For a particle under going rectilinear motion with uniform acceleration, the magnitude of displacement is one third the distance covered in some time interval. The magnitude of final velocity is less than magnitude of initial velocity for this time interval. Then the ratio of initial speed velocity to the final speed for this time interval is :

Options

A. `sqrt(2)`
B. 2
C. `sqrt(3)`
D. 3

Explanation

To solve the problem, we need to analyze the motion of a particle undergoing rectilinear motion with uniform acceleration. Here’s a step-by-step solution:

### Step 1: Understand the problem
We are given that the magnitude of displacement (s) is one third of the distance covered (d) in a certain time interval. This means:
\[ s = \frac{1}{3} d \]

### Step 2: Relate displacement and distance
If we denote the distance covered as \( d \) and the displacement as \( s \), we can express this relationship as:
\[ d = 3s \]

### Step 3: Analyze the motion
Since the final velocity is less than the initial velocity, the particle must be decelerating. This suggests that the particle may come to a stop and then reverse direction.

### Step 4: Use kinematic equations
We can use the second equation of motion:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) = final velocity
- \( u \) = initial velocity
- \( a \) = acceleration
- \( s \) = displacement

### Step 5: Substitute for distance
From the problem, we know that the distance \( d \) covered can be expressed in terms of \( s \):
\[ d = 3s \]
In our scenario, if the particle travels a distance of \( 2s \) before coming to a stop, we can set \( s = 2s \) in the equation:
\[ 0 = u^2 + 2a(2s) \]
This simplifies to:
\[ 0 = u^2 + 4as \]
Thus, we can find \( a \):
\[ a = -\frac{u^2}{4s} \]

### Step 6: Find final velocity
Now, we can find the final velocity \( v \) after the particle has moved \( s \):
Using the equation:
\[ v^2 = u^2 + 2as \]
Substituting \( a \):
\[ v^2 = u^2 + 2 \left(-\frac{u^2}{4s}\right) s \]
\[ v^2 = u^2 - \frac{u^2}{2} \]
\[ v^2 = \frac{u^2}{2} \]
Taking the square root gives:
\[ v = \frac{u}{\sqrt{2}} \]

### Step 7: Calculate the ratio
Now we can find the ratio of initial speed \( u \) to final speed \( v \):
\[ \frac{u}{v} = \frac{u}{\frac{u}{\sqrt{2}}} = \sqrt{2} \]

### Final Answer
The ratio of initial speed to final speed for this time interval is:
\[ \text{Ratio} = \sqrt{2} \]

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