If the displacement ($s$) of a particle is directly proportional to the square root of time ($t$), we can write this relationship as: $s \propto \sqrt{t}$ or $s = k \sqrt{t}$ where $k$ is a constant of proportionality.

To find the relationship between acceleration and time, we need to derive the expressions for velocity and acceleration.

1. Velocity ($v$) is the first derivative of displacement with respect to time: $v = \frac{ds}{dt}$

Given $s = k \sqrt{t}$, we find: $v = \frac{d}{dt}(k \sqrt{t}) = k \cdot \frac{d}{dt}(\sqrt{t}) = k \cdot \frac{1}{2\sqrt{t}} = \frac{k}{2\sqrt{t}}$

2. Acceleration ($a$) is the first derivative of velocity with respect to time: $a = \frac{dv}{dt}$

Given $v = \frac{k}{2\sqrt{t}}$, we find: $a = \frac{d}{dt} \left( \frac{k}{2\sqrt{t}} \right) = \frac{k}{2} \cdot \frac{d}{dt} (t^{-\frac{1}{2}}) = \frac{k}{2} \cdot \left( -\frac{1}{2} t^{-\frac{3}{2}} \right) = -\frac{k}{4} t^{-\frac{3}{2}}$

Thus, the acceleration $a$ is proportional to $t^{-\frac{3}{2}}$.

In summary, the acceleration of the particle is inversely proportional to the cube of the square root of time, which can be expressed as: $a \propto t^{-\frac{3}{2}}$

Would you like more details or have any questions?

Here are 5 questions you might consider next:

1. How do you derive the second derivative of a function?
2. What is the physical significance of acceleration being inversely proportional to time raised to a power?
3. Can you explain the relationship between displacement, velocity, and acceleration in more detail?
4. How does this example apply to real-world scenarios?
5. What are some other examples where displacement is proportional to different functions of time?

Tip: Remember, understanding how to take derivatives and interpret their physical meaning is crucial in kinematics and dynamics.