Answer
$y=\dfrac{1}{x}-1$; $v(\dfrac{-1}{2})=4i-4j$ and $a(\dfrac{-1}{2})=-16i-16j$
Work Step by Step
Since, we have $x=\dfrac{t}{t+1}; y=\dfrac{1}{t}$
This can be written as:
$y=\dfrac{1}{x}-1$
Now, velocity $v(t)=r'(t)=\dfrac{1}{(t+1)^2}i-\dfrac{1}{t^2}j$
and $v(\dfrac{-1}{2})=4i-4j$
As we know acceleration $a(t)=v'(t)=\dfrac{-2}{(t+1)^3}i+\dfrac{2}{t^3}j$
and $a(\dfrac{-1}{2})=-16i-16j$
Hence, we have $y=\dfrac{1}{x}-1$; $v(\dfrac{-1}{2})=4i-4j$ and $a(\dfrac{-1}{2})=-16i-16j$
