Assume you have a vector function $\left(x(t),y(t)\right)$ that describes the position of some object.

Open  View and select CAS.

Velocity Vector $\overrightarrow{v}(t)$

1.

Enter the functions $x(t)$ and $y(t)$. Beware that GeoGebra doesn’t allow you to use the letters $x(t)$ and $y(t)$ as names for functions, so give them other names.

2.

Enter Derivative(<Expression>) and then enter the name of $x(t)$. Press Enter. You get the $x$-coordinate of the velocity vector.

3.

Enter Derivative(<Expression>) and then enter the name of $y(t)$. Press Enter. You get the $y$-coordinate of the velocity vector.

Speed $\left|\overrightarrow{v}(t)\right|$

1.

First, find the velocity vector as described in the previous step.

2.

Enter

sqrt(<x-coordinate of the velocity vector n>^2

+ <y-coordinate of the velocity vector>^2)

3.

Press Enter. You get the function $\left|\overrightarrow{v}(t)\right|$ for the speed.

Acceleration Vector $\overrightarrow{a}(t)$

1.

First, find the velocity vector.

2.

Enter Derivative(<expression>), and then enter the expression you have for the $x$-coordinate of the velocity vector. Press Enter. You get the $x$-coordinate of the acceleration vector.

3.

Enter Derivative(<expression>), and enter the expression you have for the $y$-coordinate of the velocity vector. Press Enter. You now get the $y$-coordinate of the acceleration vector $\overrightarrow{a}(t)$.

Acceleration $\left|\overrightarrow{a}(t)\right|$

1.

First, find the acceleration vector as above.

2.

Enter

sqrt(<x-coordinate of acceleration vector>^2

+ <y-coordinate of acceleration vector>^2)

3.

Press Enter. You now get the acceleration $|\overrightarrow{a}(t)|$.