To express the acceleration in terms of velocity, let's start from the given relationships.

We know: $s = k \sqrt{t}$

From this, we derived: $v = \frac{k}{2\sqrt{t}}$

Rearrange this equation to solve for $\sqrt{t}$: $\sqrt{t} = \frac{k}{2v}$

Next, substitute $\sqrt{t}$ into the expression for acceleration: $a = -\frac{k}{4} t^{-\frac{3}{2}}$

Since $t = \left(\frac{k}{2v}\right)^2$, we get: $t^{-\frac{3}{2}} = \left(\frac{k}{2v}\right)^{-3} = \left(\frac{2v}{k}\right)^3$

Therefore, $a = -\frac{k}{4} \left(\frac{2v}{k}\right)^3$ $a = -\frac{k}{4} \cdot \frac{8v^3}{k^3}$ $a = -\frac{8v^3}{4k^2}$ $a = -\frac{2v^3}{k^2}$

In summary, the acceleration $a$ is proportional to the cube of the velocity $v$: $a \propto v^3$

Would you like more details or have any questions?

Here are 5 questions you might consider next:

1. How do you relate different orders of derivatives (displacement, velocity, acceleration) in general?
2. Can you provide an example where acceleration is directly proportional to time?
3. What are the implications of acceleration being proportional to the cube of velocity in a physical system?
4. How do you interpret the negative sign in the acceleration expression?
5. What are the practical applications of understanding acceleration in terms of velocity?

Tip: When working with different physical quantities, always ensure units are consistent to maintain the accuracy of your calculations and interpretations.