A moving particle has position $(x(t), y(t))$ at time $t$. At $t = 1$, its position is $\left(6, 0\right)$ and the velocity vector at any time $t > 0$ is $\langle 1 - \frac { 2 } { t ^ { 2 } } , 1 + \frac { 1 } { t ^ { 2 } } \rangle$.

1. What is the acceleration vector at $t = 2$?

   Hint (a):

   The acceleration vector can be obtained by taking the derivative of each component of the velocity vector.

2. For what time $t > 0$ does the velocity vector have a slope of $5$?

   Hint (b):

   Divide the vertical component by the horizontal component of the velocity vector. This should equal $5$.

   Step (b):

   $\frac{1+(1/t)^2}{1-2/t^2}=5$