Answer
$a(t)=-\frac{1}{2t^\frac{3}{2}}$
$x(t) = 2\sqrt t+2$
Work Step by Step
Derivate $v(t)$ to find $a(t)$:
$v(t)=\frac{1}{\sqrt t}$
$v'(t)=(-\frac{1}{2})t^{-\frac{1}{2}-1} = -\frac{1}{2t^\frac{3}{2}}$
$a(t)=v'(t)=-\frac{1}{2t^\frac{3}{2}}$
Integrate $v(t)$ to find $x(t)$:
$v(t)=\frac{1}{\sqrt t}$
$\int f(t) dt=\int \frac{1}{\sqrt t} dt$
Apply power rule:
$\int t^n dt = \frac{t^{n+1}}{n+1}$ with $n =-\frac{1}{2}$
$x(t)= 2\sqrt{t} + C$
At $t=1$, position have been given as $x=4$. Use this in the equation above to find C:
$4= 2\sqrt{1} + C$
$C = 4 - 2\sqrt{1}=2$
Final equation:
$x(t) = 2\sqrt t+2$
