\frac{5}{\sqrt{t}}\\\\\therefore a(t)=\frac{d(\frac{5}{\sqrt{t}})}{dt}\\\\a(t)=5\times \frac{dt^{-\frac{1}{2}}}{dt}\\\\=\frac{-5}{2}t^{\frac{-1}{2}-1}\\\\\therefore a(t)=x''(t)=\frac{-2.5}{t^{\frac{3}{2}}}[/tex]
Now by definition of velocity we have
[tex]v(t)=\frac{dx(t)}{dt}\\\\\Rightarrow dx(t)=v(t)dt[/tex]
Integrating on both sides we get
[tex]v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int v(t)dt[/tex]
Applying values we get
[tex]v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int \frac{5}{t^{\frac{1}{2}}}dt\\\\\therefore x(t)=\frac{5}{0.5}\int t^{-0.5}dt\\\\x(t)=10\sqrt{t}+c[/tex]
To find the constant we note that at t=1 the particle is at x=12 Thus applying values in the above equation we get
[tex]c=12-10\\\therefore c=2\\\\x(t)=10\sqrt{t}+2[/tex]
The acceleration function of the particle is [tex]a(t) = \frac{-2.5}{t^{3/2}}[/tex].
The position of the particle is [tex]x (t) = 10\sqrt{t} + 2[/tex].
Acceleration of the particle
The acceleration of the particle is detemine from the differentiation of the velocity.
a = dv/dt
[tex]v= \frac{5}{\sqrt{t} } = \frac{5}{t^{1/2}} = 5t^{-1/2}\\\\a = \frac{dv}{dt} = -0.5 \times 5t^{-3/2}= -2.5t^{-3/2}= \frac{-2.5}{t^{3/2}}[/tex]
Position of the particle
The position of the particle is determined from the integration of the velocity;
[tex]x = \int\limits {v} \, dt\\\\v = 5t^{-1/2}\\\\x = 10t^{1/2} + C\\\\12 = 10(1)^{1/2} + C\\\\12 = 10 + C\\\\C = 2\\\\x (t) = 10t^{1/2} + 2\\\\ x (t) = 10\sqrt{t} } + 2[/tex]
Learn more about acceleration here: https://brainly.com/question/14344386
