Answer
Please see the figure attached. Dashed arrows indicate the direction of motions.
Velocity vectors at $t=0$ and $t=1$:
${\bf{v}}\left( 0 \right) = \left( {0, - 1} \right)$
${\bf{v}}\left( 1 \right) = \left( { - 2, - 1} \right)$
Acceleration vectors at $t=0$ and $t=1$:
${\bf{a}}\left( 0 \right) = {\bf{a}}\left( 1 \right) = \left( { - 2,0} \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {1 - {t^2},1 - t} \right)$ for $ - 2 \le t \le 2$.
The velocity vector:
${\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) = \left( { - 2t, - 1} \right)$
So, ${\bf{v}}\left( 0 \right) = \left( {0, - 1} \right)$ and ${\bf{v}}\left( 1 \right) = \left( { - 2, - 1} \right)$.
The acceleration vector:
${\bf{a}}\left( t \right) = {\bf{r}}{\rm{''}}\left( t \right) = \left( { - 2,0} \right)$
Notice that the acceleration vector is constant. So, ${\bf{a}}\left( 0 \right) = {\bf{a}}\left( 1 \right) = \left( { - 2,0} \right)$.
As can be seen in the figure, the dashed arrows indicate the direction of motions.
