Which of the following best represents the graph of velocity (v) versus displacement (s) for a particle travelling with an acceleration (uniform) and some non zero initial velocity?

Options

A.
B.
C.
D.

Explanation

To solve the problem of determining the graph of velocity (v) versus displacement (s) for a particle traveling with uniform acceleration and some non-zero initial velocity, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Motion**:
The particle is moving with uniform acceleration (a) and has an initial velocity (u). The relationship between velocity, displacement, and acceleration is given by the equation of motion.

2. **Use the Equation of Motion**:
The relevant equation of motion is:
\[ v^2 = u^2 + 2as \]
Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the displacement.

3. **Rearranging the Equation**:
We can rearrange the equation to express \(v\) in terms of \(s\):
\[ v^2 = 2as + u^2 \]
This can be rewritten as:
\[ v^2 - u^2 = 2as \]

4. **Identify the Form of the Equation**:
The equation \(v^2 = 2as + u^2\) is similar to the standard form of a parabola \(y^2 = 4px\). In our case, if we let \(y = v\) and \(x = s\), we can see that this is a quadratic relationship.

5. **Determine the Characteristics of the Graph**:
- The graph will be a parabola.
- Since \(a\) (acceleration) is positive, the parabola opens upwards.
- The vertex of the parabola is at the point where \(s = -\frac{u^2}{2a}\) (this is derived from setting \(v = 0\) to find the displacement at which the particle stops if it were to decelerate).

6. **Conclusion**:
The graph of velocity (v) versus displacement (s) for a particle with uniform acceleration and a non-zero initial velocity is a parabola that opens upwards.

### Final Answer:
The best representation of the graph of velocity versus displacement for the given conditions is **Option D**, which is a parabola symmetric about the x-axis and opening in the positive direction.