Answer
$y=\dfrac{1}{x}-1$; $v(\dfrac{-1}{2})=4i-4j$ and $a(\dfrac{-1}{2})=-16i-16j$
Work Step by Step
We know that the velocity and acceleration is defined as: velocity is given as: $v(t)=r'(t)$ and acceleration is given as: $a(t)=v'(t)$
In the given problem, we have $x=\dfrac{t}{t+1}; y=\dfrac{1}{t}$
Re-write as follows: $y=(\dfrac{1}{x})-1$
$v(t)=r'(t)=\dfrac{1}{(t+1)^2}i-\dfrac{1}{t^2}j \\ \implies v(\dfrac{-1}{2})=4i-4j$
Next, we have $a(t)=v'(t)=\dfrac{-2}{(t+1)^3}i+\dfrac{2}{t^3}j \\ \implies a(\dfrac{-1}{2})=-16i-16j$
