The displacement (s)- velocity ( v ) graph of a particle if it starts moving from rest with a uniform acceleration which is parallel to its instantaneous direction of motion is

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Answer

To solve the problem of determining the displacement (s) - velocity (v) graph of a particle that starts moving from rest with uniform acceleration parallel to its instantaneous direction of motion, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Initial Conditions**:
- The particle starts from rest, which means at time \( t = 0 \), the initial velocity \( v_0 = 0 \) and the initial displacement \( s_0 = 0 \).

2. **Using the Equation of Motion**:
- Since the particle is moving with uniform acceleration \( a \), the equations of motion can be applied. The velocity \( v \) at any time \( t \) can be expressed as:
\[ v = u + at \]
where \( u = 0 \) (initial velocity). Thus, we have:
\[ v = at \]

3. **Relating Displacement and Velocity**:
- The displacement \( s \) can be expressed as:
\[ s = ut + \frac{1}{2} a t^2 \]
Again, since \( u = 0 \):
\[ s = \frac{1}{2} a t^2 \]

4. **Expressing Time in Terms of Velocity**:
- From the equation \( v = at \), we can express time \( t \) in terms of velocity \( v \):
\[ t = \frac{v}{a} \]

5. **Substituting Time into Displacement Equation**:
- Now, substitute \( t \) into the displacement equation:
\[ s = \frac{1}{2} a \left(\frac{v}{a}\right)^2 \]
Simplifying this gives:
\[ s = \frac{1}{2} a \cdot \frac{v^2}{a^2} = \frac{v^2}{2a} \]

6. **Identifying the Graph**:
- The relationship \( s = \frac{1}{2a} v^2 \) indicates that displacement \( s \) is proportional to the square of the velocity \( v \). This is a parabolic relationship, which means the graph of displacement versus velocity will be a parabola opening upwards.

### Conclusion:
The displacement (s) - velocity (v) graph of the particle is a parabola that opens upwards.