A

Step 1: Identify the Radius Vector:
The radius vector $ extbf{r}(t)$ is given as:
$$ extbf{r}(t) = a t extbf{i} - b t^2 extbf{j} $$
where $ extbf{i}$ and $ extbf{j}$ are the unit vectors in the x and y directions respectively.

Step 2: Determine the Trajectory:
To eliminate time $t$, we express $t$ from the x-component of the radius vector:
$$ x = a t \Rightarrow t = \frac{x}{a} $$
Substituting $t$ in the y-component:
$$ y = -b t^2 \Rightarrow y = -b \left(\frac{x}{a}\right)^2 = -\frac{b}{a^2} x^2 $$
This gives the trajectory equation: $$ y(x) = -\frac{b}{a^2} x^2 $$

Step 3: Plot the Function:
The trajectory is a parabola that opens downwards.

Step 4: Calculate Velocity and Acceleration:
The velocity vector $ extbf{v}(t)$ is the derivative of the position vector with respect to time:
$$ extbf{v}(t) = \frac{d extbf{r}}{dt} = a extbf{i} - 2bt extbf{j} $$
The acceleration vector $ extbf{a}(t)$ is the derivative of the velocity vector:
$$ extbf{a}(t) = \frac{d extbf{v}}{dt} = -2b extbf{j} $$

Step 5: Modulus of Velocity and Acceleration:
The modulus of the velocity is:
$$ |\textbf{v}| = \sqrt{a^2 + (2bt)^2} = \sqrt{a^2 + 4b^2t^2} $$
The modulus of the acceleration is constant:
$$ |\textbf{a}| = 2b $$
Therefore, the results are: trajectory $y(x) = -\frac{b}{a^2} x^2$ with its plot as a downward parabola, and velocity and acceleration vectors depend on time and constants $a$, $b$.