The velocity of a particle moving in a straight line is directly proportional to `3//4th` power of time elapsed. How does its displacement and acceleration depend on time?

Explanation

To solve the problem, we need to analyze how the velocity, displacement, and acceleration of a particle depend on time when the velocity is directly proportional to the \( \frac{3}{4} \) power of time elapsed.

### Step-by-Step Solution:

1. **Understanding the relationship of velocity with time:**
Given that the velocity \( v \) is directly proportional to \( t^{\frac{3}{4}} \):
\[ v = k \cdot t^{\frac{3}{4}} \]
where \( k \) is a constant of proportionality.

2. **Finding acceleration:**
Acceleration \( a \) is the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Now, differentiating \( v \):
\[ a = \frac{d}{dt}(k \cdot t^{\frac{3}{4}}) = k \cdot \frac{3}{4} t^{\frac{3}{4} - 1} = k \cdot \frac{3}{4} t^{-\frac{1}{4}} \]
Thus, we can express acceleration as:
\[ a \propto t^{-\frac{1}{4}} \]

3. **Finding displacement:**
Displacement \( x \) is the integral of velocity with respect to time:
\[ x = \int v \, dt = \int (k \cdot t^{\frac{3}{4}}) \, dt \]
Performing the integration:
\[ x = k \cdot \int t^{\frac{3}{4}} \, dt = k \cdot \left( \frac{t^{\frac{3}{4} + 1}}{\frac{3}{4} + 1} \right) + C \]
Simplifying:
\[ x = k \cdot \left( \frac{t^{\frac{7}{4}}}{\frac{7}{4}} \right) + C = \frac{4k}{7} t^{\frac{7}{4}} + C \]
Ignoring the constant \( C \) for proportionality, we have:
\[ x \propto t^{\frac{7}{4}} \]

### Summary of Relationships:
- Velocity \( v \propto t^{\frac{3}{4}} \)
- Acceleration \( a \propto t^{-\frac{1}{4}} \)
- Displacement \( x \propto t^{\frac{7}{4}} \)