Step 1: Set up the positions of the bus and the student.
Let the student start at position $x = 0$ and the bus starts at position $x = 16$ m. The bus starts from rest and accelerates at $a = 9$ m/s^2. After time $t$, the bus position is: \[ x_B = 16 + \frac{1}{2}(9)t^2 = 16 + 4.5t^2 \] The student runs at constant speed $v$, so: \[ x_S = vt \]
Step 2: Write the catching condition.
For the student to just catch the bus (minimum speed), the student's position equals the bus position at exactly one value of $t$ (tangent condition): \[ vt = 16 + 4.5t^2 \] Rearranging: \[ 4.5t^2 - vt + 16 = 0 \]
Step 3: Apply the tangent (minimum speed) condition.
For a unique solution (the student just barely catches the bus), the discriminant of this quadratic in $t$ must equal zero: \[ D = v^2 - 4(4.5)(16) = 0 \] \[ v^2 = 4 \times 4.5 \times 16 = 288 \] \[ v = \sqrt{288} = 12\sqrt{2} \text{ m/s} \]
Step 4: Compare with the given form.
The problem states the minimum velocity is $\alpha\sqrt{2}$ m/s. Comparing: \[ \alpha\sqrt{2} = 12\sqrt{2} \] Therefore $\alpha = 12$.
Step 5: Verify the calculation.
$v^2 = 288$, $v = \sqrt{288} = \sqrt{144 \times 2} = 12\sqrt{2}$ m/s. Yes, $\alpha = 12$.
Step 6: State the result.
The minimum speed for the student to catch the bus corresponds to $\alpha = 12$. \[ \boxed{\alpha = 12} \]