A particle starts from rest with constant acceleration. The ratio of space-average velocity to the time average velocity is :-

Options

A. `1/2`
B. `3/4`
C. `4/3`
D. `3/2`

Explanation

To solve the problem of finding the ratio of space-average velocity to time-average velocity for a particle starting from rest with constant acceleration, we can follow these steps:

### Step 1: Define the Variables
- Let the initial velocity \( u = 0 \) (since the particle starts from rest).
- Let the constant acceleration be \( a \).
- Let the time be \( t \).

### Step 2: Find the Final Velocity
Using the equation of motion:
\[ v = u + at \]
Substituting \( u = 0 \):
\[ v = at \]

### Step 3: Find the Displacement
Using the equation of motion for displacement:
\[ x = ut + \frac{1}{2} a t^2 \]
Substituting \( u = 0 \):
\[ x = \frac{1}{2} a t^2 \]

### Step 4: Calculate Space-Average Velocity
The space-average velocity \( V_s \) is defined as:
\[ V_s = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{x}{t} \]
Substituting for \( x \):
\[ V_s = \frac{\frac{1}{2} a t^2}{t} = \frac{1}{2} a t \]

### Step 5: Calculate Time-Average Velocity
The time-average velocity \( V_t \) is defined as:
\[ V_t = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{\int v \, dt}{t} \]
Substituting \( v = at \):
\[ V_t = \frac{\int_0^t (at) \, dt}{t} \]
Calculating the integral:
\[ \int_0^t (at) \, dt = a \int_0^t t \, dt = a \left[ \frac{t^2}{2} \right]_0^t = a \frac{t^2}{2} \]
Thus:
\[ V_t = \frac{a \frac{t^2}{2}}{t} = \frac{a t}{2} \]

### Step 6: Find the Ratio of Space-Average Velocity to Time-Average Velocity
Now we can find the ratio \( R \):
\[ R = \frac{V_s}{V_t} = \frac{\frac{1}{2} a t}{\frac{a t}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \]

### Conclusion
The ratio of space-average velocity to time-average velocity is:
\[ \boxed{1} \]