### Home > APCALC > Chapter 12 > Lesson 12.1.2 > Problem12-23

12-23.

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 $t = 2$?

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 $5$?

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

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