The retardation of a particle moving in a straight line is proportional to its displacement proportionallity constant being unity. Find the total distance convered by the particle till it comes to rest. Given that velocity of particle is `v_(0)` at zero displacement.

Options

A. Uniform velocity
B. Uniform speed
C. Uniform acceleration
D. Data insufficient for generalization

Explanation

To solve the problem, we need to find the total distance covered by a particle until it comes to rest, given that its retardation is proportional to its displacement. Let's break down the solution step by step.

### Step 1: Understand the relationship between retardation and displacement
The problem states that the retardation (negative acceleration) of the particle is proportional to its displacement \( s \). We can express this mathematically as:
\[ a = -s \]
where \( a \) is the retardation.

### Step 2: Relate acceleration to velocity and displacement
We know that acceleration can also be expressed in terms of velocity \( v \) and displacement \( s \) using the chain rule:
\[ a = v \frac{dv}{ds} \]
Thus, we can set the two expressions for acceleration equal to each other:
\[ v \frac{dv}{ds} = -s \]

### Step 3: Rearranging the equation
Rearranging the equation gives us:
\[ v \, dv = -s \, ds \]

### Step 4: Integrate both sides
Now, we will integrate both sides. The left side will be integrated with respect to \( v \) from the initial velocity \( v_0 \) to \( 0 \) (when the particle comes to rest), and the right side will be integrated with respect to \( s \) from \( 0 \) to \( s_0 \) (the total distance covered):
\[ \int_{v_0}^{0} v \, dv = -\int_{0}^{s_0} s \, ds \]

### Step 5: Perform the integration
The left side integrates to:
\[ \left[ \frac{v^2}{2} \right]_{v_0}^{0} = 0 - \frac{v_0^2}{2} = -\frac{v_0^2}{2} \]
The right side integrates to:
\[ -\left[ \frac{s^2}{2} \right]_{0}^{s_0} = -\frac{s_0^2}{2} \]

### Step 6: Set the integrals equal to each other
Now we can set the results of the integrals equal to each other:
\[ -\frac{v_0^2}{2} = -\frac{s_0^2}{2} \]

### Step 7: Solve for \( s_0 \)
Multiplying through by -2 gives:
\[ v_0^2 = s_0^2 \]
Taking the square root of both sides, we find:
\[ s_0 = v_0 \]

### Conclusion
The total distance covered by the particle until it comes to rest is:
\[ \boxed{v_0} \]